Three identity families obtained without any new proof. Setting the two angles equal in the sum identities gives the double angle formulas, including the three equivalent forms of the cosine one; solving those backwards gives the power reduction formulas that trade a square for a doubled angle; and substituting theta over two into those and taking square roots gives the half angle formulas, whose plus-or-minus is settled by the quadrant of the half angle rather than of the original.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 10 — Foundations of Trigonometry
§10.4 Trigonometric Identities, pp. 776-780
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 776-780 — the pages these objectives are drawn from
Warm-up
You proved the sum identities last lesson. This lesson asks one question of them and gets three families of answers.
Discussion prompt
Take the sum identity for sine and put beta equal to alpha. What do you get, and what does the left-hand side become?
Hint: Just substitute. Do not try to be clever.
Answer:
\[ \sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta \]
\[ \beta = \alpha: \quad \sin(2\alpha) = \sin\alpha\cos\alpha + \cos\alpha\sin\alpha = 2\sin\alpha\cos\alpha \]
That is the double angle formula for sine, and it took one substitution. Every formula in this lesson arrives the same way — by substituting into something already proved, or by rearranging it. There is no new theorem in this lesson at all, which is worth knowing before you start memorising.
Concept
The sum identities, with the two angles set equal, give the double angle formulas. Those, solved for the squared term, give the power reduction formulas. Those, with the angle halved and a square root taken, give the half angle formulas.
So the three families in this lesson are three views of one identity, and the reason there are so many formulas is not that there is a lot to know but that one relationship is useful in several directions.
Figure (svg): A flow diagram showing the sum identity for cosine becoming the double angle identity by setting beta equal to alpha, then being solved backwards for the power reduction formulas, then having theta replaced by theta over two to give the half angle formulas
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 776-779
Section
Section 1
Concept
Putting beta equal to alpha in each sum identity collapses it into a formula for the doubled angle. The cosine one then comes in three equivalent forms, obtained by trading squares using the Pythagorean identity.
Theorem 10.17 — The double angle identities. The cosine of twice an angle has three equivalent forms; the sine of twice an angle is twice the product of the sine and cosine; the tangent of twice an angle is twice the tangent over one minus the tangent squared.
\[ \cos(2\theta) = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta \]
To find the cosine of a doubled angle you need only one piece of information. To find the sine of a doubled angle you appear to need both, though the third worked example below shows that the tangent alone will do.
Figure (svg): The three equivalent forms of the cosine double angle identity, each obtained from the first by trading a square using the Pythagorean identity
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 776-776
Picture it
These two have only one form each, and the sine formula is the one used most often.
Figure (svg): The double angle identity for sine and for tangent, shown as boxed statements with the note that sine appears to need both values while tangent needs only one
Notice that the sine formula is a product of two different functions, which is why it cannot be rewritten to need only one of them the way the cosine formula can.
Worked example
Example 10.4.3, part 1. Find both values at the doubled angle and say where it lands.
\[ P(-3, 4) \text{ lies on the terminal side of } \theta. \text{ Find } \cos(2\theta), \; \sin(2\theta), \text{ and the quadrant of } 2\theta. \]
Find the cosine and sine of theta
Why: The radius is five, so divide each coordinate by it.
\[ \cos = -\frac{3}{5}, \sin = \frac{4}{5} \]
Apply the cosine double angle formula
Why: Use the first form, since both values are known.
\[ \cos 2 t = \frac{9}{25} - \frac{16}{25} = -\frac{7}{25} \]
Apply the sine double angle formula
Why: Twice the product of the two values.
\[ \sin 2 t = 2(\frac{4}{5}) (-\frac{3}{5}) = -\frac{24}{25} \]
Read the quadrant of the doubled angle
Why: Both values negative means quadrant three.
\[ 2 \theta\text{ in QIII} \]
Figure (svg): The solution to Worked example doubling from a point shown as a ladder of expressions, one row per legal move
\[ \cos(2\theta) = -\tfrac{7}{25}, \quad \sin(2\theta) = -\tfrac{24}{25}, \quad 2\theta \in \text{QIII} \]
Verify: check the identity
Why: Forty-nine plus five hundred seventy-six over six hundred twenty-five is 625 over 625, which is 1. And theta itself was in quadrant two, so doubling it carries the terminal side past the negative x-axis into quadrant three — consistent with the signs.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 776-777
Matching
The three forms of the cosine double angle formula exist so that one value is always enough.
Match the pairs
Why: Each form is designed around what you have. If only the cosine is known, use the form containing only cosines; if only the sine, the form containing only sines. Knowing both makes the raw form the quickest, since no rearrangement happened. And with only a tangent, the tangent double angle formula avoids reconstructing either value — which is often the fastest route of all.
Worked example
The three forms exist so that one value is always enough. Pick the one matching what you were given.
\[ \text{If } \sin(\theta) = \tfrac{2}{3}, \text{ find } \cos(2\theta) \text{ without finding } \cos(\theta). \]
Choose the form containing only the sine
Why: The third form is one minus twice the sine squared.
\[ \cos 2 t = 1 - 2 \sin ^{2} t \]
Substitute
Why: Two thirds squared is four ninths.
\[ = 1 - 2(\frac{4}{9}) \]
Simplify
Why: One minus eight ninths.
\[ = \frac{1}{9} \]
Note what was avoided
Why: The cosine of theta is plus or minus root five over three, and the quadrant was never given — but it never mattered.
Figure (svg): The solution to Worked example choose the form that fits shown as a ladder of expressions, one row per legal move
\[ \cos(2\theta) = 1 - 2\left(\tfrac{2}{3}\right)^2 = \tfrac{1}{9} \]
Verify: try the other route
Why: With cosine of theta equal to plus or minus root five over three, the first form gives five ninths minus four ninths, which is one ninth, for either sign — because the cosine appears squared. So the answer really is independent of the quadrant, which is exactly why the third form was the right choice.
Trap
\[ \sin\left(\tfrac{\pi}{6}\right) = \tfrac{1}{2} \quad\Longrightarrow\quad \sin\left(\tfrac{\pi}{3}\right) = 2\left(\tfrac{1}{2}\right) = 1 \]
Double the value because the angle doubled
Why: The notation sine of 2 theta looks like a coefficient multiplying the whole expression.
But the sine of pi over three is root three over two, about 0.866, not 1. Doubling the angle does not double the value, and the number 2 in that expression is inside the function, not outside it.
\[ \sin\left(2 \cdot \tfrac{\pi}{6}\right) = 2\sin\left(\tfrac{\pi}{6}\right)\cos\left(\tfrac{\pi}{6}\right) = 2\left(\tfrac{1}{2}\right)\left(\tfrac{\sqrt{3}}{2}\right) = \tfrac{\sqrt{3}}{2} \]
Use the double angle identity, which is what the identity is for
Why: The whole reason these formulas exist is that the naive doubling is wrong.
A free check that catches this instantly: doubling a sine could take it above 1, which is impossible. Any time an operation on an angle is confused with the same operation on the value, the range check exposes it.
Fill the middle
Derive the second form of the cosine double angle identity from the first.
Fill in the blanks
\cos 2\theta = \cos^2\theta - \sin^2\theta = \cos^2\theta - (1 - cos^2 theta) = 2cos^2 theta - 1
Why: The Pythagorean identity rearranged says sine squared equals one minus cosine squared. Substituting that in and collecting gives two cosine squared minus one. The third form comes the same way by substituting for cosine squared instead. All three are the same identity, and deriving them beats memorising them.
Prediction
An angle theta lies in quadrant one.
Predict first
Which quadrant does 2 theta lie in?
Correct: It could be quadrant one, two, or on an axis.
Why: Quadrant one runs from 0 to pi over two, so twice such an angle runs from 0 to pi — which covers quadrants one and two entirely, plus the positive y-axis at exactly pi over four doubled. So doubling does not preserve the quadrant and can even land on an axis. This is why the quadrant of a doubled angle must be determined from the computed signs rather than assumed, as the first worked example did.
Socratic
The sine and tangent double angle formulas each have exactly one form. Cosine has three.
Discussion prompt
Explain what makes cosine different, and why having three forms is a genuine convenience rather than clutter.
Hint: Look at what appears in the raw form of each.
Answer:
The raw cosine form contains only squares — cosine squared and sine squared — and the Pythagorean identity lets a square of one be traded for a square of the other. So each square can be eliminated, giving two more forms.
The sine formula contains cosine and sine to the first power, as a product. There is no identity trading a first power of one for a first power of the other, so nothing can be eliminated and only one form exists.
The convenience is real: it means the cosine of a doubled angle can always be found from a single value, with no quadrant information needed, because the value only ever appears squared. That is exactly what the second worked example exploited.
Section
Section 2
Concept
A common calculus-facing question gives you the sine of an angle as a variable and asks for another quantity in terms of that variable. The method is unchanged; only the notation is unfamiliar.
The book flags this explicitly: if your first reaction to seeing the sine of theta equal x is that x should be the cosine, you have learned the earlier material well, but here x is only a name.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 776-777
Picture it
Everything in this lesson is a substitution into something proved earlier. Keep the chain in view.
Figure (svg): A flow diagram showing the sum identity for cosine becoming the double angle identity by setting beta equal to alpha, then being solved backwards for the power reduction formulas, then having theta replaced by theta over two to give the half angle formulas
When a problem looks unfamiliar, the question to ask is which link of this chain it is standing on.
Worked example
Example 10.4.3, part 2. The interval given is what resolves the sign.
\[ \text{If } \sin(\theta) = x \text{ for } -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}, \text{ find } \sin(2\theta) \text{ in terms of } x. \]
Write the identity you need
Why: The sine double angle formula needs both the sine and the cosine.
\[ \sin 2 t = 2 \sin t \cos t \]
Recover the cosine from the Pythagorean identity
Why: Substituting the given sine leaves a square root with a plus-or-minus.
\[ \cos t = +- \sqrt{1 - x ^{2}} \]
Resolve the sign from the stated interval
Why: Between negative pi over two and pi over two the terminal side is on the right half, so the cosine is non-negative.
\[ \cos t = \sqrt{1 - x ^{2}} \]
Substitute into the identity
Why: Twice x times the square root.
\[ \sin 2 t = 2 x \sqrt{1 - x ^{2}} \]
Figure (svg): The solution to Worked example express a double angle in terms of a variable shown as a ladder of expressions, one row per legal move
\[ \sin(2\theta) = 2x\sqrt{1 - x^2} \]
Verify: test at a known angle
Why: Take theta equal to pi over six, so x is one half. The formula gives 2 times one half times the root of three quarters, which is root three over two. And the sine of pi over three is indeed root three over two.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 777-777
Faded example
If the cosine of theta is x with theta between 0 and pi, express the cosine of 2 theta in terms of x.
Fill in the blanks
\cos(2\theta) = 2 \cos^2\theta - 1 = 1x^2 - ___
Why: The form containing only the cosine is the natural choice, since the cosine is exactly what was given. No square root appears, so the interval is never needed and the answer is 2x squared minus 1 regardless of quadrant. Compare this with the sine version, where a square root does appear and the interval becomes essential — the difference is entirely down to which form was available.
Worked example
Example 10.4.3, part 3. This identity shows that one value really is enough for sine too.
\[ \text{Verify that } \sin(2\theta) = \frac{2\tan(\theta)}{1 + \tan^2(\theta)}. \]
Start on the right and recognise the denominator
Why: One plus tangent squared is secant squared.
\[ = 2 \tan t / \sec ^{2} t \]
Convert to cosine and sine
Why: Tangent is sine over cosine and secant squared is one over cosine squared.
\[ = 2(\sin / \cos) (\cos ^{2}) \]
Cancel one cosine
Why: The cosine squared in the numerator meets the cosine in the denominator.
\[ = 2 \sin t \cos t \]
Recognise the result
Why: That is the double angle formula for sine.
\[ = \sin 2 t \]
Figure (svg): The solution to Worked example the sine of a double angle from the tangent alone shown as a ladder of expressions, one row per legal move
\[ \frac{2\tan\theta}{1 + \tan^2\theta} = \frac{2\tan\theta}{\sec^2\theta} = 2\sin\theta\cos\theta = \sin(2\theta) \]
Verify: test at pi over four
Why: The tangent there is 1, so the right side is 2 over 2, which is 1. And the sine of pi over two is 1. The identity is also genuinely useful: it computes the sine of a doubled angle from the tangent alone, with no quadrant information and no square roots.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 777-777
Error analysis
A student is told the cosine of theta is x with theta between pi over two and pi, and asked for the sine of 2 theta.
Annotate
On: \( \sin(\theta) = \sqrt{1 - x^2} \;\Longrightarrow\; \sin(2\theta) = 2x\sqrt{1 - x^2} \)
An interval is never decoration in these problems. It appears precisely because a square root has to be signed, and skipping the check gives the right answer only half the time.
Prediction
You are asked for the cosine of 2 theta given only the sine of theta, with no quadrant stated.
Predict first
Can you answer?
Correct: Yes, because the sine appears only squared.
Why: The third form of the cosine double angle identity is one minus twice the sine squared, and squaring destroys any sign information, so the answer is the same whichever quadrant the angle is in. This is a genuine feature of that form rather than a coincidence. Asking for the sine of 2 theta instead would need the quadrant, because the cosine would have to be recovered with a square root.
Reverse engineer
Here is a complete solution with the question removed.
Fill in the blanks
\cos\theta = \pm\sqrt2, \; \text___ \cos\theta = -\sqrt___ \;\Longrightarrow\; \text___ = ___x\sqrt___ \cdot (-1) = -___x\sqrt___
Why: The sine of theta must have been given as x, since the Pythagorean identity was used to recover the cosine. A quadrant two restriction made that cosine negative. The doubled quantity being sought is the sine of 2 theta, which is twice the sine times the cosine. The question was: given sine theta equals x with theta in quadrant two, find the sine of 2 theta.
Edge cases
The expression 2x times the root of one minus x squared gives the sine of a doubled angle.
Discussion prompt
What is the largest value that expression can take as x ranges over its possible values, and what does that tell you?
Hint: The result is a sine, so what must it satisfy? Then check whether the bound is attained.
Answer:
Since the result is a sine, it must lie between negative one and one. Setting x equal to root two over two gives 2 times root two over two times the root of one half, which is exactly 1 — so the bound is attained.
\[ x = \tfrac{\sqrt{2}}{2}: \quad 2 \cdot \tfrac{\sqrt{2}}{2} \cdot \tfrac{\sqrt{2}}{2} = 1 \]
And that value of x is the sine of pi over four, whose double is pi over two, where the sine really is 1. So the algebra and the geometry agree at the boundary, which is a strong check that the expression is right. An expression claiming to be a sine that could exceed one would be wrong, and testing its maximum is a cheap way to catch that.
Section
Section 3
Concept
The cosine of any whole multiple of an angle can be written as a polynomial in the cosine of that angle. The method is to split the multiple as a sum, expand, and then use the double angle formulas and the Pythagorean identity to eliminate every sine.
That last point is why the method works at all. In the cosine expansion the sines always occur in even powers, so the Pythagorean identity can always finish the job. The resulting polynomials are the Chebyshev polynomials, which recur throughout numerical analysis.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 777-778
Picture it
This section runs the identity in the direction that raises powers. The next runs it the other way.
Figure (svg): Two columns contrasting the direction of the double angle formulas with the direction of the power reduction formulas
Writing the cosine of three theta as a cubic in the cosine of theta trades a large angle for a high power. Power reduction trades a high power for a large angle. They are the same trade in opposite directions.
Worked example
Example 10.4.3, part 4. Split, expand, then eliminate every sine.
\[ \text{Express } \cos(3\theta) \text{ as a polynomial in } \cos(\theta). \]
Split the multiple as a sum
Why: Three theta is two theta plus theta.
\[ \cos(2 t + t) \]
Expand with the sum identity
Why: Matched terms, sign reversed.
\[ \cos 2 t \cos t - \sin 2 t \sin t \]
Substitute the double angle formulas
Why: Use the cosine form with only cosines, and the sine formula.
\[ (2 \cos ^{2} t - 1) \cos t - 2 \sin t \cos t \sin t \]
Eliminate the sine squared
Why: Sine squared is one minus cosine squared; expand and collect.
\[ 4 \cos ^{3} t - 3 \cos t \]
Figure (svg): The solution to Worked example the cosine of three theta shown as a ladder of expressions, one row per legal move
\[ \cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta) \]
Verify: test at pi over three
Why: The cosine there is one half, so the polynomial gives 4 times one eighth minus 3 times one half, which is one half minus three halves, namely negative one. And the cosine of pi, which is three times pi over three, is indeed negative one.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 777-778
Ranking
Express the cosine of three theta as a polynomial in the cosine.
Put in order
Why: Splitting first is what makes the sum identity applicable. Expanding introduces double angles, which the double angle formulas remove. That leaves a sine squared, which the Pythagorean identity converts to cosines, and collecting finishes. Each step removes one obstacle, and doing them out of order leaves nothing to work on.
Worked example
The same method one step further, and this time the doubling can be done twice.
\[ \text{Express } \cos(4\theta) \text{ as a polynomial in } \cos(\theta). \]
Recognise four theta as twice two theta
Why: Doubling twice is easier than splitting as a sum here.
\[ \cos(2(2 t)) \]
Apply the cosine double angle formula
Why: Use the form with only cosines, applied to the angle 2 theta.
\[ = 2 \cos ^{2}(2 t) - 1 \]
Substitute the inner double angle formula
Why: Cosine of 2 theta is two cosine squared minus one.
\[ = 2(2 \cos ^{2} t - 1) ^{2} - 1 \]
Expand and collect
Why: Square the bracket, distribute the 2, subtract 1.
\[ = 8 \cos ^{4} t - 8 \cos ^{2} t + 1 \]
Figure (svg): The solution to Worked example the cosine of four theta shown as a ladder of expressions, one row per legal move
\[ \cos(4\theta) = 8\cos^4(\theta) - 8\cos^2(\theta) + 1 \]
Verify: test at pi over four
Why: The cosine there is root two over two, so cosine squared is one half and cosine to the fourth is one quarter. The polynomial gives 8 times a quarter minus 8 times a half plus one, which is 2 minus 4 plus 1, namely negative one. And the cosine of pi, which is four times pi over four, is negative one.
Trap
\[ \cos(3\theta) = (2\cos^2\theta - 1)\cos\theta - 2\sin^2\theta\cos\theta \]
Stop once the double angles have been expanded
Why: Every double angle has been replaced, so the expansion looks complete.
But the question asked for a polynomial in the cosine, and a sine squared is still present. The expression is correct but it is not an answer to the question asked.
\[ = (2\cos^2\theta - 1)\cos\theta - 2(1 - \cos^2\theta)\cos\theta = 4\cos^3\theta - 3\cos\theta \]
Trade every sine squared for cosines and collect
Why: The Pythagorean identity turns the remaining sine squared into one minus cosine squared.
This always works for cosine of a multiple angle, because the sines always appear in even powers. If you ever find yourself with a lone sine to the first power that will not go away, the expansion has gone wrong somewhere earlier.
Faded example
Complete the derivation of the cosine of three theta.
Fill in the blanks
(2\cos^2\theta - 1)\cos\theta - 2(1 - cos^2 theta)\cos\theta = 2\cos^3\theta - \cos\theta - 2\cos\theta + 2\cos^3\theta = 4cos^3 theta - 3cos theta
Why: The sine squared becomes one minus cosine squared. Distributing gives two cosine cubed minus cosine from the first bracket, and negative two cosine plus two cosine cubed from the second. Collecting the cubes gives four cosine cubed and collecting the linear terms gives negative three cosine.
Prediction
You are asked to express the sine of three theta as a polynomial in the sine of theta.
Predict first
Will this work the same way?
Correct: Yes, but the cosines appear in even powers so they can be eliminated.
Why: Expanding the sine of two theta plus theta gives sine of 2t times cosine of t plus cosine of 2t times sine of t. Substituting the double angle formulas produces two sine times cosine squared plus a term with cosine squared in it, and the cosines do come out in even powers. Eliminating them gives three sine minus four sine cubed. So it works, but only because of the parity — and asking for the cosine of three theta as a polynomial in the sine would genuinely fail.
Explain it to yourself
The polynomials produced this way are the Chebyshev polynomials, which are important well beyond trigonometry.
Discussion prompt
Explain why writing the cosine of a multiple angle as a polynomial is useful at all. What has been gained by trading an angle for a power?
Hint: Which is easier for a computer to evaluate, a cosine or a cubic?
Answer:
A polynomial can be evaluated with nothing but multiplication and addition, while a cosine requires an infinite process. So the identity converts a transcendental computation into an arithmetic one: knowing the cosine of theta, the cosine of three theta costs one cube and one multiplication.
This is exactly how numerical libraries and older calculators worked, and it is why Chebyshev polynomials sit at the centre of approximation theory. The general principle is worth carrying: an identity that trades one kind of difficulty for another is useful whenever the second kind is cheaper, and which is cheaper depends on who is doing the computing.
Section
Section 4
Concept
Solving the second form of the cosine double angle identity for the cosine squared, and the third for the sine squared, gives two formulas that trade a square for a first power at twice the angle.
Theorem 10.18 — The power reduction formulas: a squared cosine or sine rewritten as a constant plus or minus half the cosine of the doubled angle.
\[ \cos^2\theta = \frac{1 + \cos 2\theta}{2}, \qquad \sin^2\theta = \frac{1 - \cos 2\theta}{2} \]
These are not new identities. They are the double angle formulas with the squared term isolated, which is why they need no separate proof and why the only difference between them is a sign.
Figure (svg): The two power reduction formulas, each expressing a squared function in terms of the cosine of the doubled angle to the first power
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 778-778
Picture it
Double angle raises powers and lowers angles. Power reduction lowers powers and raises angles.
Figure (svg): Two columns contrasting the direction of the double angle formulas with the direction of the power reduction formulas
Neither direction is more correct. Which you want depends entirely on whether the square or the large angle is the thing standing in your way.
Worked example
Example 10.4.4. The formulas are applied twice, because the first application leaves a square behind.
\[ \text{Rewrite } \sin^2(\theta)\cos^2(\theta) \text{ as a sum of cosines to the first power.} \]
Apply both power reduction formulas
Why: One to each squared factor.
\[ (\frac{1 - \cos 2 t}{2}) (\frac{1 + \cos 2 t}{2}) \]
Multiply out as a difference of squares
Why: The product of a difference and a sum.
\[ (\frac{1}{4}) (1 - \cos ^{2} 2 t) \]
Notice a square remains
Why: Cosine squared of 2 theta still has a power to strip.
\[ = \frac{1}{4} - (\frac{1}{4}) \cos ^{2} 2 t \]
Apply power reduction again, to the doubled angle
Why: This time the angle doubles from 2 theta to 4 theta.
\[ = \frac{1}{8} - (\frac{1}{8}) \cos 4 t \]
Figure (svg): The solution to Worked example reduce a product of squares shown as a ladder of expressions, one row per legal move
\[ \sin^2\theta\cos^2\theta = \frac{1}{8} - \frac{1}{8}\cos(4\theta) \]
Verify: test at pi over four
Why: The left side is one half times one half, which is one quarter. The right side is one eighth minus one eighth times the cosine of pi, which is one eighth plus one eighth, namely one quarter. They agree.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 778-778
Fill the middle
Reduce sine squared of 3 theta.
Fill in the blanks
\sin^2(3\theta) = \frac6 theta})}___
Why: The formula says sine squared of an angle equals one minus the cosine of twice that angle, over two. The angle here is 3 theta, so twice it is 6 theta. The commonest slip is to leave the inner angle unchanged, which produces an identity that fails at almost every value — testing at a single convenient angle catches it immediately.
Worked example
Same method, one more round of stripping.
\[ \text{Rewrite } \cos^4(\theta) \text{ in terms of cosines to the first power.} \]
Write the fourth power as a square of a square
Why: So that the power reduction formula applies.
\[ (\cos ^{2} t) ^{2} \]
Apply power reduction to the inner square
Why: Then square the result.
\[ (\frac{1 + \cos 2 t}{2}) ^{2} \]
Expand the square
Why: One plus twice the cosine plus the cosine squared, all over four.
\[ \frac{1 + 2 \cos 2 t + \cos ^{2} 2 t}{4} \]
Reduce the remaining square
Why: Cosine squared of 2 theta becomes one plus cosine of 4 theta over two.
\[ = \frac{3}{8} + (\frac{1}{2}) \cos 2 t + (\frac{1}{8}) \cos 4 t \]
Figure (svg): The solution to Worked example reduce a fourth power shown as a ladder of expressions, one row per legal move
\[ \cos^4\theta = \frac{3}{8} + \frac{1}{2}\cos(2\theta) + \frac{1}{8}\cos(4\theta) \]
Verify: test at zero
Why: The left side is one to the fourth, which is 1. The right side is three eighths plus one half plus one eighth, which is 1. They agree, and the sum of the coefficients being exactly 1 is a useful general check for these reductions.
Error analysis
A student reduces cosine squared of 2 theta.
Annotate
On: \( \cos^2(2\theta) = \frac{1 + \cos(2\theta)}{2} \)
The formula's variable is whatever sits inside the square, and everything else doubles relative to that. Writing the formula with a fresh letter — cosine squared of u equals one plus cosine of 2u over two — makes the substitution unambiguous.
Sorting
Some problems want powers removed and some want angles removed.
Sort into buckets
Sort each task by which family it needs.
Prediction
You reduce cosine to the sixth power using the power reduction formulas repeatedly.
Predict first
What is the largest angle that will appear in the final answer?
Correct: 6 theta.
Why: Each round of reduction halves the power and doubles the angle. Starting from a sixth power, reducing gives terms up to cosine cubed of 2 theta; expanding and reducing again gives terms up to cosine of 6 theta. In general reducing the nth power of a cosine produces angles up to n theta, which is the same bound the multiple-angle polynomials have — the two directions really are inverse to each other.
Real world
The average power delivered by an alternating current is proportional to the average of the square of the voltage, and the voltage is a sinusoid, V times the sine of omega t.
Discussion prompt
Use power reduction to explain why the average of the square of the voltage over a whole cycle is exactly half the square of the peak voltage, and say what that has to do with the figure quoted on domestic mains supplies.
Hint: Reduce the square, then think about what the cosine term averages to over a full cycle.
Answer:
\[ V^2\sin^2(\omega t) = V^2 \cdot \frac{1 - \cos(2\omega t)}{2} = \frac{V^2}{2} - \frac{V^2}{2}\cos(2\omega t) \]
Over a whole cycle the cosine term averages to zero, since it spends as long positive as negative. What is left is the constant, so the average of the squared voltage is exactly half the square of the peak.
The square root of that average is the peak divided by root two, and it is called the root-mean-square voltage. It is the figure quoted on mains supplies — a 230 volt supply actually peaks at about 325 volts. The whole convention exists because power depends on the square, and power reduction is what turns that square into something with an obvious average.
Section
Section 5
Concept
Replace theta by theta over two in the power reduction formulas and take a square root. The result expresses the cosine, sine or tangent of a half angle in terms of the cosine of the whole angle, with a sign that has to be supplied separately.
Theorem 10.19 — The half angle formulas, with the choice of sign determined by the quadrant in which the terminal side of the half angle lies.
\[ \cos\left(\tfrac{\theta}{2}\right) = \pm\sqrt{\frac{1 + \cos\theta}{2}}, \qquad \sin\left(\tfrac{\theta}{2}\right) = \pm\sqrt{\frac{1 - \cos\theta}{2}} \]
The sign is the whole difficulty. It cannot be recovered from the formula, because the square root destroyed it, and it must be read off the quadrant of the half angle — which is usually not the quadrant of the original.
Figure (svg): A diagram showing that halving an angle moves it into a different quadrant, with an angle in quadrant three whose half lies in quadrant two
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 778-779
Picture it
Halving an angle moves it, and usually into a different quadrant.
Figure (svg): A diagram showing that halving an angle moves it into a different quadrant, with an angle in quadrant three whose half lies in quadrant two
An angle in quadrant three has its half in quadrant two, where the sine is positive and the cosine negative — the opposite pattern from the original. Reading the sign off the original angle would get both wrong.
Worked example
Example 10.4.5, part 1. The same value found a different way last lesson, and it looks completely different.
\[ \text{Use a half angle formula to find the exact value of } \cos(15^\circ). \]
Recognise 15 as half of a special angle
Why: Thirty degrees is special and 15 is half of it.
\[ 15 = \frac{30}{2} \]
Decide the sign
Why: Fifteen degrees is in quadrant one, so its cosine is positive.
Substitute into the formula
Why: The cosine of 30 degrees is root three over two.
\[ \sqrt{\frac{1 + \sqrt{3} / 2}{2}} \]
Clear the compound fraction
Why: Multiply top and bottom inside the root by 2.
\[ \sqrt{\frac{2 + \sqrt{3}}{4}} \]
Figure (svg): The solution to Worked example the cosine of 15 degrees, again shown as a ladder of expressions, one row per legal move
\[ \cos(15^\circ) = \frac{\sqrt{2 + \sqrt{3}}}{2} \]
Verify: compare with the other route
Why: Last lesson the difference formula gave root six plus root two over four. Numerically that is about 0.9659, and this expression is the root of about 3.732, which is about 1.9319, over 2 — also about 0.9659. The two expressions are equal, though it takes some work to show it algebraically.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 779-779
Matching
Take the angle to lie in the standard interval from zero to two pi.
Match the pairs
Why: Halving maps the whole interval from 0 to two pi onto the interval from 0 to pi, so every half angle lands in quadrant one or two. Quadrants one and two halve into quadrant one; quadrants three and four halve into quadrant two. That means the sine of a half angle is always positive for an angle in this standard range — a fact worth knowing, though it fails once the original angle is allowed outside that interval.
Worked example
Example 10.4.5, part 2. Halve the interval before choosing the sign.
\[ \text{If } -\pi \le \theta \le 0 \text{ with } \cos(\theta) = -\tfrac{3}{5}, \text{ find } \sin\left(\tfrac{\theta}{2}\right). \]
Halve the interval
Why: Dividing every part of the inequality by 2.
\[ -\frac{\pi}{2} \le \frac{t}{2} \le 0 \]
Read the sign from that interval
Why: Between negative pi over two and zero the angle is in quadrant four, where the sine is negative.
Substitute into the formula
Why: One minus negative three fifths is eight fifths.
\[ -\sqrt{\frac{\frac{8}{5}}{2}} \]
Simplify and rationalise
Why: Eight fifths over two is four fifths, whose root is two over root five.
\[ -2 \sqrt{5} / 5 \]
Figure (svg): The solution to Worked example the sign comes from the half angle shown as a ladder of expressions, one row per legal move
\[ \sin\left(\frac{\theta}{2}\right) = -\frac{2\sqrt{5}}{5} \]
Verify: check the size
Why: Two root five over five is about 0.894, comfortably inside the interval from negative one to one. And the original angle is in quadrant two, since its cosine is negative and it lies between negative pi and zero — meaning it is between negative pi and negative pi over two — so its half really is in quadrant four, where a negative sine is right.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 779-779
Trap
\[ \theta \in \text{QIII}, \; \text{so } \sin\left(\tfrac{\theta}{2}\right) < 0 \]
Read the sign off the quadrant of theta
Why: The quadrant of theta is what the problem stated, so it is the one in view.
But an angle in quadrant three lies between pi and three pi over two, so its half lies between pi over two and three pi over four — which is quadrant two, where the sine is POSITIVE.
\[ \pi < \theta < \tfrac{3\pi}{2} \;\Longrightarrow\; \tfrac{\pi}{2} < \tfrac{\theta}{2} < \tfrac{3\pi}{4} \;\Longrightarrow\; \sin\left(\tfrac{\theta}{2}\right) > 0 \]
Halve the interval first, then read the quadrant of the result
Why: The formula's sign belongs to the half angle, so the half angle's quadrant is what decides it.
Divide the whole inequality by two before doing anything else. It costs one line and it is the only reliable way to get these signs right, because the quadrant of the half is genuinely unrelated to the quadrant of the original — quadrant three always halves into quadrant two, which is a different sign pattern entirely.
Faded example
Find the exact sine of 22.5 degrees.
Fill in the blanks
22.5^\circ = \frac45 deg}sqrt2/2, \text___ \;\Longrightarrow\; \sin(22.5^\circ) = \sqrt___}}___}
Why: Twenty-two and a half degrees is half of 45 degrees, whose cosine is root two over two. The half angle is in quadrant one so the sine is positive. Simplifying, the answer is the root of two minus root two, all over two, which is about 0.3827 — and the sine of 22.5 degrees is indeed about 0.3827.
Prediction
An angle theta lies strictly between zero and two pi.
Predict first
What can you say about the sign of the sine of theta over two?
Correct: It is always positive.
Why: Halving an interval from 0 to two pi gives an interval from 0 to pi, which is the entire upper half of the circle — and the sine is positive throughout it. So for any angle in one standard revolution, the sine of its half is positive, whichever quadrant the original was in. This is a genuinely useful shortcut, but note carefully that it depends on the stated interval: an angle between two pi and four pi has its half in the lower half plane, where the sine is negative.
Counterexample
A student proposes: the half angle formula for tangent always requires a plus-or-minus, so you can never determine the sign without knowing the quadrant.
Discussion prompt
Find a form of the tangent half angle formula that has no square root, and explain why it settles the sign automatically.
Hint: The book derives one in Example 10.4.5 part 3.
Answer:
\[ \tan\left(\frac{\theta}{2}\right) = \frac{\sin\theta}{1 + \cos\theta} \]
This form has no square root at all, so no sign was ever discarded. The sign of the answer is simply the sign of the sine of theta, since the denominator is never negative — the cosine is at least negative one, so one plus it is at least zero.
So the claim is false: the square-root form needs a quadrant, but this equivalent form does not. There is a second such form, one minus the cosine over the sine, obtained the same way. When a formula's plus-or-minus is inconvenient, it is worth asking whether an equivalent form avoids it — the square root is the cause of the ambiguity, not the mathematics.
Comparison
Fill the blanks from memory. Each row is one substitution away from the row above it.
Comparison matrix
| Family | What it does | Where it came from |
|---|---|---|
| Double angle | one angle in, twice the angle out | the sum identities with the two angles set equal |
| Cosine, three forms | so that a single value is always enough | trading squares with the Pythagorean identity |
| Power reduction | a square in, a first power out | the double angle formula solved for the square |
| Half angle | one angle in, half the angle out | power reduction with theta halved, then a root |
| Multiple angle | a large multiple as a polynomial | repeated sum and double angle formulas |
The right-hand column is a single chain, and nothing on it was proved from scratch. That is worth knowing before deciding how much of this lesson to memorise.
Pattern
Whether the question doubles an angle, halves one, or strips a power, the same five moves cover it.
The half angle sign is the single most-lost mark in this section, and it costs one line to get right. Divide the inequality by two, in writing, every time.
OpenStax Algebra and Trigonometry 2e, §9.3 Double-Angle, Half-Angle, and Reduction Formulas §9.3
Check
A double angle. Choose the form that fits what you have.
Check your understanding
If the cosine of theta is one third, what is the cosine of 2 theta?
Answer: B
Why: Use the form containing only the cosine: two cosine squared minus one. That gives two times one ninth minus one, which is two ninths minus one, namely negative seven ninths. No quadrant is needed because the cosine appears only squared.
Check
Power reduction. Watch which angle doubles.
Check your understanding
Which expression equals sine squared of 5 theta?
Answer: B
Why: The power reduction formula for sine says sine squared of an angle equals one minus the cosine of twice that angle, over two. The angle here is 5 theta, so twice it is 10 theta, and the sign is a minus because it is a sine.
Check
A half angle. Halve the interval before choosing the sign.
Check your understanding
If theta lies between pi and three pi over two, what is the sign of the cosine of theta over two?
Answer: B
Why: Halving the inequality gives theta over two between pi over two and three pi over four, which is quadrant two. There the cosine is negative, so the minus sign is taken in the half angle formula.
Real world
A projectile launched at speed v and angle theta above the horizontal lands, on level ground, at a range given by two v squared times the sine of theta times the cosine of theta, all over g.
Discussion prompt
Rewrite the range using a double angle identity, and use the new form to find the launch angle giving the greatest range. Explain why the rewritten form makes the answer obvious.
Hint: The numerator contains twice a sine times a cosine.
Answer:
\[ R = \frac{2v^2\sin\theta\cos\theta}{g} = \frac{v^2\sin(2\theta)}{g} \]
The range is now a single sine, and a sine is largest when its argument is pi over two. So 2 theta equals pi over two, giving theta equal to pi over four, or 45 degrees.
The original form hides this completely: it is a product of two functions that pull in opposite directions as theta grows, and finding the maximum would need calculus. The rewritten form makes it a one-line observation, because a bounded function's maximum is known without any work. The identity did not change the physics; it changed which fact about the expression was visible — and that is what identities are for.
It also predicts something less obvious: since the sine of 2 theta equals the sine of pi minus 2 theta, launch angles of 30 and 60 degrees give exactly the same range, as do any pair summing to 90 degrees.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
To find the cosine of 2 theta, how much do you need to know about theta?
Correct: Either the cosine or the sine of theta, and nothing else.
\[ \cos(2\theta) = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta \]
Why: The three forms of the cosine double angle identity include one containing only cosines and one containing only sines, so whichever value you have, a form exists that uses only it. And because that value appears squared, no quadrant is needed either — squaring destroys the sign information that a quadrant would have supplied. This is a genuine asymmetry with the sine of 2 theta, which does require both values and therefore does require a quadrant when only one is given.
Explain it
They have been handed the double angle, power reduction and half angle formulas as three separate lists and are trying to memorise all eight.
Discussion prompt
In no more than five sentences, show them how the three lists collapse into one chain. Then give them the single thing they must not forget in the half angle formulas.
Hint: Start from the sum identity they already know.
Answer:
A usable answer: start from the cosine sum formula, which you already have, and set the two angles equal — that is the double angle formula, and the three versions of it are just the Pythagorean identity swapping a cosine squared for a sine squared. Now solve that for the squared term and you have the power reduction formulas. Now put theta over two wherever theta was and take a square root, and you have the half angle formulas. Three lists, one chain, and nothing new to prove.
The one thing not to forget: the plus-or-minus in a half angle formula is decided by where theta over two lands, not where theta lands. Halve the interval on paper before you choose, because an angle in quadrant three has its half in quadrant two, where the signs are completely different.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: The form choice is fixed by asking which value you were given and picking the form containing only that. The half angle sign is fixed by halving the interval in writing before choosing. Full reduction is fixed by expecting to apply the formula more than once and checking whether any square survives each round. Multiple angles are fixed by remembering the order: split as a sum, expand, substitute double angles, then eliminate the sine squares. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of the page write the cosine sum identity in a box. Draw an arrow down from it labelled put beta equal to alpha, and write the three forms of the cosine double angle identity, plus the sine and tangent versions. Draw a second arrow down labelled solve for the square, and write the two power reduction formulas. Draw a third arrow labelled halve the angle and take roots, and write the three half angle formulas with their plus-or-minus. To the right of the half angle formulas, draw the unit circle and mark which quadrant each of the four quadrants halves into, so the sign question can be answered by looking rather than reasoning. In the bottom left, derive the cosine of three theta as a polynomial, showing every step. In the bottom right, reduce cosine to the fourth power completely, and check your answer by putting theta equal to zero. Finally, circle the one identity on the page that was proved in the previous lesson rather than derived in this one.
The circled identity should be the cosine sum formula at the top. Everything below it on the page is a consequence, and there is no second proof anywhere in this lesson.
Recap
Five things, and the third one is where the marks are usually lost.
| If the question says | Your first move is |
|---|---|
| Find cos 2t given cos t | Use the form with only cosines; no quadrant needed |
| Find sin 2t given one value | Recover the other with the Pythagorean identity and the quadrant |
| Rewrite as first powers | Apply power reduction, then check whether a square survives |
| Find a value at half a special angle | Halve the interval, choose the sign, then substitute |
| Express cos of a multiple as a polynomial | Split as a sum and expand |
One family of identities remains: the ones that convert between products and sums. They are the last of the chapter and, like everything in this lesson, they are derived from the sum and difference formulas rather than proved anew.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 776-780 — everything on these slides traces back here
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