Derives the double-angle identities by setting the two angles equal in the sum formulas, gives the cosine's three equivalent forms and says when to use each, rearranges them into the reduction formulas, and runs them backwards to reach the half-angle formulas — where the sign is decided by the halved angle's quadrant.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 7 — Trigonometric Identities and Equations
§7.3 Double-Angle, Half-Angle, and Reduction Formulas, pp. 821-834
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §7.3 Double-Angle, Half-Angle, and Reduction Formulas §7.3, pp. 821-834 — the pages these objectives are drawn from
Warm-up
The sum formulas take two angles. Nothing stops them being equal.
Discussion prompt
Put B equal to A in the sine sum formula. What do you get?
Hint: Write out the formula and substitute.
Answer:
The sine of a sum is sine A cosine B plus cosine A sine B. With B equal to A both products become sine A cosine A.
So the sine of twice A is twice the sine of A times the cosine of A. A brand new identity, from one substitution.
This is the whole method of the section. Every double-angle formula is a sum formula with the angles set equal, so there is nothing new to memorise — only a substitution to perform.
Concept
Setting the two angles equal in the sum formulas gives the double-angle identities. Solving those for the squared terms gives the reduction formulas, and replacing the angle by half of itself gives the half-angle formulas.
\[ \sin 2A=2\sin A\cos A, \quad \cos 2A=\cos^2A-\sin^2A \]
Three families of identities from one substitution and two rearrangements. The only genuinely new decision in the whole section is which sign to take when the half-angle formula's square root is undone.
Figure (svg): Three cards giving the double-angle identities for the sine, the cosine in its three equivalent forms, and the tangent
OpenStax, Precalculus, §7.3 Double-Angle, Half-Angle, and Reduction Formulas §7.3, pp. 821-825
Section
Section 1
Concept
Putting the two angles equal in each sum formula gives a double-angle identity for the sine, the cosine and the tangent.
The tangent formula inherits its denominator's sign from the sum formula, where the denominator was one minus a product. With the angles equal that product becomes a square, which is why the denominator reads one minus tangent squared.
Figure (svg): Three cards giving the double-angle identities for the sine, the cosine in its three equivalent forms, and the tangent
OpenStax, Precalculus, §7.3 Double-Angle, Half-Angle, and Reduction Formulas §7.3, pp. 821-826
Picture it
Note that the cosine alone has multiple forms.
Figure (svg): Three cards giving the double-angle identities for the sine, the cosine in its three equivalent forms, and the tangent
The sine and tangent each have one form; the cosine has three, because the Pythagorean identity can replace either squared term. That flexibility is used constantly.
Worked example
One substitution.
\[ \text{Derive } \sin 2A \text{ from the sum formula.} \]
Write the sum formula
Why: For the sine.
\[ \sin A \cos B + \cos A \sin B \]
Set B equal to A
Why: Both angles the same.
\[ \sin A \cos A + \cos A \sin A \]
Combine like terms
Why: Two identical products.
\[ 2 \sin A \cos A \]
State the result
Why: The double-angle identity.
\[ \sin 2 A \]
Figure (svg): Three cards giving the double-angle identities for the sine, the cosine in its three equivalent forms, and the tangent
\[ \sin 2A=2\sin A\cos A \]
Verify: test at a convenient angle
Why: At thirty degrees the right side is twice one half times root three over two, which is root three over two — and that is the sine of sixty degrees. The substitution is valid for every angle since nothing about A was assumed.
OpenStax, Precalculus, §7.3 Double-Angle, Half-Angle, and Reduction Formulas §7.3, pp. 821-823
Prediction
The sine of a doubled angle is twice the sine times something.
Predict first
What is the missing factor?
Correct: The cosine of the angle.
Why: Setting the two angles equal in the sine sum formula gives sine times cosine twice over, so the result is twice their product. The cosine factor is what makes the identity true and its omission is the commonest error here.
Worked example
Find both functions first, then substitute.
\[ \text{Given } \sin A=\tfrac{3}{5} \text{ with } A \text{ in QII, find } \sin 2A. \]
Find the cosine
Why: By the Pythagorean identity.
\[ | \cos A | = \frac{4}{5} \]
Apply the quadrant
Why: The cosine is negative in QII.
\[ \cos A = -\frac{4}{5} \]
Apply the double-angle formula
Why: Twice the product.
\[ 2(\frac{3}{5}) (-\frac{4}{5}) \]
Compute
Why: Multiply out.
\[ -\frac{24}{25} \]
Figure (svg): The solution to Worked example compute a doubled angle's values shown as a ladder of expressions, one row per legal move
\[ \sin 2A=-\frac{24}{25} \]
Verify: check the quadrant of the doubled angle
Why: A is between ninety and a hundred and eighty degrees, so twice A is between a hundred and eighty and three hundred and sixty — where the sine is negative. The negative answer is consistent, which is a genuine check on the sign of the cosine used.
OpenStax, Precalculus, §7.3 Double-Angle, Half-Angle, and Reduction Formulas §7.3, pp. 823-826
Trap
\[ \sin 2A=2\sin A \]
Move the 2 outside the function
Why: The coefficient is treated as though it factored out of the sine.
At thirty degrees this claims 1, but the sine of sixty is about 0.87.
Doubling the angle is not doubling the value. The sine is a function, and a coefficient inside its argument cannot be moved outside.
The real formula is twice the sine times the cosine — the extra factor is exactly what the false version omits.
This is the same category error as distributing over a sum, and the same one-line numerical test disposes of it. It survives only when it is never checked.
Faded example
Set B equal to A in the tangent sum formula.
Fill in the blanks
\tan 2A=\frac22=\frac___\tan A}___}}A}
Why: The numerator's two identical terms combine into twice the tangent and the denominator's product becomes a square. The whole derivation is one substitution and one collection of like terms.
Sorting
Check it against the formulas.
Sort into buckets
Sort each statement.
Explain it to yourself
Every double-angle formula comes from a sum formula.
Discussion prompt
Explain why that means there is almost nothing new to learn.
Hint: What is the only new step?
Answer:
The sum formulas take two independent angles, and nothing in their derivation required the angles to be different. So substituting one for the other is always legal.
That substitution turns each sum formula directly into the corresponding double-angle formula, with only like terms to collect afterwards.
So the only new content is the substitution itself, which takes a line. Anyone who knows §7.2 can reconstruct all of §7.3's double-angle identities on demand, which is far safer than recalling three more formulas.
Section
Section 2
Concept
The Pythagorean identity can replace either squared term in the cosine's basic form, producing two more versions. Each is written in a single function, which makes it useful in a different situation.
All three are the same identity and give the same answer. Choosing between them is purely a matter of convenience, and choosing well saves computing a value you do not need.
| form | written in | use it when |
|---|---|---|
| cos squared minus sin squared | both | you know both values |
| 2 cos squared minus 1 | the cosine | only the cosine is known |
| 1 minus 2 sin squared | the sine | only the sine is known |
Figure (svg): A diagram showing how the three forms of the cosine double-angle identity are obtained from one another using the Pythagorean identity
OpenStax, Precalculus, §7.3 Double-Angle, Half-Angle, and Reduction Formulas §7.3, pp. 825-829
Picture it
One Pythagorean substitution each way.
Figure (svg): A diagram showing how the three forms of the cosine double-angle identity are obtained from one another using the Pythagorean identity
The arrows are substitutions, not new derivations. Any of the three can be turned into either of the others in one step, so remembering one is enough.
Worked example
Only the sine is given, so use the sine-only form.
\[ \text{Given } \sin A=\tfrac{1}{3}, \text{ find } \cos 2A. \]
Choose the form
Why: The one written in the sine.
\[ 1 - 2 \sin ^{2} A \]
Substitute
Why: The sine is one third.
\[ 1 - 2(\frac{1}{9}) \]
Compute
Why: Two ninths subtracted.
\[ 1 - \frac{2}{9} \]
Simplify
Why: A common denominator.
\[ \frac{7}{9} \]
Figure (svg): A diagram showing how the three forms of the cosine double-angle identity are obtained from one another using the Pythagorean identity
\[ \cos 2A=\frac{7}{9} \]
Verify: check no quadrant was needed
Why: This form uses only the sine squared, so the sign of the cosine of A never entered — and the answer is the same whichever quadrant A is in. Choosing this form made the quadrant information unnecessary, which is the practical value of having three versions.
OpenStax, Precalculus, §7.3 Double-Angle, Half-Angle, and Reduction Formulas §7.3, pp. 826-828
Matching
Choose the form that needs nothing extra.
Match the pairs
Why: Each form is written in a single function precisely so it can be used without computing the other. Matching the form to the given information avoids both extra work and the quadrant decision that finding the other value would require.
Worked example
Any form becomes any other in one substitution.
\[ \text{Turn } \cos^2A-\sin^2A \text{ into a form in the cosine only.} \]
Replace the sine squared
Why: By the Pythagorean identity.
\[ \cos ^{2} - (1 - \cos ^{2}) \]
Distribute the minus
Why: Both terms change sign.
\[ \cos ^{2} - 1 + \cos ^{2} \]
Collect
Why: Two cosine squared terms.
\[ 2 \cos ^{2} - 1 \]
State the result
Why: The cosine-only form.
Figure (svg): The solution to Worked example convert between forms shown as a ladder of expressions, one row per legal move
\[ \cos 2A=2\cos^2A-1 \]
Verify: test at a convenient angle
Why: At sixty degrees the cosine is one half, so this gives twice one quarter minus one, which is negative one half — and the cosine of a hundred and twenty degrees is indeed negative one half. The conversion preserved the identity.
OpenStax, Precalculus, §7.3 Double-Angle, Half-Angle, and Reduction Formulas §7.3, pp. 828-829
Error analysis
A student is given only the cosine and picks a form.
Annotate
On: \( \cos A=\tfrac{2}{5} \;\Longrightarrow\; \cos 2A=\cos^2A-\sin^2A \)
The three forms exist precisely so that one of them matches the information available. Picking the basic form by habit creates work that the right choice avoids entirely, including a quadrant decision that can go wrong.
Prediction
You use the form written in the sine squared.
Predict first
Do you need to know A's quadrant?
Correct: No, since only the square of the sine appears.
Why: Squaring destroys the sign, so the same answer results whether the sine is positive or negative. Choosing a form that uses only the squared value of what you were given sidesteps the quadrant question altogether.
Faded example
Replace the cosine squared this time.
Fill in the blanks
\cos^2A-\sin^2A=(1-\sin^2}A)-\sin^2A=1-2\sin^2A
Why: Substituting one minus sine squared for the cosine squared and collecting gives one minus twice the sine squared. The two conversions are mirror images of each other, replacing whichever squared term you want to eliminate.
Step zero
You are asked for the cosine of a doubled angle.
Discussion prompt
What do you decide before writing anything?
Hint: Three forms are available.
Answer:
Decide which of the three forms to use, based on what you have been given rather than on which one you remember best.
If only the sine is known, use the sine-only form; if only the cosine, the cosine-only form. This avoids computing the other value, and with it avoids a quadrant decision.
Only if both are already known is the basic form the natural choice. The decision costs nothing and can save half the problem, which is why the three forms are given rather than just one.
Section
Section 3
Concept
Solving the cosine's second and third forms for the squared term expresses a square in terms of a cosine of the doubled angle — lowering the power at the cost of doubling the angle.
This trade is what makes several standard calculus integrals possible: a squared trigonometric function cannot be integrated directly, but a first power of a cosine can. The section's name comes from this reduction of power.
Figure (svg): A card showing the reduction formulas that rewrite a squared trigonometric function in terms of a cosine of the doubled angle
OpenStax, Precalculus, §7.3 Double-Angle, Half-Angle, and Reduction Formulas §7.3, pp. 829-831
Picture it
Squares on the left, first powers on the right.
Figure (svg): A card showing the reduction formulas that rewrite a squared trigonometric function in terms of a cosine of the doubled angle
The only difference between the two is the sign in the numerator: minus for the sine and plus for the cosine. That matches the cosine of a doubled angle being large when the cosine is large.
Worked example
Solve one of the cosine forms.
\[ \text{Derive } \sin^2A=\frac{1-\cos 2A}{2}. \]
Start from the sine-only form
Why: The cosine of twice A.
\[ \cos 2 A = 1 - 2 \sin ^{2} A \]
Isolate the squared term
Why: Move it across.
\[ 2 \sin ^{2} A = 1 - \cos 2 A \]
Divide by two
Why: Isolate the square.
\[ \sin ^{2} A = \frac{1 - \cos 2 A}{2} \]
State the result
Why: The reduction formula.
Figure (svg): A card showing the reduction formulas that rewrite a squared trigonometric function in terms of a cosine of the doubled angle
\[ \sin^2A=\frac{1-\cos 2A}{2} \]
Verify: test at a convenient angle
Why: At thirty degrees the left side is one quarter, and the right side is one minus the cosine of sixty, over two — that is one minus one half over two, which is one quarter. The rearrangement preserved the identity.
OpenStax, Precalculus, §7.3 Double-Angle, Half-Angle, and Reduction Formulas §7.3, pp. 829-830
Faded example
The sign differs from the sine's.
Fill in the blanks
\cos^2A=\frac+}\cos 2A}___
Why: The cosine's version takes a plus and the sine's a minus. Checking at zero settles which is which: the cosine squared is 1 there, and only the plus version gives that.
Worked example
Apply the formula twice.
\[ \text{Express } \cos^4A \text{ without powers above one.} \]
Write as a square of a square
Why: The fourth power.
\[ (\cos ^{2} A) ^{2} \]
Apply the reduction formula
Why: To the inner square.
\[ (\frac{1 + \cos 2 A}{2}) ^{2} \]
Expand
Why: A binomial square.
\[ \frac{1 + 2 \cos 2 A + \cos ^{2} 2 A}{4} \]
Reduce the remaining square
Why: Now with angle 4A.
\[ \cos ^{2} 2 A = \frac{1 + \cos 4 A}{2} \]
Figure (svg): The solution to Worked example reduce a fourth power shown as a ladder of expressions, one row per legal move
\[ \cos^4A=\frac{3+4\cos 2A+\cos 4A}{8} \]
Verify: test at a convenient angle
Why: At zero the left side is 1, and the right side is three plus four plus one, over eight — which is 1. Each application of the formula halves the power and doubles the angle, so a fourth power takes two rounds.
OpenStax, Precalculus, §7.3 Double-Angle, Half-Angle, and Reduction Formulas §7.3, pp. 830-831
Trap
\[ \sin^2A=\frac{1-\cos A}{2} \]
Apply the formula with the original angle
Why: The doubling inside the cosine is dropped.
At thirty degrees this gives one minus root three over two, over two — about 0.067, not 0.25.
The cosine on the right takes twice the angle, which is the whole point of the trade: the power drops and the angle rises.
Without the doubling the statement is simply false, and a single numerical check exposes it.
Name the trade when you use the formula — 'power down, angle up' — and the doubling is much harder to lose.
Prediction
You apply a reduction formula to a squared function of A.
Predict first
What angle appears on the right?
Correct: Twice A.
Why: The formulas are rearrangements of the cosine double-angle identity, so the doubled angle they contain carries over. Lowering the power always costs a doubling of the angle, which is the trade the name refers to.
Sorting
The numerator's sign differs.
Sort into buckets
Sort each expression by which reduction formula handles it.
Explain it
The formulas are called reduction formulas.
Discussion prompt
Explain to a classmate what is being reduced and what it costs.
Hint: Something goes down and something goes up.
Answer:
The power is reduced — a square becomes a first power. That is what the name refers to.
The cost is that the angle doubles. So the expression is not simpler in every sense; it is simpler in the one way that matters for integration and for further manipulation.
A good explanation names why anyone wants this: a first power of a cosine is easy to integrate and a square is not, so this trade is what makes several standard calculus problems tractable. It is a technique borrowed forward, not an end in itself.
Section
Section 4
Concept
Replacing the angle by half of itself in the reduction formulas and taking a square root gives the half-angle formulas — with a sign that must be decided separately.
The tangent's alternative forms — the sine over one plus the cosine, and one minus the cosine over the sine — carry their sign automatically. When a half-angle tangent is wanted, those forms are strictly better because they remove the decision.
Figure (svg): A card showing that the half-angle formula's sign is decided by the quadrant of the half angle, not of the original angle
OpenStax, Precalculus, §7.3 Double-Angle, Half-Angle, and Reduction Formulas §7.3, pp. 831-834
Picture it
The original angle and its half can be in different quadrants.
Figure (svg): A card showing that the half-angle formula's sign is decided by the quadrant of the half angle, not of the original angle
This is the one genuinely new judgement in the section. Halve first, then read the quadrant of the result — the original angle's quadrant is irrelevant to the sign.
Worked example
Fifteen degrees is half of thirty.
\[ \text{Find } \sin 15^\circ \text{ using a half-angle formula.} \]
Identify the whole angle
Why: Twice fifteen.
\[ 30 ^\circ \]
Determine the sign
Why: Fifteen degrees is in QI, where the sine is positive.
Apply the formula
Why: One minus the cosine of thirty, over two.
\[ \sqrt{\frac{1 - r 3 / 2}{2}} \]
Simplify
Why: Clear the inner fraction.
\[ \sqrt{\frac{2 - r 3}{4}} \]
Figure (svg): A card showing that the half-angle formula's sign is decided by the quadrant of the half angle, not of the original angle
\[ \sin 15^\circ=\frac{\sqrt{2-\sqrt{3}}}{2} \]
Verify: compare with the §7.2 answer
Why: Section 7.2 gave root six minus root two over four for this value, about 0.2588. This expression evaluates to the same 0.2588, so the two forms agree — nested radicals often hide an equivalence that a numerical check reveals immediately.
OpenStax, Precalculus, §7.3 Double-Angle, Half-Angle, and Reduction Formulas §7.3, pp. 831-833
Prediction
The angle A is 250 degrees.
Predict first
Which quadrant is half of A in?
Correct: The second.
Why: Half of 250 is 125 degrees, which lies between ninety and a hundred and eighty. The original angle was in the third quadrant, so the two disagree — which is exactly why the half angle must be computed rather than assumed.
Worked example
Here the two quadrants disagree.
\[ \text{Given } \cos A=\tfrac{1}{2} \text{ with } A=300^\circ, \text{ find } \sin\tfrac{A}{2}. \]
Halve the angle
Why: Compute it explicitly.
\[ 150 ^\circ \]
Read that quadrant
Why: The second.
Determine the sign
Why: The sine is positive in QII.
Apply the formula
Why: One minus one half, over two.
\[ \sqrt{\frac{1}{4}} \]
Figure (svg): The solution to Worked example the sign requires care shown as a ladder of expressions, one row per legal move
\[ \sin\tfrac{A}{2}=\tfrac{1}{2} \]
Verify: check against the original angle's quadrant
Why: A is in the fourth quadrant, where the sine is negative — so using A's quadrant would have given the wrong sign. The half angle at a hundred and fifty degrees has a positive sine, and the sine of a hundred and fifty degrees is indeed one half. Halving first is what makes this right.
OpenStax, Precalculus, §7.3 Double-Angle, Half-Angle, and Reduction Formulas §7.3, pp. 833-834
Error analysis
A student computes a half-angle sine.
Annotate
On: \( A=200^\circ \text{ is in QIII, so } \sin\tfrac{A}{2}<0 \)
Halving moves an angle into a different quadrant more often than not, since it compresses the whole circle into a half. The habit of computing the half angle as a number first makes the error impossible.
Faded example
The sign in the numerator differs from the sine's.
Fill in the blanks
\cos\frac+2=\pm\sqrt___}\cos A}___}}
Why: The cosine version takes a plus, matching its reduction formula. Checking at A equal to zero settles it: the cosine of zero is 1, and only the plus version gives that.
Sorting
Some forms carry their sign automatically.
Sort into buckets
Sort each formula.
Explain it to yourself
The half angle's quadrant decides, not the original's.
Discussion prompt
Explain why that is the right rule.
Hint: Which angle's sine is the formula computing?
Answer:
The formula computes the sine of the half angle, so it is that angle whose quadrant determines whether the value is positive or negative. The original angle appears only inside the expression under the root.
The confusion arises because the original angle is what the problem states, so it is the one in mind. But it is not the angle the answer describes.
And the two often disagree, since halving compresses the whole circle into a half turn. Computing the half angle as an explicit number first removes the ambiguity, which is why it is worth doing even when it seems unnecessary.
Section
Section 5
Concept
Each family answers a different structural signal: a doubled argument, a squared function, or a halved argument.
The last two rows matter as much as the first three. These identities are used in both directions, and recognising the right side of a formula in an expression is often what unlocks a verification.
| what you see | what to use |
|---|---|
| a doubled argument | a double-angle identity |
| a squared function | a reduction formula |
| a halved argument | a half-angle formula |
| sinA cosA together | the sine double-angle identity, backwards |
| 1 plus or minus cos 2A | a reduction formula, backwards |
Figure (svg): Three cards giving the double-angle identities for the sine, the cosine in its three equivalent forms, and the tangent
OpenStax, Precalculus, §7.3 Double-Angle, Half-Angle, and Reduction Formulas §7.3, pp. 821-834
Picture it
Reading these left to right or right to left are both standard moves.
Figure (svg): Three cards giving the double-angle identities for the sine, the cosine in its three equivalent forms, and the tangent
Seeing twice a sine times a cosine and recognising it as a single sine of a doubled angle is the reverse reading, and it is what makes many verifications collapse in one step.
Worked example
Recognise the right side of a formula.
\[ \text{Verify } \frac{\sin 2A}{\sin A}=2\cos A. \]
Expand the numerator
Why: By the double-angle identity.
\[ 2 \sin A \cos A \]
Write the quotient
Why: Over the sine.
\[ 2 \sin A \cos A / \sin A \]
Cancel
Why: One factor of the sine.
\[ 2 \cos A \]
Compare
Why: It matches the right side.
Figure (svg): The solution to Worked example verify using a double-angle identity shown as a ladder of expressions, one row per legal move
\[ \frac{\sin 2A}{\sin A}=2\cos A \]
Verify: note the domain
Why: The cancellation requires the sine to be nonzero, which is exactly where the left side is undefined anyway. So the identity holds wherever both sides are defined, which is all that is ever claimed.
OpenStax, Precalculus, §7.3 Double-Angle, Half-Angle, and Reduction Formulas §7.3, pp. 826-829
Prediction
You see twice the sine of 5A times the cosine of 5A.
Predict first
What does it simplify to?
Correct: The sine of 10A.
Why: This is the right side of the sine double-angle identity with the inner angle being 5A, so the identity read backwards gives the sine of twice that. Recognising the pattern twice-sine-times-cosine is one of the most useful reverse readings in the chapter.
Worked example
The expression is the right side of an identity.
\[ \text{Simplify } 1-2\sin^2 3A. \]
Recognise the pattern
Why: The sine-only cosine double-angle form.
\[ 1 - 2 \sin ^{2} x = \cos 2 x \]
Identify the inner angle
Why: Here it is 3A.
\[ x = 3 A \]
Apply the identity backwards
Why: Double the inner angle.
\[ \cos 6 A \]
State the result
Why: One term.
\[ \cos 6 A \]
Figure (svg): Three cards giving the double-angle identities for the sine, the cosine in its three equivalent forms, and the tangent
\[ 1-2\sin^2 3A=\cos 6A \]
Verify: test at a convenient angle
Why: At A equal to ten degrees the left side is one minus twice the sine squared of thirty, which is one minus one half, or one half. The right side is the cosine of sixty degrees, also one half — confirming that the inner angle doubles rather than the outer one.
OpenStax, Precalculus, §7.3 Double-Angle, Half-Angle, and Reduction Formulas §7.3, pp. 829-832
Trap
\[ 1-2\sin^2 3A=\cos 3A \]
Keep the inner angle unchanged
Why: The identity is applied without doubling the argument.
At ten degrees this claims the cosine of thirty, about 0.87, but the value is 0.5.
The identity doubles whatever angle appears inside. With 3A inside the sine, the cosine gets 6A.
Reading a formula backwards keeps every structural feature, including the doubling — it is the same identity, used right to left.
Substituting a name for the inner angle makes this safe. Call it x, apply the identity to get the cosine of 2x, then put 3A back in for x.
Matching
Each is the right side of an identity.
Match the pairs
Why: Two of these collapse to the same thing, which is the point of the cosine having three forms. Recognising any of them in an expression lets a verification take one step instead of several.
Faded example
An expression matching a reduction formula.
Fill in the blanks
\frac24=\cos^___}(___A)
Why: The reduction formula's right side has the doubled angle, so recovering the squared form halves it — eight becomes four. The plus sign in the numerator identifies this as the cosine's version rather than the sine's.
Explain it
These identities are used both ways.
Discussion prompt
Explain to a classmate why reading them backwards matters as much as forwards.
Hint: What does a verification usually need?
Answer:
Forwards, they expand a compound argument into something in a single angle. That is the obvious use and the one every example shows first.
Backwards, they collapse several terms into one. An expression like twice a sine times a cosine is three symbols that become one, which is often the whole of a verification.
And backwards is harder, because it requires recognising a pattern rather than following a rule. A good explanation suggests reading each identity aloud right to left once, so the right-hand shapes become as familiar as the left-hand ones.
Comparison
Fill the blanks from memory. Each answers a different structural signal.
Comparison matrix
| double-angle | reduction | half-angle | |
|---|---|---|---|
| signal | a doubled argument | a squared function | a halved argument |
| effect on the angle | halves it | doubles it | doubles it inside |
| effect on the power | raises it to two | lowers it to one | introduces a root |
| needs a sign decision | no | no | yes, from the half angle's quadrant |
The last row is the only place in the section where a genuine judgement is required, and it is where the errors are.
Pattern
Five steps, and the second is the one that is skipped.
Step 2 looks unnecessary and is the reason the error in this section is so common. Writing the half angle down turns a judgement into a lookup.
OpenStax Algebra and Trigonometry 2e, §9.3 Double-Angle, Half-Angle, and Reduction Formulas §9.3
Check
The sine double-angle identity.
Check your understanding
What does the sine of a doubled angle equal?
Answer: A
Why: Setting the two angles equal in the sine sum formula gives sine times cosine twice over. The cosine factor is essential, and dropping it is the standard error.
Check
The half-angle sign.
Check your understanding
For an angle A of 300 degrees, what determines the sign of the half angle's sine?
Answer: A
Why: The formula computes the sine of the half angle, so the half angle's own quadrant decides. Here 150 degrees is in the second quadrant, where the sine is positive — even though 300 degrees is in the fourth, where it is negative.
Check
The reduction formulas.
Check your understanding
What does the sine squared of A equal?
Answer: A
Why: This is the cosine double-angle identity in its sine-only form, solved for the squared term. The angle doubles, which is the price paid for lowering the power.
Real world
Squaring a signal is what a power meter does, and reduction formulas describe the result.
Discussion prompt
An alternating current varies as a sine wave. Why does its power involve a cosine of twice the frequency?
Hint: Power is proportional to the square of the current.
Answer:
Power goes as the square of the current, so a sinusoidal current gives a squared sine for the power.
The reduction formula rewrites that square as a constant minus a cosine of the doubled angle. So the power has a steady average part plus a ripple at twice the original frequency.
Which is why mains-powered equipment hums at twice the supply frequency rather than at the supply frequency itself. The reduction formula predicts the pitch of the hum, and the constant term it isolates is exactly the average power the meter reports.
Commit first
State your confidence along with your answer.
Predict first
Why does the cosine have three double-angle forms?
Correct: The Pythagorean identity can replace either squared term.
Why: The basic form contains both a cosine squared and a sine squared, and either can be rewritten in terms of the other. That gives two more versions, each written in a single function, and all three are the same identity.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
A classmate used the original angle's quadrant to pick a half-angle sign. Explain the mistake.
Hint: Which angle does the answer describe?
Answer:
The formula gives the sine of the half angle, so it is that angle whose quadrant decides the sign. The original angle only supplies the cosine value inside.
And halving usually moves the angle into a different quadrant. Three hundred degrees is in the fourth, but its half at a hundred and fifty is in the second — opposite signs for the sine.
The fix is mechanical: write the half angle down as a number before deciding anything. A good explanation stresses that this turns a judgement into a lookup, which is why it eliminates the error rather than merely reducing it.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The fourth is the only genuinely new judgement in the section and where nearly every error occurs. The second saves the most work once it becomes automatic.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Start from the cosine sum formula and derive the double-angle identity, then show both Pythagorean substitutions that give the other two forms. From those, derive both reduction formulas by solving for the squared term. Finally write one half-angle formula and, beside it, a worked instance where the original angle and its half are in different quadrants.
If every formula on your page arrived by derivation rather than recall, and your half-angle instance shows the two quadrants disagreeing, the section is genuinely yours rather than memorised.
Recap
Five things, and the first makes the rest almost free.
| if you remember one thing | it should be this |
|---|---|
| about double angles | they are the sum formulas with the two angles equal |
| about the cosine's forms | pick the one written in the function you were given |
| about reduction | power down, angle up — that is the whole trade |
| about half angles | compute the half angle first, then read its quadrant |
Section 7.4 completes the identity toolkit with the sum-to-product and product-to-sum formulas, which convert between the two forms that acoustics and signal processing each prefer.
OpenStax, Precalculus, §7.3 Double-Angle, Half-Angle, and Reduction Formulas §7.3, pp. 821-834 — everything on these slides traces back here
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