The double-angle formulas including the three equivalent forms for cosine, the half-angle formulas and the quadrant rule for their signs, finding exact values by halving a special angle, working from a given function value, and simplifying expressions and deriving a projectile range model.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 14 — Trigonometric Graphs, Identities, and Equations
Apply Double-Angle and Half-Angle Formulas
Objectives
Five outcomes. Two angles made equal, and then one angle halved.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.7 Apply Double-Angle and Half-Angle Formulas §14.7, pp. 955-961 — the lesson these objectives are drawn from
Warm-up
Lesson 14.6 gave you the sine of a sum: sin a cos b plus cos a sin b.
Discussion prompt
What happens to that formula if the two angles are the same? Set b equal to a and simplify.
Hint: Both terms become the same product.
Answer:
\[ \sin(a+a) = \sin a\cos a+\cos a\sin a \]
\[ \sin 2a = 2\sin a\cos a \]
Two identical terms collapsed into one. Doing the same to the cosine and tangent formulas gives the rest of the double-angle formulas — no new work, just a special case of what you already have.
Concept
Setting the two angles equal in the sum formulas gives the double-angle formulas. Solving a cosine double-angle formula for the squared function and taking a square root gives the half-angle formulas.
double-angle formulas — Identities giving the sine, cosine and tangent of twice an angle in terms of the functions of the angle itself. The half-angle formulas run the cosine version backwards and introduce a square root.
\[ \sin 2a = 2\sin a\cos a; \qquad \sin\frac{a}{2} = \pm\sqrt{\frac{1-\cos a}{2}} \]
The square root in the half-angle formulas is what forces a sign choice, and the quadrant of the HALF angle is what settles it.
Figure (svg): Two columns comparing the double-angle formulas with the half-angle ones
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.7 Apply Double-Angle and Half-Angle Formulas §14.7, pp. 955-955
Section
Section 1
Concept
Setting both angles equal gives sin 2a as twice sin a cos a, and cos 2a in three equivalent forms. The tangent formula follows the same way.
\[ \cos 2a = \cos^2a-\sin^2a = 2\cos^2a-1 = 1-2\sin^2a \]
The three cosine forms differ only by a substitution of the Pythagorean identity, and the useful one is whichever leaves only the function already in the problem.
Figure (svg): The double-angle formulas, with three equivalent forms for the cosine
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.7 Apply Double-Angle and Half-Angle Formulas §14.7, pp. 955-955 — Double-Angle and Half-Angle Formulas
Picture it
The double-angle formulas in full.
Figure (svg): The double-angle formulas, with three equivalent forms for the cosine
Each cosine form is reached from the first by replacing one squared function using the Pythagorean identity, so all three always agree.
Worked example
Setting both angles equal in the sum formulas.
\[ \text{Derive } \sin 2a, \; \cos 2a \text{ and } \tan 2a \text{ from the sum formulas.} \]
Sine of a sum with b equal to a
Why: The two terms become identical.
\[ 2 \sin a \cos a \]
Cosine of a sum with b equal to a
Why: Like with like.
\[ \cos ^{2} a - \sin ^{2} a \]
Rewrite the cosine two more ways
Why: Substitute the Pythagorean identity for one square.
\[ 2 \cos ^{2} a - 1; 1 - 2 \sin ^{2} a \]
Tangent of a sum with b equal to a
Why: Numerator doubles; denominator gains a square.
\[ 2 \tan a / (1 - \tan ^{2} a) \]
Figure (svg): The double-angle formulas, with three equivalent forms for the cosine
\[ \sin 2a = 2\sin a\cos a; \; \cos 2a = \cos^2a-\sin^2a \]
Verify: test at a equal to 30 degrees
Why: Sin 60 is root 3 over 2, and twice sin 30 cos 30 is twice one half times root 3 over 2, also root 3 over 2. Cos 60 is one half, and cos squared 30 minus sin squared 30 is three quarters minus one quarter, also one half. Both check.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.7 Apply Double-Angle and Half-Angle Formulas §14.7, pp. 955-955
Sorting
Pick the one written in the function you have.
Sort into buckets
Sort each situation by the cosine form it calls for.
All three are the same identity, so no choice is ever wrong — but the right choice can turn three lines of substitution into none.
Worked example
Why three forms are kept rather than one.
\[ \text{Given only } \sin a, \text{ find } \cos 2a. \text{ Given only } \cos a, \text{ find } \cos 2a. \]
Only the sine is known
Why: Use the form written entirely in sine.
\[ 1 - 2 \sin ^{2} a \]
Only the cosine is known
Why: Use the form written entirely in cosine.
\[ 2 \cos ^{2} a - 1 \]
Both are known
Why: Any of the three works; the first is shortest.
\[ \cos ^{2} a - \sin ^{2} a \]
Note the point
Why: The forms are equal, so the choice is about convenience.
Figure (svg): The solution to Worked example choose the right cosine form shown as a ladder of expressions, one row per algebraic move
\[ 1-2\sin^2a; \qquad 2\cos^2a-1 \]
Verify: check the two forms agree
Why: Take a with sine three fifths and cosine four fifths. The first form gives 1 minus 2 times nine twenty-fifths, which is seven twenty-fifths. The second gives 2 times sixteen twenty-fifths minus 1, also seven twenty-fifths. They must agree, since each comes from the other by one substitution.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.7 Apply Double-Angle and Half-Angle Formulas §14.7, pp. 955-955
Trap
\[ \sin 2a \]
Write it as twice the sine
Why: The 2 is treated as a multiplier on the whole expression.
\[ = 2\sin a \quad \text{(wrong)} \]
At a equal to 60 degrees the left side is sin 120, about 0.866, and the right side is about 1.73 — larger than any sine can be.
\[ \sin 2a = 2\sin a\cos a \]
The 2 multiplies the ANGLE, not the function
Why: Doubling an angle is not doubling its sine.
\[ 2\sin 60\degree\cos 60\degree = 2\left(\tfrac{\sqrt3}{2}\right)\left(\tfrac{1}{2}\right) = \tfrac{\sqrt3}{2} \quad \checkmark \]
The bound on sine catches this instantly: any claimed sine above 1 is wrong before the formula is even consulted.
Fill the middle
The warm-up.
Fill in the blanks
\sin(a+a) = \sin a\cos a+\cos a\sin a = 2\sin a\cos a
Why: The two terms are the same product, so they add to twice it. That single collapse is the whole derivation of the sine double-angle formula.
Matching
Set both angles equal.
Match the pairs
Why: The two cosine forms differ by more than a sign: one starts at 1 and subtracts, the other starts at negative 1 and adds. Mixing them up produces an answer that is wrong by exactly 2, which is easy to spot with a test value.
Prediction
Commit before reasoning.
Predict first
How is 2 cos squared a minus 1 obtained from cos squared a minus sin squared a?
Correct: By replacing sin squared a with 1 minus cos squared a.
\[ \cos^2a-(1-\cos^2a) = 2\cos^2a-1 \]
Why: Substituting gives cos squared a minus the quantity 1 minus cos squared a, which is 2 cos squared a minus 1. Replacing cos squared a instead gives the third form. So there is really one identity and two substitutions, which is why they can never disagree — and why forgetting one costs nothing, since it can be rebuilt in a line.
Section
Section 2
Concept
Solving a cosine double-angle formula for the squared function and taking a square root gives the half-angle formulas. The root introduces a sign, and the quadrant of the half angle decides it.
\[ \cos\frac{a}{2} = \pm\sqrt{\frac{1+\cos a}{2}} \]
The two tangent forms carry no square root, so they need no sign choice — a real convenience when only the tangent is wanted.
Figure (svg): The half-angle formulas with the quadrant rule for choosing the sign
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.7 Apply Double-Angle and Half-Angle Formulas §14.7, pp. 955-955 — Double-Angle and Half-Angle Formulas
Picture it
The half-angle formulas and the rule for their signs.
Figure (svg): The half-angle formulas with the quadrant rule for choosing the sign
The sine and cosine formulas need a quadrant check; the two tangent forms do not, because no root was taken.
Worked example
Running the double-angle formula backwards.
\[ \text{Derive } \cos\frac{a}{2} \text{ from } \cos 2u = 2\cos^2u-1. \]
Solve for the squared cosine
Why: Add 1 and halve.
\[ \cos ^{2} u = \frac{1 + \cos 2 u}{2} \]
Take a square root
Why: A root introduces two signs.
\[ \cos u = +- \sqrt{\frac{1 + \cos 2 u}{2}} \]
Rename
Why: Let a be 2u, so u is a over 2.
\[ \cos(\frac{a}{2}) = +- \sqrt{\frac{1 + \cos a}{2}} \]
Note where the sign comes from
Why: Only the quadrant of a over 2 can settle it.
Figure (svg): The half-angle formulas with the quadrant rule for choosing the sign
\[ \cos\frac{a}{2} = \pm\sqrt{\frac{1+\cos a}{2}} \]
Verify: test at a equal to 60 degrees
Why: The formula gives plus or minus the root of 1 plus one half, over 2, which is the root of three quarters, or root 3 over 2. And cos 30 degrees is root 3 over 2 — with the plus sign, since 30 degrees is in the first quadrant. The derivation and the known value agree.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.7 Apply Double-Angle and Half-Angle Formulas §14.7, pp. 955-955
Fill the middle
Example 2b.
Fill in the blanks
\frac4___ < a < 2\pi \;\Longrightarrow\; \frac______} < \frac______ < \pi
Why: Half of three halves of pi is three quarters of pi, and half of 2 pi is pi. The half angle therefore lies between them, in the second quadrant.
Worked example
The book's margin note on Example 2.
\[ \text{Given } \frac{3\pi}{2} < a < 2\pi, \text{ find the quadrant of } \frac{a}{2}. \]
Halve every part of the inequality
Why: Multiply through by one half.
\[ 3 \pi / 4 < \frac{a}{2} < \pi \]
Identify the quadrant
Why: Between three quarters of pi and pi.
Read the signs there
Why: Sine positive, cosine negative.
\[ \sin(\frac{a}{2}) > 0 \]
Compare with a itself
Why: A is in the fourth quadrant, where sine is negative.
Figure (svg): One angle in the fourth quadrant whose half lies in the second
\[ \frac{3\pi}{4} < \frac{a}{2} < \pi \]
Verify: check with a concrete angle
Why: Take a equal to 350 degrees, in the fourth quadrant. Half of it is 175 degrees, in the second — where sine is positive even though the sine of 350 is negative. The half angle's quadrant genuinely has to be computed rather than inherited.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.7 Apply Double-Angle and Half-Angle Formulas §14.7, pp. 956-956
Error analysis
A student finds the sine of half an angle lying in the fourth quadrant.
Annotate
On: \( a \text{ in QIV, so } \sin a < 0, \text{ hence } \sin\frac{a}{2} < 0 \)
Halving an interval moves it, often into a different quadrant. The rule is to halve first and read the quadrant afterwards.
Sorting
Halve the interval, then read the quadrant.
Sort into buckets
Sort each original quadrant by where its half angle falls.
Any angle in a full turn has a half angle in the first or second quadrant, so the sine of a half angle is always positive — but the cosine can be either sign.
Two truths and a lie
All three are about the half-angle formulas.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. An angle of 350 degrees is in the fourth quadrant while its half, 175 degrees, is in the second. Halving compresses the whole circle into the upper half plane, so the half angle is always in the first or second quadrant regardless of where the original one was.
Prediction
Commit before reasoning.
Predict first
The sine and cosine half-angle formulas carry a plus-or-minus but the tangent ones do not. Why?
Correct: Because the tangent forms contain no square root, so the arithmetic produces the sign itself.
\[ \tan\frac{a}{2} = \frac{1-\cos a}{\sin a}: \text{ no root, no ambiguity} \]
Why: A square root loses the sign of what was squared, which is exactly why the sine and cosine versions need a quadrant to recover it. The tangent formulas are quotients of two quantities whose signs are already known, so the division delivers the correct sign automatically. Where a formula involves no root, no sign ambiguity can arise.
Section
Section 3
Concept
Treating an angle as half of a familiar one gives exact values for angles that no sum or difference of special angles can produce, such as 7.5, 15 and 22.5 degrees.
\[ 165\degree = \tfrac{1}{2}(330\degree) \]
The result usually contains a nested radical, which is the price of the square root in the formula and is a perfectly acceptable exact form.
Figure (svg): Two exact values found by treating an angle as half of a familiar one
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.7 Apply Double-Angle and Half-Angle Formulas §14.7, pp. 955-955 — Evaluate trigonometric expressions
Picture it
Example 1: the cosine of 165 degrees and the tangent of pi over 12.
Figure (svg): Two exact values found by treating an angle as half of a familiar one
The first needed a sign from the second quadrant; the second needed none, since it used a tangent formula.
Worked example
Example 1a.
\[ \text{Find the exact value of } \cos 165\degree. \]
Write the angle as a half
Why: Three hundred thirty halved is 165.
\[ \cos\text{ of half of } 330 \]
Decide the sign
Why: One hundred sixty-five degrees is in the second quadrant, where cosine is negative.
Substitute into the formula
Why: The cosine of 330 degrees is root 3 over 2.
\[ -\sqrt{\frac{1 + \sqrt{3} / 2}{2}} \]
Simplify the nested fraction
Why: Combine over 4 and take the root.
\[ -\sqrt{2 + \sqrt{3}} / 2 \]
Figure (svg): Two exact values found by treating an angle as half of a familiar one
\[ \cos 165\degree = -\frac{\sqrt{2+\sqrt3}}{2} \]
Verify: check against a decimal
Why: Root 3 is about 1.732, so 2 plus root 3 is 3.732 and its root is about 1.932, giving negative 0.966. A calculator gives cos 165 degrees as negative 0.9659. The sign is right because 165 degrees is obtuse, where the cosine must be negative.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.7 Apply Double-Angle and Half-Angle Formulas §14.7, pp. 955-955
Fill the middle
Example 1a.
Fill in the blanks
\sqrt4___}___} = \sqrt______}}
Why: Writing 1 as two halves gives 2 plus root 3, all over 2, and dividing by 2 again gives a denominator of 4. The root of 4 is 2, which comes out of the radical.
Worked example
Example 1b and Guided Practice 1 to 3.
\[ \text{Find } \tan\frac{\pi}{12}, \; \tan\frac{\pi}{8}, \; \sin\frac{5\pi}{8} \text{ and } \cos 15\degree. \]
Tangent of pi over 12
Why: Half of pi over 6; use the tangent form, no sign needed.
\[ 2 - \sqrt{3} \]
Tangent of pi over 8
Why: Half of pi over 4; again no sign needed.
\[ \sqrt{2} - 1 \]
Sine of 5 pi over 8
Why: Half of 5 pi over 4; the angle is in QII, so positive.
\[ \sqrt{2 + \sqrt{2}} / 2 \]
Cosine of 15 degrees
Why: Half of 30 degrees; QI, so positive.
\[ \sqrt{2 + \sqrt{3}} / 2 \]
Figure (svg): The solution to Worked example three more by halving shown as a ladder of expressions, one row per algebraic move
\[ 2-\sqrt3, \; \sqrt2-1, \; \frac{\sqrt{2+\sqrt2}}{2}, \; \frac{\sqrt{2+\sqrt3}}{2} \]
Verify: compare the last with Lesson 14.6's answer
Why: The difference formula gave cos 15 degrees as root 6 plus root 2, over 4. Squaring that gives 8 plus 4 root 3, over 16, which is 2 plus root 3, over 4 — and the square of the halving answer is the same. Two different routes to the same exact value, written differently.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.7 Apply Double-Angle and Half-Angle Formulas §14.7, pp. 955-956
Trap
\[ \cos 165\degree = \cos\tfrac{1}{2}(330\degree) \]
Use a plus because cos 330 degrees is positive
Why: The sign of the original angle's cosine is carried across.
\[ = +\frac{\sqrt{2+\sqrt3}}{2} \quad \text{(wrong)} \]
One hundred sixty-five degrees is obtuse, so its cosine must be negative. The sign belongs to the half angle, not to the original.
\[ 165\degree \text{ is in QII, where cosine is negative} \]
Read the sign from the half angle's own quadrant
Why: The formula's plus-or-minus refers to the angle being computed.
\[ = -\frac{\sqrt{2+\sqrt3}}{2} \approx -0.966 \]
Sketching the half angle before choosing the sign takes a few seconds and settles it without any reasoning about the original.
Sorting
Read the quadrant of the angle being computed.
Sort into buckets
Sort each half-angle value by its sign.
Every sine here is positive because a half angle always lands in the first or second quadrant. Only the cosines can go either way.
Matching
Halve a special angle.
Match the pairs
Why: The two tangent answers came out without radicals nested inside radicals, because the tangent formula involves no square root. The sine and cosine answers both carry a nested radical, which is the signature of a half-angle computation.
Prediction
Commit before reasoning.
Predict first
Sum and difference formulas reach every multiple of 15 degrees. What does halving add?
Correct: The multiples of 7.5 degrees, since halving 15 gives 7.5.
\[ 30 \to 15 \to 7.5 \to 3.75 \to \dots \]
Why: Halving any angle already on the list produces one halfway between two of its entries, so 15 becomes 7.5 and 30 becomes 15. Halving again would reach 3.75, and so on indefinitely — each round doubles the number of exactly known angles at the cost of another nested radical. Twenty degrees, however, is never reached, since no amount of halving turns a multiple of 15 into one.
Section
Section 4
Concept
Given one function value and an interval for an angle, the Pythagorean identity supplies the other value and the double-angle formulas give the doubled angle. The half angle needs its own quadrant.
\[ \sin 2a = 2\sin a\cos a \]
The double angle needs no new sign work, since the formula multiplies known values. Only the half angle requires the extra quadrant step.
Figure (svg): One angle in the fourth quadrant whose half lies in the second
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.7 Apply Double-Angle and Half-Angle Formulas §14.7, pp. 956-956 — Evaluate trigonometric expressions
Picture it
Example 2: a in the fourth quadrant, a over 2 in the second.
Figure (svg): One angle in the fourth quadrant whose half lies in the second
The sine of a is negative and the sine of a over 2 is positive. The two quadrants have to be handled separately.
Worked example
Example 2, both parts.
\[ \text{Given } \cos a = \tfrac{5}{13} \text{ with } \tfrac{3\pi}{2} < a < 2\pi, \text{ find } \sin 2a \text{ and } \sin\tfrac{a}{2}. \]
Find the missing value for a
Why: The identity gives twelve thirteenths; QIV makes it negative.
\[ \sin a = -\frac{12}{13} \]
Apply the double-angle formula
Why: Twice negative twelve thirteenths times five thirteenths.
\[ -\frac{120}{169} \]
Locate the half angle
Why: Halving the interval gives three quarters of pi to pi.
Apply the half-angle formula
Why: Positive root of 1 minus five thirteenths, over 2.
\[ 2 \sqrt{13} / 13 \]
Figure (svg): One angle in the fourth quadrant whose half lies in the second
\[ \sin 2a = -\frac{120}{169}; \quad \sin\frac{a}{2} = \frac{2\sqrt{13}}{13} \]
Verify: check both are valid sines
Why: Negative 120 over 169 is about negative 0.710 and 2 root 13 over 13 is about 0.555, both within negative 1 and 1. And the signs match their quadrants: 2a lies in the third or fourth quadrant where sine is negative, and a over 2 in the second where it is positive.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.7 Apply Double-Angle and Half-Angle Formulas §14.7, pp. 956-956
Fill the middle
Example 2a.
Fill in the blanks
\sin 2a = 2\left(-\tfrac120___\right)\left(\tfrac______\right) = -\frac___}___
Why: Two times 12 times 5 is 120, and thirteen squared is 169. The sign is negative because one factor is negative and the other positive.
Worked example
Guided Practice 4 and 5.
\[ \text{Given } \sin a = \tfrac{\sqrt2}{2} \text{ in QI, find } \cos 2a \text{ and } \tan\tfrac{a}{2}; \text{ given } \cos a = -\tfrac{3}{5} \text{ in QIII, find } \sin 2a \text{ and } \sin\tfrac{a}{2}. \]
First: use the sine-only cosine form
Why: One minus twice one half.
\[ \cos 2 a = 0 \]
First: the tangent half angle
Why: One minus root 2 over 2, over root 2 over 2.
\[ \sqrt{2} - 1 \]
Second: find sin a
Why: The identity gives four fifths; QIII makes it negative.
\[ \sin a = -\frac{4}{5} \]
Second: double and halve
Why: Two negatives multiply to a positive; a over 2 is in QII.
\[ \frac{24}{25}\text{ and } 2 \sqrt{5} / 5 \]
Figure (svg): The solution to Worked example two more from given values shown as a ladder of expressions, one row per algebraic move
\[ 0, \; \sqrt2-1; \qquad \tfrac{24}{25}, \; \tfrac{2\sqrt5}{5} \]
Verify: check the first against a known angle
Why: A sine of root 2 over 2 in the first quadrant means a is 45 degrees, so 2a is 90 degrees and its cosine is 0 — matching. And a over 2 is 22.5 degrees, whose tangent is root 2 minus 1, about 0.414. Recognising the angle turns the whole problem into a check.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.7 Apply Double-Angle and Half-Angle Formulas §14.7, pp. 956-956
Error analysis
A student finds the sine of twice an angle knowing only its cosine.
Annotate
On: \( \cos a = \tfrac{5}{13} \;\Longrightarrow\; \sin 2a = 2\left(\tfrac{5}{13}\right) = \tfrac{10}{13} \)
The formula is a product of two different functions, so having one of them is never enough. Finding the missing value is the first step, not an optional extra.
Ranking
Finding a half angle from a given value.
Put in order
Why: Steps three and four are the ones with no counterpart in the double-angle case, and they are where the errors happen. Halving the interval before reading any quadrant is what keeps the two angles' signs separate.
Comparison
Fill the blanks. One needs an extra step.
Comparison matrix
| Question | Double angle | Half angle |
|---|---|---|
| Which values are needed | both sine and cosine of a | only the cosine of a |
| Is a square root involved? | no | yes, for sine and cosine |
| Whose quadrant decides the sign | the arithmetic decides it | the half angle's own quadrant |
| Extra step | none | halve the interval first |
The double angle is the easier of the two precisely because no root is taken, so nothing about the sign has to be recovered separately.
Prediction
Commit before reasoning.
Predict first
If a lies in the fourth quadrant, where can 2a lie?
Correct: The third or fourth, since doubling an angle between 270 and 360 degrees gives one between 540 and 720.
\[ 270\degree < a < 360\degree \;\Longrightarrow\; 180\degree < 2a-360\degree < 360\degree \]
Why: Subtracting a full turn puts 2a between 180 and 360 degrees, so it lies in the third or fourth quadrant — in either case with a negative sine, which matches the answer of negative 120 over 169. Doubling an interval widens it, so the doubled angle can straddle two quadrants; halving narrows it, so a half angle is always pinned to one.
Section
Section 5
Concept
Substituting a double-angle formula often turns a complicated quotient into a single function. In a model it can compress a product of a sine and a cosine into one sine of a doubled angle, revealing what the model does.
\[ x = \frac{v^2\sin 2\theta}{32} \]
Reading a formula backwards is what makes it useful here: spotting 2 sin a cos a inside an expression and replacing it with sin 2a.
Figure (svg): A projectile's horizontal range plotted against its launch angle, peaking at forty-five degrees
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.7 Apply Double-Angle and Half-Angle Formulas §14.7, pp. 956-957 — Simplify an expression; Derive a trigonometric model
Picture it
Example 4: a projectile's horizontal distance.
Figure (svg): A projectile's horizontal range plotted against its launch angle, peaking at forty-five degrees
Written as a sine of twice the angle, the model's peak at 45 degrees is immediate — and 30 and 60 degrees give equal ranges because their doubles are supplementary.
Worked example
Example 3, a multiple-choice item.
\[ \text{Simplify } \frac{1-\cos 2\theta}{\sin 2\theta}. \]
Choose the useful cosine form
Why: The one in sine only, so the 1 will cancel.
\[ 1 - 2 \sin ^{2} t \]
Substitute both formulas
Why: Numerator and denominator both expand.
\[ \frac{1 - (1 - 2 \sin ^{2} t)}{2 \sin t \cos t} \]
Simplify the numerator
Why: The ones cancel.
\[ 2 \sin ^{2} t \]
Cancel and finish
Why: Divide out 2 sin t.
\[ \sin t / \cos t = \tan t \]
Figure (svg): The solution to Worked example simplify a quotient shown as a ladder of expressions, one row per algebraic move
\[ \frac{1-\cos 2\theta}{\sin 2\theta} = \tan\theta \]
Verify: check at 30 degrees
Why: The numerator is 1 minus cos 60, which is one half, and the denominator is sin 60, about 0.866. Their quotient is about 0.577, and tan 30 degrees is 0.5774. Choosing the sine-only cosine form was what made the ones cancel; either other form would have worked but left more to tidy up.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.7 Apply Double-Angle and Half-Angle Formulas §14.7, pp. 956-956
Fill the middle
Example 3.
Fill in the blanks
\fractan t___ = \frac______ = ___
Why: Sine over cosine is the tangent. Dividing out the common factor of 2 sin theta is what leaves the quotient identity exposed.
Worked example
Example 4.
\[ \text{From } -\frac{16}{v^2\cos^2\theta}x^2+(\tan\theta)x = 0, \text{ find the horizontal range.} \]
Factor out x and discard the trivial root
Why: The launch point itself is x equal to zero.
Solve for x
Why: Multiply both sides by the coefficient's reciprocal.
\[ x = \frac{v ^{2} \cos ^{2} t \tan t}{16} \]
Use cos t tan t equals sin t
Why: One cosine cancels against the tangent's denominator.
\[ x = \frac{v ^{2} \cos t \sin t}{16} \]
Rewrite as a double angle
Why: One sixteenth is twice one thirty-second.
\[ x = \frac{v ^{2} \sin 2 t}{32} \]
Figure (svg): A projectile's horizontal range plotted against its launch angle, peaking at forty-five degrees
\[ x = \frac{v^2\sin 2\theta}{32} \]
Verify: read the maximum off the formula
Why: The sine reaches 1 when twice the angle is 90 degrees, so the range is greatest at 45 degrees — a fact that is invisible in the original form and obvious in this one. That is the whole payoff of the rewriting.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.7 Apply Double-Angle and Half-Angle Formulas §14.7, pp. 957-957
Trap
\[ \frac{1-\cos 2\theta}{\sin 2\theta}, \text{ using } \cos 2\theta = 2\cos^2\theta-1 \]
Substitute the cosine-only form
Why: Any of the three forms is chosen at random.
\[ = \frac{2-2\cos^2\theta}{2\sin\theta\cos\theta} \quad \text{(more work)} \]
This is correct but now needs the Pythagorean identity to turn 2 minus 2 cos squared into 2 sin squared before anything cancels.
\[ \text{use } \cos 2\theta = 1-2\sin^2\theta \]
Choose the form whose constant cancels the one already there
Why: The numerator's 1 pairs with the form starting at 1.
\[ = \frac{2\sin^2\theta}{2\sin\theta\cos\theta} = \tan\theta \]
Both routes reach the same answer, so this is about economy rather than correctness — but scanning for the form that cancels a constant usually saves two lines.
Matching
Substitute a double-angle formula and cancel.
Match the pairs
Why: The second used the cosine form as a difference of squares, which factors against the denominator. Each one worked because a chosen double-angle form produced a factor that cancelled, so the choice of form was doing the work each time.
Sorting
Pick the one that cancels something.
Sort into buckets
Sort each expression by the cosine form that simplifies it fastest.
The third case is the one people miss: writing cos 2t as a difference of squares makes it factor, which no other form does.
Prediction
Commit before reasoning.
Predict first
The range formula gives the same distance at 30 degrees and at 60 degrees. Why?
Correct: Because their doubles, 60 and 120 degrees, are supplementary and have equal sines.
\[ \sin 60\degree = \sin 120\degree \;\Longrightarrow\; \text{equal range at } 30\degree \text{ and } 60\degree \]
Why: The range depends only on the sine of twice the angle, and sin 60 equals sin 120 since supplementary angles share a sine. So any two launch angles adding to 90 degrees give the same range — one a low flat shot and one a high lob. That pairing is invisible in the original formula and immediate once the double angle appears, which is exactly why the rewriting was worth doing.
Comparison
Fill the blanks. One doubles, one halves.
Comparison matrix
| Function | Double angle | Half angle |
|---|---|---|
| sine | 2 sin a cos a | +- sqrt((1 - cos a)/2) |
| cosine | cos^2 a - sin^2 a, and two other forms | +- sqrt((1 + cos a)/2) |
| tangent | 2 tan a/(1 - tan^2 a) | (1 - cos a)/sin a |
| Sign choice needed? | no | yes, for sine and cosine only |
The half-angle column carries square roots because it was reached by undoing a squaring, and every square root brings a sign question with it.
Pattern
Decide which family, then which form.
The tangent half-angle formulas need no sign choice, since they contain no square root.
OpenStax Algebra and Trigonometry 2e, §9.3 Double-Angle, Half-Angle, and Reduction Formulas §9.3
Check
The 2 multiplies the angle.
Check your understanding
Given sin a = -12/13 and cos a = 5/13, what is sin 2a?
Answer: A
Why: The formula is 2 sin a cos a, so it is 2 times -12/13 times 5/13.
Check
The half angle has its own quadrant.
Check your understanding
If a is between 3pi/2 and 2pi, where is a/2?
Answer: A
Why: Halving every part of the inequality gives 3pi/4 to pi.
Check
Choose the form that cancels.
Check your understanding
Simplify (1 - cos 2t)/(sin 2t).
Answer: A
Why: Using 1 - 2 sin squared t leaves 2 sin squared t over 2 sin t cos t.
Real world
An athlete throws a javelin from ground level at 28 metres per second. In metric units the range is v squared times the sine of twice the launch angle, all over 9.8.
Discussion prompt
Find the maximum range and the angle that achieves it, then find the two angles giving a range of 60 metres.
Hint: The sine of twice the angle is the only part that varies.
Answer:
\[ R = \frac{28^2\sin 2\theta}{9.8} = 80\sin 2\theta \]
\[ \text{maximum } 80 \text{ m at } 2\theta = 90\degree, \text{ so } \theta = 45\degree \]
\[ 60 = 80\sin 2\theta \;\Longrightarrow\; \sin 2\theta = 0.75 \;\Longrightarrow\; 2\theta \approx 48.6\degree \text{ or } 131.4\degree \]
The maximum range is 80 metres at 45 degrees, and 60 metres is reached at about 24.3 or 65.7 degrees — a flat throw or a lofted one.
Real javelin throwers use around 33 to 36 degrees rather than 45, because a javelin is a wing rather than a point mass: it generates lift, and the model here neglects both that and air resistance. The mathematics still explains the shape of the answer — two angles for every achievable distance, converging as the distance approaches the maximum — and the double-angle rewriting is what makes that structure visible in a single line.
Commit first
Answer, then rate your confidence honestly.
Predict first
An angle a lies in the fourth quadrant. Is the sine of a over 2 negative?
Correct: No — halving the interval puts a over 2 in the second quadrant, where sine is positive.
\[ \tfrac{3\pi}{2} < a < 2\pi \;\Longrightarrow\; \tfrac{3\pi}{4} < \tfrac{a}{2} < \pi \]
Why: An interval from three halves of pi to 2 pi halves to one from three quarters of pi to pi, which is the second quadrant — where the sine is positive. A concrete case makes it plain: 350 degrees is in the fourth quadrant and its half, 175 degrees, is in the second. In fact halving compresses the whole circle into the upper half plane, so the sine of a half angle is always positive. The plus-or-minus in the formula refers to the angle being computed, never to the one you started from.
Explain it
They keep writing sin 2a as 2 sin a.
Discussion prompt
In four sentences or fewer, explain why that is wrong and what the formula actually is.
Hint: Try it at 60 degrees.
Answer:
Try a equal to 60 degrees. Twice the sine is about 1.73, but no sine can be bigger than 1, so that cannot be right.
The 2 is doubling the ANGLE, not the sine. The real formula is sin 2a equals 2 sin a cos a, and at 60 degrees that gives root 3 over 2, which is exactly sin 120 degrees.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the cosine forms, pick the one written in whichever function the problem already has. For half-angle signs, halve the interval before reading any quadrant. For nested radicals, combine over a common denominator before taking the root. For spotting doubles, watch for 2 sin a cos a and for a difference of two squared functions.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a formulas page. Top left: derive the sine and cosine double-angle formulas from the sum formulas, then produce the other two cosine forms by substitution, showing each line. Top right: derive one half-angle formula by solving a cosine double-angle formula for the squared function, and mark where the plus-or-minus enters. Middle: work Example 1 in full, including the sign reasoning, and check your answer against a decimal. Bottom left: work Example 2 in full, drawing both a and a over 2 on a circle so the two quadrants are visible side by side. Bottom right: derive the projectile range formula, and use it to explain in one sentence why 30 and 60 degrees give the same distance.
If any half-angle sine came out negative, recheck the interval: halving always lands the angle in the first or second quadrant, where sine is positive.
Recap
Five things, and Chapter 14 is complete.
| If you see | Then |
|---|---|
| sin 2a | Write 2 sin a cos a, never 2 sin a |
| cos 2a | Choose the form in whichever function you already have |
| A half angle | Halve the interval and read its own quadrant |
| A tangent half angle | No sign choice is needed |
| 2 sin a cos a inside an expression | Replace it with sin 2a |
| An angle like 15 or 22.5 degrees | Halve a special angle |
That completes the Algebra 2 course: every lesson of all fourteen chapters, from expressions and equations through to trigonometric identities.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.7 Apply Double-Angle and Half-Angle Formulas §14.7, pp. 955-961 — everything on these slides traces back here
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