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Precalculus · Lesson 6.2

Angle Sum and Double-Angle Identities

How it works

Angle-sum formulas give exact values by combining familiar angles. A double-angle formula is a special case. Substitution should keep the subtraction signs and the quadrant signs of every sine and cosine.

sin⁡(a+b)=sin⁡acos⁡b+cos⁡asin⁡b,cos⁡(2a)=1−2sin⁡2a\sin(a+b)=\sin a\cos b+\cos a\sin b,\quad\cos(2a)=1-2\sin^2a

Worked examples

Example 1

Find sin⁡(π/12)\displaystyle \sin(\pi/12) exactly.

Use π/12=π/4−π/6\displaystyle \pi/12=\pi/4-\pi/6. Then sin⁡(π/12)=(6−2)/4\displaystyle \sin(\pi/12)=(\sqrt{6}-\sqrt{2})/4.

Example 2

If sin⁡x=3/5\displaystyle \sin x=3/5 and x is acute, find sin⁡(2x)\displaystyle \sin(2x).

Cosine is 4/5\displaystyle 4/5. Thus sin⁡(2x)=2(3/5)(4/5)=24/25\displaystyle \sin(2x)=2(3/5)(4/5)=24/25.

Problems

Problem 6.2.1

Find cos⁡(π/12)\displaystyle \cos(\pi/12) exactly.

Show worked solution

Use the difference formula: cos⁡(π/4−π/6)=6/4+2/4\displaystyle \cos(\pi/4-\pi/6)=\sqrt{6}/4+\sqrt{2}/4.

Problem 6.2.2

If sin⁡x=3/5\displaystyle \sin x=3/5, find cos⁡(2x)\displaystyle \cos(2x).

Show worked solution

1−2sin⁡2x=1−18/25=7/25\displaystyle 1-2\sin^2x=1-18/25=7/25, regardless of which allowed quadrant contains x.

Problem 6.2.3

Explain why sin⁡(a+b)=sin⁡a+sin⁡b\displaystyle \sin(a+b)=\sin a+\sin b is false.

Show worked solution

Take a=b=π/2\displaystyle a=b=\pi/2. The left side is 0\displaystyle 0, but the right side is 2\displaystyle 2.

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