2.3, 3.3 - Compound-angle formulae

2.3, 3.3 - Compound-angle formulae

Knowing sine and cosine of two angles lets you find sine and cosine of their sum, but you cannot simply add the ratios. This shared lesson for 2.3 and 3.3 develops the compound-angle formulae, uses them for exact values and proofs, and solves equations whose angles have been shifted.

Sine and cosine of a sum

Can you find sin⁡75∘\sin75^\circ exactly from 45∘45^\circ and 30∘30^\circ? Adding their sine values cannot work: sin⁡45∘+sin⁡30∘>1\sin45^\circ+\sin30^\circ>1. Instead, both sine and cosine of each angle contribute.

Here is why a cosine difference formula is plausible and how to establish it. On a unit circle, points at angles AA and BB have coordinates (cos⁡A,sin⁡A)(\cos A,\sin A) and (cos⁡B,sin⁡B)(\cos B,\sin B). The squared distance between them, found from their coordinates, is

d2=(cos⁡A−cos⁡B)2+(sin⁡A−sin⁡B)2=2−2(cos⁡Acos⁡B+sin⁡Asin⁡B)\begin{aligned} d^2&=(\cos A-\cos B)^2+(\sin A-\sin B)^2\\ &=2-2(\cos A\cos B+\sin A\sin B) \end{aligned}

using sin⁡2t+cos⁡2t=1\sin^2t+\cos^2t=1. The cosine rule in the triangle with two unit sides gives d2=2−2cos⁡(A−B)d^2=2-2\cos(A-B). For a reflex angle difference, the triangle's interior angle has the same cosine. Equating the expressions proves the cosine difference formula. Replacing BB by −B-B, and using the symmetry and complementary-angle relationships of sine and cosine, gives the other forms.

For example, substituting 90∘−A90^\circ-A and BB into the cosine difference formula gives

sin⁡(A+B)=cos⁡((90∘−A)−B)=sin⁡Acos⁡B+cos⁡Asin⁡B.\sin(A+B)=\cos((90^\circ-A)-B) =\sin A\cos B+\cos A\sin B.

This uses the complementary relationships cos⁡(90∘−A)=sin⁡A\cos(90^\circ-A)=\sin A and sin⁡(90∘−A)=cos⁡A\sin(90^\circ-A)=\cos A, which also follow from the unit circle.

Sine and cosine compound-angle formulae

For real angles AA and BB, in the same units,

sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡Bsin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡Bcos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡Bcos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B.\begin{aligned} \sin(A+B)&=\sin A\cos B+\cos A\sin B\\ \sin(A-B)&=\sin A\cos B-\cos A\sin B\\ \cos(A+B)&=\cos A\cos B-\sin A\sin B\\ \cos(A-B)&=\cos A\cos B+\sin A\sin B. \end{aligned}

In the formula booklet.

Sine keeps the sign between the angles; cosine changes it. For instance, cos⁡(45∘+30∘)\cos(45^\circ+30^\circ) uses a minus between its two products. Angle units must agree, but the identities work equally in degrees and radians.

Worked example 1: an exact sine of a sum

Question. Find the exact value of sin⁡75∘\sin75^\circ.

Plan. Split 75∘75^\circ into 45∘+30∘45^\circ+30^\circ, whose sine and cosine values are known exactly.

sin⁡75∘=sin⁡45∘cos⁡30∘+cos⁡45∘sin⁡30∘=2232+2212=6+24.\begin{aligned} \sin75^\circ&=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ\\ &=\frac{\sqrt2}{2}\frac{\sqrt3}{2}+\frac{\sqrt2}{2}\frac12\\ &=\frac{\sqrt6+\sqrt2}{4}. \end{aligned}

Answer. (6+2)/4(\sqrt6+\sqrt2)/4.

Check. The exact form is approximately 0.9660.966, matching a calculator value for sin⁡75∘\sin75^\circ, and it lies below 11.

Keep the two known angles and change only the plus to a minus. This isolates the sign change in the formula.

Worked example 2: an exact sine of a difference

Question. Find the exact value of sin⁡15∘\sin15^\circ.

Plan. Write 15∘=45∘−30∘15^\circ=45^\circ-30^\circ and subtract the second product.

sin⁡15∘=sin⁡45∘cos⁡30∘−cos⁡45∘sin⁡30∘=64−24=6−24.\begin{aligned} \sin15^\circ&=\sin45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ\\ &=\frac{\sqrt6}{4}-\frac{\sqrt2}{4}\\ &=\frac{\sqrt6-\sqrt2}{4}. \end{aligned}

Answer. (6−2)/4(\sqrt6-\sqrt2)/4.

Check. The result is positive and approximately 0.2590.259, smaller than sin⁡30∘=1/2\sin30^\circ=1/2, as the first-quadrant graph requires.

The cosine formulae use the same exact values. For a new sine or cosine value, first look for a sum or difference of familiar angles.

Watch out

cos⁡(A−B)\cos(A-B) is cos⁡Acos⁡B+sin⁡Asin⁡B\cos A\cos B+\sin A\sin B. Writing a minus between these products gives cos⁡(A+B)\cos(A+B) instead. Checking A=BA=B exposes the error: cos⁡0=1\cos0=1.

Apply the cosine formula to the same kind of exact-value calculation.

Tangent of a sum

A tangent compound-angle formula comes from dividing the sine expansion by the cosine expansion. For a sum,

tan⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡Bcos⁡Acos⁡B−sin⁡Asin⁡B.\tan(A+B)=\frac{\sin A\cos B+\cos A\sin B}{\cos A\cos B-\sin A\sin B}.

If cos⁡Acos⁡B≠0\cos A\cos B\ne0, divide numerator and denominator by cos⁡Acos⁡B\cos A\cos B. This produces a formula involving only tangent.

Tangent compound-angle formulae

tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B,tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B.\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B},\qquad \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}.

These forms require tan⁡A\tan A and tan⁡B\tan B to exist and the displayed denominator to be nonzero. The target tangent must also be defined.

In the formula booklet.

The conditions matter. Although tan⁡(90∘+30∘)=tan⁡120∘\tan(90^\circ+30^\circ)=\tan120^\circ exists, the right side using tan⁡90∘\tan90^\circ does not. Use sine and cosine expansions, or a different angle split, for such a case. An identity equates expressions on their common domain; it cannot make an undefined intermediate ratio meaningful.

Worked example 3: tangent of a sum

Question. Find the exact value of tan⁡75∘\tan75^\circ.

Plan. Use 45∘+30∘45^\circ+30^\circ; both tangents exist and their product is not 11.

tan⁡75∘=1+1/31−1/3=3+13−1=(3+1)23−1rationalise=2+3.\begin{aligned} \tan75^\circ&=\frac{1+1/\sqrt3}{1-1/\sqrt3}\\ &=\frac{\sqrt3+1}{\sqrt3-1}\\ &=\frac{(\sqrt3+1)^2}{3-1} &&\text{rationalise}\\ &=2+\sqrt3. \end{aligned}

Answer. 2+32+\sqrt3.

Check. Dividing the exact sine and cosine of 75∘75^\circ gives the same value. Numerically it is 3.732…3.732\ldots, consistent with the steep positive tangent graph near 90∘90^\circ.

Change only the sum to a difference. The numerator changes to subtraction, while the denominator changes to addition.

Worked example 4: tangent of a difference

Question. Find the exact value of tan⁡15∘\tan15^\circ.

Plan. Use 45∘−30∘45^\circ-30^\circ and check the denominator is nonzero.

tan⁡15∘=1−1/31+1/3=3−13+1=(3−1)23−1=2−3.\begin{aligned} \tan15^\circ&=\frac{1-1/\sqrt3}{1+1/\sqrt3}\\ &=\frac{\sqrt3-1}{\sqrt3+1}\\ &=\frac{(\sqrt3-1)^2}{3-1}=2-\sqrt3. \end{aligned}

Answer. 2−32-\sqrt3.

Check. 15∘15^\circ and 75∘75^\circ are complementary, so their tangents have product 11. Indeed, (2−3)(2+3)=1(2-\sqrt3)(2+\sqrt3)=1.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account