10G - Compound-angle formulae

10G - Compound-angle formulae

Can you find cos⁡15∘\cos15^\circ exactly using only familiar angles? You will use the supplied sine, cosine and tangent formulae for sums and differences of angles. These connect known ratios to new exact values and let you establish short identities by expanding the angles carefully.

Sine and cosine of sums or differences

Why is sin⁡(30∘+30∘)\sin(30^\circ+30^\circ) not sin⁡30∘+sin⁡30∘\sin30^\circ+\sin30^\circ? The left side is 32\frac{\sqrt3}{2}, while the right side is 11: adding angles is different from adding their heights on the unit circle.

A compound angle is an angle written as a sum or difference, such as 45∘−30∘45^\circ-30^\circ. The formulae below combine products of ratios, rather than simply adding the ratios.

Sine and cosine addition formulae

For any real angles AA and BB, measured in the same unit,

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}

On the formulae sheet.

For sine, the sign between the products matches the sign inside the angle. For cosine, it reverses. All four formulae apply in degrees or radians, including negative angles. There is no denominator restriction because sine and cosine exist for every real angle.

Worked example 1: adding two angles

Question. Angles AA and BB are acute, with sin⁡A=35\sin A=\frac35 and sin⁡B=513\sin B=\frac5{13}. Find sin⁡(A+B)\sin(A+B) and cos⁡(A+B)\cos(A+B) exactly.

Plan. Find the positive cosines, then substitute into the two sum formulae.

The Pythagorean identity gives cos⁡A=45\cos A=\frac45 and cos⁡B=1213\cos B=\frac{12}{13}; both are positive because the angles are acute.

sin⁡(A+B)=351213+45513sine sum formula=36+2065=5665cos⁡(A+B)=451213−35513cosine sum formula=48−1565=3365\begin{aligned} \sin(A+B)&=\frac35\frac{12}{13}+\frac45\frac5{13} &&\text{sine sum formula}\\ &=\frac{36+20}{65}=\frac{56}{65}\\ \cos(A+B)&=\frac45\frac{12}{13}-\frac35\frac5{13} &&\text{cosine sum formula}\\ &=\frac{48-15}{65}=\frac{33}{65} \end{aligned}

Answer. sin⁡(A+B)=5665\sin(A+B)=\frac{56}{65} and cos⁡(A+B)=3365\cos(A+B)=\frac{33}{65}.

Check. 562+332=65256^2+33^2=65^2, as the Pythagorean identity requires. Also, A≈36.87∘A\approx36.87^\circ and B≈22.62∘B\approx22.62^\circ; the sum is acute, matching the two positive answers.

Keep both angles but change their sum to their difference. This reverses the sign between the products in each calculation.

Worked example 2: subtracting the same angles

Question. With the same acute angles, sin⁡A=35\sin A=\frac35 and sin⁡B=513\sin B=\frac5{13}, find sin⁡(A−B)\sin(A-B) and cos⁡(A−B)\cos(A-B).

Plan. Use the same four ratios in the difference formulae.

sin⁡(A−B)=351213−45513sine difference formula=36−2065=1665cos⁡(A−B)=451213+35513cosine difference formula=48+1565=6365\begin{aligned} \sin(A-B)&=\frac35\frac{12}{13}-\frac45\frac5{13} &&\text{sine difference formula}\\ &=\frac{36-20}{65}=\frac{16}{65}\\ \cos(A-B)&=\frac45\frac{12}{13}+\frac35\frac5{13} &&\text{cosine difference formula}\\ &=\frac{48+15}{65}=\frac{63}{65} \end{aligned}

Answer. sin⁡(A−B)=1665\sin(A-B)=\frac{16}{65} and cos⁡(A−B)=6365\cos(A-B)=\frac{63}{65}.

Check. 162+632=65216^2+63^2=65^2. The difference is about 14.25∘14.25^\circ, so a small positive sine and a cosine close to 11 are sensible.

The cosine difference uses a plus sign even though the angle contains a minus sign.

Watch out

cos⁡(A−B)≠cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A-B)\ne\cos A\cos B-\sin A\sin B. That expression is cos⁡(A+B)\cos(A+B). Select the formula before inserting the numbers.

Find the missing ratios before applying the next difference formula.

Tangent of a sum or difference

When the given information consists of tangents, a tangent formula can avoid finding four separate sine and cosine values.

For example, if tan⁡A=12\tan A=\frac12 and tan⁡B=13\tan B=\frac13, the sum formula uses those two fractions directly. It still needs a denominator: some angle sums have undefined tangent.

Tangent addition formulae

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

Both require cos⁡A≠0\cos A\ne0 and cos⁡B≠0\cos B\ne0. Also require 1−tan⁡Atan⁡B≠01-\tan A\tan B\ne0 for the sum, or 1+tan⁡Atan⁡B≠01+\tan A\tan B\ne0 for the difference.

On the formulae sheet.

The individual cosines must be nonzero because the right side uses tan⁡A\tan A and tan⁡B\tan B. Once those exist, a zero formula denominator corresponds to cos⁡(A+B)=0\cos(A+B)=0 or cos⁡(A−B)=0\cos(A-B)=0, so the requested tangent is undefined too.

Worked example 3: a tangent sum

Question. Acute angles AA and BB satisfy tan⁡A=12\tan A=\frac12 and tan⁡B=13\tan B=\frac13. Find tan⁡(A+B)\tan(A+B) exactly.

Plan. Check the sum denominator, then substitute the tangents.

1−tan⁡Atan⁡B=1−16=56nonzero, so the formula appliestan⁡(A+B)=12+131−1213use the sum formula=5/65/6=1\begin{aligned} 1-\tan A\tan B&=1-\frac16=\frac56 &&\text{nonzero, so the formula applies}\\ \tan(A+B)&=\frac{\frac12+\frac13}{1-\frac12\frac13} &&\text{use the sum formula}\\ &=\frac{5/6}{5/6}=1 \end{aligned}

Answer. tan⁡(A+B)=1\tan(A+B)=1.

Check. Right triangles for the two angles give sine/cosine pairs (15,25)\left(\frac1{\sqrt5},\frac2{\sqrt5}\right) and (110,310)\left(\frac1{\sqrt{10}},\frac3{\sqrt{10}}\right). The sine and cosine sum formulae both give 550\frac5{\sqrt{50}}; their quotient is 11.

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