Transformation formulae

The trans­for­ma­tion for­mu­lae change a prod­uct of sines and cosines into a sum, or a sum into a prod­uct. That is often exactly what is needed: a prod­uct of awk­ward angles may turn into a sum of stan­dard angles, and a sum may fac­torise so that terms can­cel. This les­son restates both groups of for­mu­lae, uses them to prove the stan­dard iden­ti­ties for the angles of a tri­an­gle, and works through a set of typ­i­cal eval­u­a­tions and proofs.

Prod­ucts into sums#

$$\begin{aligned} 2\sin A\cos B &= \sin(A + B) + \sin(A - B) \\ 2\cos A\sin B &= \sin(A + B) - \sin(A - B) \\ 2\cos A\cos B &= \cos(A + B) + \cos(A - B) \\ 2\sin A\sin B &= \cos(A - B) - \cos(A + B) \end{aligned}$$

(The last can also be writ­ten $\cos(A + B) - \cos(A - B) = -2\sin A\sin B$.)

Sums into prod­ucts#

$$\displaystyle \begin{aligned} \sin C + \sin D &= 2\sin\frac{C + D}{2}\cos\frac{C - D}{2} \\ \sin C - \sin D &= 2\cos\frac{C + D}{2}\sin\frac{C - D}{2} \\ \cos C + \cos D &= 2\cos\frac{C + D}{2}\cos\frac{C - D}{2} \\ \cos C - \cos D &= -2\sin\frac{C + D}{2}\sin\frac{C - D}{2} \end{aligned}$$

Iden­ti­ties for the angles of a tri­an­gle#

Results. If $A + B + C = 180°$ (or $\pi$), then

1. $\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C$
2. $\displaystyle \sin A + \sin B + \sin C = 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$
3. $\cos 2A + \cos 2B + \cos 2C = -1 - 4\cos A\cos B\cos C$
4. $\displaystyle \cos A + \cos B + \cos C = 1 + 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$

The method in every proof: apply a for­mula to the first two terms, replace the angle sums using $A + B = 180° - C$, sim­plify by tak­ing out a com­mon fac­tor, replace again so that the bracket becomes a sum or dif­fer­ence, and fin­ish with a for­mula.

The replace­ments used are

$$\sin(A + B) = \sin(180° - C) = \sin C, \qquad \cos C = \cos(180° - (A + B)) = -\cos(A + B)$$

$$\displaystyle \cos\frac{A + B}{2} = \cos\left(90° - \frac{C}{2}\right) = \sin\frac{C}{2}, \qquad \sin\frac{A + B}{2} = \cos\frac{C}{2}$$

Proof of 1.

$$\begin{aligned} \sin 2A + \sin 2B + \sin 2C &= 2\sin(A + B)\cos(A - B) + \sin 2C \\ &= 2\sin C\cos(A - B) + 2\sin C\cos C \\ &= 2\sin C\left[\cos(A - B) + \cos C\right] \\ &= 2\sin C\left[\cos(A - B) - \cos(A + B)\right] \\ &= 2\sin C\left[2\sin A\sin B\right] \\ &= 4\sin A\sin B\sin C \end{aligned}$$

Proof of 4.

$$\displaystyle \begin{aligned} \cos A + \cos B + \cos C &= 2\cos\frac{A + B}{2}\cos\frac{A - B}{2} + \cos C \\ &= 2\sin\frac{C}{2}\cos\frac{A - B}{2} + 1 - 2\sin^2\frac{C}{2} \\ &= 1 + 2\sin\frac{C}{2}\left[\cos\frac{A - B}{2} - \sin\frac{C}{2}\right] \\ &= 1 + 2\sin\frac{C}{2}\left[\cos\frac{A - B}{2} - \cos\frac{A + B}{2}\right] \\ &= 1 + 2\sin\frac{C}{2}\left[2\sin\frac{A}{2}\sin\frac{B}{2}\right] \\ &= 1 + 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} \end{aligned}$$

Results 2 and 3 are proved the same way, and are left as prac­tice.

Exam­ple 1#

Prove that $\displaystyle \cos 20°\cos 40° - \sin 5°\sin 25° = \frac{\sqrt{3} + 1}{4}$.

Turn both prod­ucts into sums:

$$\displaystyle \begin{aligned} \text{LHS} &= \frac{1}{2}\left[2\cos 40°\cos 20° - 2\sin 25°\sin 5°\right] \\ &= \frac{1}{2}\left[(\cos 60° + \cos 20°) - (\cos 20° - \cos 30°)\right] \\ &= \frac{1}{2}\left[\frac{1}{2} + \frac{\sqrt{3}}{2}\right] = \frac{\sqrt{3} + 1}{4} \end{aligned}$$

Exam­ple 2#

Prove that $4(\cos 66° + \sin 84°) = \sqrt{3} + \sqrt{15}$.

First make both terms cosines: $\sin 84° = \cos 6°$. Then use $\cos C + \cos D$:

$$\displaystyle 4(\cos 66° + \cos 6°) = 4 \cdot 2\cos 36°\cos 30° = 8 \cdot \frac{\sqrt{5} + 1}{4} \cdot \frac{\sqrt{3}}{2} = \sqrt{15} + \sqrt{3}$$

Exam­ple 3#

Prove that $\displaystyle \frac{1 + \sin 2\theta - \cos 2\theta}{1 + \sin 2\theta + \cos 2\theta} = \tan\theta$.

Group $1 - \cos 2\theta = 2\sin^2\theta$ and $1 + \cos 2\theta = 2\cos^2\theta$:

$$\displaystyle \frac{2\sin^2\theta + 2\sin\theta\cos\theta}{2\cos^2\theta + 2\sin\theta\cos\theta} = \frac{2\sin\theta(\sin\theta + \cos\theta)}{2\cos\theta(\cos\theta + \sin\theta)} = \tan\theta$$

Exam­ple 4#

If $\displaystyle \tan\theta = \frac{b}{a}$, prove that $a\cos 2\theta + b\sin 2\theta = a$.

Use the forms in $\tan\theta$:

$$\displaystyle \begin{aligned} a\cos 2\theta + b\sin 2\theta &= a \cdot \frac{1 - \frac{b^2}{a^2}}{1 + \frac{b^2}{a^2}} + b \cdot \frac{2 \cdot \frac{b}{a}}{1 + \frac{b^2}{a^2}} \\ &= a \cdot \frac{a^2 - b^2}{a^2 + b^2} + \frac{2b^2}{a} \cdot \frac{a^2}{a^2 + b^2} \\ &= \frac{a(a^2 - b^2) + 2ab^2}{a^2 + b^2} = \frac{a(a^2 + b^2)}{a^2 + b^2} = a \end{aligned}$$

Exam­ple 5#

Prove that $\displaystyle \frac{1}{\sin 10°} - \frac{\sqrt{3}}{\cos 10°} = 4$.

The idea is to write $\cos 10° - \sqrt{3}\sin 10°$ as $2$ times a sin­gle sine, using $\displaystyle \frac{1}{2} = \sin 30°$ and $\displaystyle \frac{\sqrt{3}}{2} = \cos 30°$. (In gen­eral $\displaystyle a\cos\theta + b\sin\theta = \sqrt{a^2 + b^2}\left[\frac{a}{\sqrt{a^2 + b^2}}\cos\theta + \frac{b}{\sqrt{a^2 + b^2}}\sin\theta\right]$.)

$$\displaystyle \begin{aligned} \frac{\cos 10° - \sqrt{3}\sin 10°}{\sin 10°\cos 10°} &= \frac{2\left[\frac{1}{2}\cos 10° - \frac{\sqrt{3}}{2}\sin 10°\right]}{\frac{1}{2}\left[2\sin 10°\cos 10°\right]} \\ &= \frac{4\left[\sin 30°\cos 10° - \cos 30°\sin 10°\right]}{\sin 20°} \\ &= \frac{4\sin 20°}{\sin 20°} = 4 \end{aligned}$$

Exam­ple 6#

If $\displaystyle \sin\alpha = \frac{3}{5}$ and $\displaystyle \frac{\pi}{2} < \alpha < \pi$, find $\cos 3\alpha$ and $\tan 2\alpha$.

$\alpha$ is in the sec­ond quad­rant, so with a $3$, $4$, $5$ tri­an­gle, $\displaystyle \cos\alpha = -\frac{4}{5}$ and $\displaystyle \tan\alpha = -\frac{3}{4}$.

$$\displaystyle \cos 3\alpha = 4\cos^3\alpha - 3\cos\alpha = -\frac{4}{5}\left[4 \cdot \frac{16}{25} - 3\right] = -\frac{4}{5}\left(-\frac{11}{25}\right) = \frac{44}{125}$$

$$\displaystyle \tan 2\alpha = \frac{2\tan\alpha}{1 - \tan^2\alpha} = \frac{-\frac{3}{2}}{1 - \frac{9}{16}} = -\frac{3}{2} \times \frac{16}{7} = -\frac{24}{7}$$

Exam­ple 7#

Find $\cos 42° + \cos 78° + \cos 162°$.

$$\displaystyle 2\cos\frac{78° + 42°}{2}\cos\frac{78° - 42°}{2} + \cos(180° - 18°) = 2\cos 60°\cos 18° - \cos 18° = \cos 18° - \cos 18° = 0$$

Exam­ple 8#

Prove that $\displaystyle \cos 12° + \cos 132° + \cos 84° + \cos 156° = -\frac{1}{2}$.

Pair the terms as $(\cos 132° + \cos 12°) + (\cos 156° + \cos 84°)$:

$$\displaystyle \begin{aligned} &= 2\cos 72°\cos 60° + 2\cos 120°\cos 36° \\ &= 2\sin 18° \cdot \frac{1}{2} + 2\left(-\frac{1}{2}\right)\cos 36° \\ &= \frac{\sqrt{5} - 1}{4} - \frac{\sqrt{5} + 1}{4} = -\frac{2}{4} = -\frac{1}{2} \end{aligned}$$

Prac­tice#

1. Write $2\sin 50°\cos 10°$ as a sum.
2. Find $2\cos 75°\cos 15°$.
3. Find $2\sin 75°\sin 15°$.
4. Find $\sin 75° + \sin 15°$.
5. Find $\cos 75° - \cos 15°$.
6. Prove that $\cos 20° + \cos 100° + \cos 140° = 0$.
7. Prove that $\sin 50° - \sin 70° + \sin 10° = 0$.
8. If $\displaystyle \tan\theta = \frac{1}{2}$, find $4\cos 2\theta + 2\sin 2\theta$.
9. If $\displaystyle \cos A = -\frac{12}{13}$ and $A$ is in the third quad­rant, find $\sin 2A$ and $\cos 2A$.
10. Prove that $\sqrt{3}\csc 20° - \sec 20° = 4$.
11. In a tri­an­gle $ABC$, prove that $\sin 2A + \sin 2B - \sin 2C = 4\cos A\cos B\sin C$.

Answers#