First slide

Trigonometric transformations

Question

$\sin 47^{\circ} + \sin 61^{\circ} - \sin 11^{\circ} - \sin 25^{\circ}$ is equal to

Moderate

A

$\sin 7^{\circ}$
B

$\cos 7^{\circ}$
C

$\sin 36^{\circ}$
D

$\cos 36^{\circ}$

Solution

$\begin{matrix} \sin 47^{\circ} - \sin 25^{\circ} + \sin 61^{\circ} - \sin 11^{\circ} \\ = 2 \cos 36^{\circ} \sin 11^{\circ} + 2 \cos 36^{\circ} \sin 25^{\circ} \\ = 2 \cos 36^{\circ} [\sin 11^{\circ} + \sin 25^{\circ}] \\ = 2 \cos 36^{\circ} [2 \sin (\frac{25^{\circ} + 11^{\circ}}{2}) \cos (\frac{25^{\circ} - 11^{\circ}}{2})] \\ = 4 (\frac{\sqrt{5} + 1}{4}) (\frac{\sqrt{5} - 1}{4}) \cos 7^{\circ} \\ = \frac{5 - 1}{4} \cos 7^{\circ} \\ = \cos 7^{\circ} \end{matrix}$

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