First slide

Trigonometric transformations

Question

$\frac{\sin 3 θ + \sin 5 θ + \sin 7 θ + \sin 9 θ}{\cos 3 θ + \cos 5 θ + \cos 7 θ + \cos 9 θ}$ is equal to

Moderate

A

$\tan 3 θ$
B

$\cot 3 θ$
C

$\tan 6 θ$
D

$\cot 6 θ$

Solution

$\frac{\sin 3 θ + \sin 5 θ + \sin 7 θ + \sin 9 θ}{\cos 3 θ + \cos 5 θ + \cos 7 θ + \cos 9 θ}$
$\begin{array}{l} = \frac{(\sin 3 θ + \sin 9 θ) + (\sin 5 θ + \sin 7 θ)}{(\cos 3 θ + \cos 9 θ) + (\cos 5 θ + \cos 7 θ)} \\ = \frac{2 \sin 6 θ \cos 3 θ + 2 \sin 6 θ \cos θ}{2 \cos 6 θ \cos 3 θ + 2 \cos 6 θ \cos θ} \\ = \frac{2 \sin 6 θ (\cos 3 θ + \cos θ)}{2 \cos 6 θ (\cos 3 θ + \cos θ)} \\ = \tan 6 θ \end{array}$

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