First slide

Trigonometric transformations

Question

$\frac{\sin 7 θ + 6 \sin 5 θ + 17 \sin 3 θ + 12 \sin θ}{\sin 6 θ + 5 \sin 4 θ + 12 \sin 2 θ}$ is equal to

Moderate

A

$2 \cos θ$
B

$\cos θ$
C

$2 \sin θ$
D

$\sin θ$

Solution

We have,
$\begin{array}{l} \frac{\sin 7 θ + 6 \sin 5 θ + 17 \sin 3 θ + 12 \sin θ}{\sin 6 θ + 5 \sin 4 θ + 12 \sin 2 θ} \\ = \frac{(\sin 7 θ + \sin 5 θ) + 5 (\sin 5 θ + \sin 3 θ) + 12 (\sin 3 θ + \sin θ)}{\sin 6 θ + 5 \sin 4 θ + 12 \sin 2 θ} \end{array}$
$\begin{array}{l} = \frac{2 \sin 6 θ \cos θ + 10 \sin 4 θ \cos θ + 24 \sin 2 θ \cos θ}{\sin 6 θ + 5 \sin 4 θ + 12 \sin 2 θ} \\ = \frac{2 \cos θ (\sin 6 θ + 5 \sin 4 θ + 12 \sin 2 θ)}{\sin 6 θ + 5 \sin 4 θ + 12 \sin 2 θ} \\ = 2 \cos θ \end{array}$

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