First slide

Trigonometric transformations

Question

If $\frac{\cos x}{a} = \frac{\cos (x + θ)}{b} = \frac{\cos (x + 2 θ)}{c} = \frac{\cos (x + 3 θ)}{d}$ then $\frac{a + c}{b + d}$ is equal to

Moderate

A

$\frac{a}{d}$
B

$\frac{c}{b}$
C

$\frac{b}{c}$
D

$\frac{d}{a}$

Solution

For each of the ratio s be 1/k,
$\begin{matrix} \frac{a + c}{b + d} = \frac{kcos x + kcos (x + 2 θ)}{kcos (x + θ) + kcos (x + 3 θ)} \\ = \frac{2 \cos (x + θ) \cos θ}{2 \cos (x + 2 θ) \cos θ} \\ = \frac{\cos (x + θ)}{\cos (x + 2 θ)} = \frac{kcos (x + θ)}{kcos (x + 2 θ)} = \frac{b}{c} \end{matrix}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App