First slide

Trigonometric transformations

Question

If $\cos θ_{1} = 2 \cos θ_{2}$ then $\tan \frac{θ_{1} - θ_{2}}{2} \tan \frac{θ_{1} + θ_{2}}{2}$ is equal to

Moderate

A

$\frac{1}{3}$
B

$- \frac{1}{3}$
C

1
D

-1

Solution

$\begin{matrix} \tan (\frac{θ_{1} - θ_{2}}{2}) \tan (\frac{θ_{1} + θ_{2}}{2}) \\ = \frac{2 \sin (\frac{θ_{1} + θ_{2}}{2}) \sin (\frac{θ_{1} - θ_{2}}{2})}{2 \cos (\frac{θ_{1} + θ_{2}}{2}) \cos (\frac{θ_{1} - θ_{2}}{2})} \\ = \frac{\cos θ_{2} - \cos θ_{1}}{\cos θ_{1} + \cos θ_{2}} = \frac{- 1}{3} \end{matrix}$

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