First slide

Trigonometric equations

Question

If $\cos θ_{1} = 2 \cos θ_{2}$ and $T a n (\frac{θ_{1} - θ_{2}}{2}) T a n (\frac{θ_{1} + θ_{2}}{2}) = \frac{- p}{q}, t h e n p + q =$

Moderate

Solution

$\begin{array}{l} W e h a v e t a n (\frac{θ_{1} - θ_{2}}{2}) t a n (\frac{θ_{1} + θ_{2}}{2}) = \frac{2 \sin (\frac{θ_{1} - θ_{2}}{2}) \sin (\frac{θ_{1} + θ_{2}}{2})}{2 \cos (\frac{θ_{1} - θ_{2}}{2}) \cos (\frac{θ_{1} + θ_{2}}{2})} \end{array}$
$\begin{array}{l} = \frac{\cos θ_{2} - \cos θ_{1}}{\cos θ_{1} + \cos θ_{2}} = \frac{\cos θ_{2} - 2 \cos θ_{2}}{2 \cos θ_{2} + \cos θ_{2}} \end{array}$
$\begin{array}{l} \Rightarrow p = 1 a n d q = 3 \end{array}$
$\frac{- p}{q} = \frac{- 1}{3}$
$∴ p + q = 4$

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