First slide

Trigonometric transformations

Question

Given that $(1 + \sqrt{1 + x}) \tan y = 1 + \sqrt{1 - x}$ Then sin4y is equal to

Moderate

A

4x
B

2x
C

x
D

none of these

Solution

$\tan y = \frac{1 + \sqrt{1 - x}}{1 + \sqrt{1 + x}}$
Let x= cos $θ$ , then
$\begin{array}{l} \sqrt{1 - x} = \sqrt{2} \sin (θ / 2); \sqrt{1 + x} = \sqrt{2} \cos (θ / 2) \\ \Rightarrow \tan y = \frac{\sqrt{2} [\frac{1}{\sqrt{2}} + \sin \frac{θ}{2}]}{\sqrt{2} [\frac{1}{\sqrt{2}} + \cos \frac{θ}{2}]} = \frac{\sin \frac{π}{4} + \sin \frac{θ}{2}}{\cos \frac{π}{4} + \cos \frac{θ}{2}} \\ = \frac{2 \sin (\frac{π}{8} + \frac{θ}{4}) \cos (\frac{π}{8} - \frac{θ}{4})}{2 \cos (\frac{π}{8} + \frac{θ}{4}) \cos (\frac{π}{8} - \frac{θ}{4})} \\ = \tan (\frac{π}{8} + \frac{θ}{4}) \end{array}$
$\begin{array}{l} \Rightarrow 4 y = \frac{π}{2} + θ \\ or \sin 4 y = \cos θ = x \end{array}$

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