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Question Number 155527 by ZiYangLee last updated on 01/Oct/21

If f (\tan^{2} \frac{θ}{2}) = \frac{2}{1 + \cos θ}, find f (\sin \frac{θ}{2}) .

Answered by Ar Brandon last updated on 01/Oct/21

f (\tan^{2} \frac{ϑ}{2}) = \frac{2}{1 + \cos ϑ}

f (\frac{\sin^{2} \frac{ϑ}{2}}{1 - \sin^{2} \frac{ϑ}{2}}) = \frac{1}{1 - \sin^{2} \frac{ϑ}{2}} \Rightarrow f (\frac{x^{2}}{1 - x^{2}}) = \frac{1}{1 - x^{2}}

\frac{x^{2}}{1 - x^{2}} = y \Rightarrow (y + 1) x^{2} - y = 0 \Rightarrow x = \pm \sqrt{\frac{y}{y + 1}}

f (y) = \frac{1}{1 - \frac{y}{y + 1}} = \frac{y + 1}{1} = y + 1 \Rightarrow f (\sin \frac{ϑ}{2}) = \sin \frac{ϑ}{2} + 1

Answered by mr W last updated on 02/Oct/21

f (\tan^{2} \frac{θ}{2}) = \frac{2}{1 + \cos θ} = \frac{2}{1 + 2 \cos^{2} \frac{θ}{2} - 1}

f (\tan^{2} \frac{θ}{2}) = \frac{1}{\cos^{2} \frac{θ}{2}} = 1 + \tan^{2} \frac{θ}{2}

\Rightarrow f (x) = 1 + x

f (\sin \frac{θ}{2}) = 1 + \sin \frac{θ}{2}

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