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

$$\mathrm{If}f\left({\tan }^2\frac{\theta }{2}\right)=\frac{2}{1+\cos \theta },\mathrm{find}f\left(\sin \frac{\theta }{2}\right).$$

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

$$f\left({\tan }^2\frac{\vartheta }{2}\right)=\frac{2}{1+\cos \vartheta }$$$$f\left(\frac{{\sin }^2\frac{\vartheta }{2}}{1-{\sin }^2\frac{\vartheta }{2}}\right)=\frac{1}{1-{\sin }^2\frac{\vartheta }{2}}\Rightarrow f\left(\frac{x^2}{1-x^2}\right)=\frac{1}{1-x^2}$$$$\frac{x^2}{1-x^2}=y\Rightarrow (y+1)x^2-y=0\Rightarrow x=\pm \sqrt{\frac{y}{y+1}}$$$$f\left(y\right)=\frac{1}{1-\frac{y}{y+1}}=\frac{y+1}{1}=y+1\Rightarrow f\left(\sin \frac{\vartheta }{2}\right)=\sin \frac{\vartheta }{2}+1$$

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

$$f\left({\tan }^2\frac{\theta }{2}\right)=\frac{2}{1+\cos \theta }=\frac{2}{1+2{\cos }^2\frac{\theta }{2}-1}$$$$f\left({\tan }^2\frac{\theta }{2}\right)=\frac{1}{{\cos }^2\frac{\theta }{2}}=1+{\tan }^2\frac{\theta }{2}$$$$\Rightarrow f\left(x\right)=1+x$$$$f\left(\sin \frac{\theta }{2}\right)=1+\sin \frac{\theta }{2}$$

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