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Hi, I started doing this problem and I couldn't figure out the last two question where $cos(\frac{\theta}{2})= \frac{1-x^2}{1+x^2}$ and $sin(\frac{\theta}{2})= \frac{2x}{1+x^2}$ Could you explain to me where to start with these 2 questions. Thank you

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    $\begingroup$ Try drawing a triangle with opposite side equal to x and adjacent side 1 $\endgroup$ – Justin Stevenson Jan 24 at 19:56
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These are trigonometric identities for half-angle formulas where in this case $x=\tan(\frac{\theta}{2})$

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

$$\sin(\theta)=2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})=2\tan(\frac{\theta}{2})\cos^2(\frac{\theta}{2})==2\frac{\tan(\frac{\theta}{2})}{1+\tan^2(\frac{\theta}{2})}=\frac{2x}{1+x^2}$$

Do you know these identities?

And please take a look at this for better formatting enter link description here

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  • $\begingroup$ this does not help at all. and thank you ill read it trough $\endgroup$ – Konstantin Uvarov Jan 24 at 20:00

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