# Calculus problem with substitution

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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

• Try drawing a triangle with opposite side equal to x and adjacent side 1 – Justin Stevenson Jan 24 at 19:56

These are trigonometric identities for half-angle formulas where in this case $$x=\tan(\frac{\theta}{2})$$
$$\cos(\theta)=\cos^2(\frac{\theta}{2})-\sin^2(\frac{\theta}{2})=(1-\tan^2(\frac{\theta}{2}))\cos^2(\frac{\theta}{2})=\frac{1-\tan^2(\frac{\theta}{2})}{1+\tan^2(\frac{\theta}{2})}=\frac{1-x^2}{1+x^2}$$
$$\sin(\theta)=2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})=2\tan(\frac{\theta}{2})\cos^2(\frac{\theta}{2})==2\frac{\tan(\frac{\theta}{2})}{1+\tan^2(\frac{\theta}{2})}=\frac{2x}{1+x^2}$$