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Question Number 119015 by benjo_mathlover last updated on 21/Oct/20

$$x=\frac{\sin 1°+\sin 2°+\sin 3°+...+\sin 45°}{\cos 1°+\cos 2°+\cos 3°+...+\cos 45°}$$$$x=?$$

Commented by Dwaipayan Shikari last updated on 21/Oct/20

$$tan(1°+\frac{45}{2}-\frac{1}{2})=tan23°$$

Answered by Dwaipayan Shikari last updated on 21/Oct/20

$$sin\theta +sin(\theta +\beta )+....+sin(\theta +(n-1)\beta )=\mathrm{\Phi }$$$$=\frac{1}{sin\frac{\beta }{2}}\left(sin(\theta +\frac{n\beta }{2}-\frac{\beta }{2})sin\frac{n\beta }{2}\right)$$$$cos\theta +cos(\theta +\beta )+cos(\theta +2\beta )+...cos(\theta +(n-1)\beta )=\mathrm{\Psi }$$$$=\frac{1}{sin\frac{\beta }{2}}\left(cos(\theta +\frac{n\beta }{2}-\frac{\beta }{2})sin\frac{n\beta }{2}\right)$$$$\frac{\mathrm{\Phi }}{\mathrm{\Psi }}=tan(\theta +\frac{n\beta }{2}-\frac{\beta }{2})$$$$\theta =\frac{\pi }{180}\beta =\frac{\pi }{180}n=45$$$$\frac{\mathrm{\Phi }}{\mathrm{\Psi }}=x=tan\left(\frac{23\pi }{180}\right)$$

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