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Question Number 113781 by Ar Brandon last updated on 15/Sep/20

Answered by Dwaipayan Shikari last updated on 15/Sep/20

$$sin\frac{\pi }{12}sin\frac{7\pi }{12}=\frac{1}{2}(cos\frac{\pi }{2}-cos\frac{8\pi }{12})=\frac{1}{2}.\frac{1}{2}=\frac{1}{4}$$

Answered by Dwaipayan Shikari last updated on 15/Sep/20

$${tan}^{-1}\frac{1}{n}-{tan}^{-1}\frac{1}{n+p}$$$$={tan}^{-1}\frac{\frac{1}{n}-\frac{1}{n+p}}{1+\frac{1}{n(n+p)}}={tan}^{-1}\frac{p}{n^2+pn+1}\left(Whichistrue\right)$$$$$$

Answered by Dwaipayan Shikari last updated on 15/Sep/20

$$2({sin}^4\frac{\pi }{8}+{sin}^4\frac{3\pi }{8})sin\frac{3\pi }{8}=sin\frac{5\pi }{8},sin\frac{\pi }{8}=sin\frac{7\pi }{8}$$$$2{({sin}^2\frac{\pi }{8}+{sin}^2\frac{3\pi }{8})}^2-4{sin}^2\frac{\pi }{8}{sin}^2\frac{3\pi }{8}$$$$2{(\frac{1-cos\frac{\pi }{4}}{2}+\frac{1+cos\frac{\pi }{4}}{2})}^2-{(cos\frac{\pi }{4}+cos\frac{\pi }{2})}^{2}sin\frac{3\pi }{8}=cos\frac{\pi }{8})$$$$2-{(\frac{1}{\sqrt{2}}+0)}^2=\frac{3}{2}$$

Answered by Dwaipayan Shikari last updated on 15/Sep/20

$$\frac{sin(A+\frac{B}{2})}{sin\frac{B}{2}}=k$$$$\frac{sin(A+\frac{B}{2})+sin\frac{B}{2}}{sin(A+\frac{B}{2})-sin\frac{B}{2}}=\frac{k+1}{k-1}$$$$\frac{sin\left(\frac{A+B}{2}\right)cos\frac{A}{2}}{cos\left(\frac{A+B}{2}\right)sin\frac{A}{2}}=\frac{k+1}{k-1}$$$$\frac{sin\frac{C}{2}}{cos\frac{C}{2}}.\frac{sin\frac{A}{2}}{cos\frac{A}{2}}=\frac{k-1}{k+1}$$$$tan\frac{A}{2}.tan\frac{C}{2}=\frac{k-1}{k+1}cos\frac{A+B}{2}=sin\frac{C}{2},sin\frac{A+B}{2}=cos\frac{C}{2}$$

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