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Question Number 20091 by Tinkutara last updated on 21/Aug/17

$$\mathrm{Prove}\mathrm{that}\underset{n=0}{\sum ^3}{\tan }^2\frac{(2n+1)\pi }{16}=28.$$

Answered by ajfour last updated on 21/Aug/17

$$\frac{\sin {}^{2}A}{\cos {}^{2}A}+\frac{\sin {}^{2}B}{\cos {}^{2}B}=\frac{\sin {}^{2}A\cos {}^{2}B+\cos {}^{2}A\sin {}^{2}B}{\cos {}^{2}A\cos {}^{2}B}$$$$=\frac{{[\sin (A+B)+\sin (A-B)]}^2+{[\sin (A+B)-\sin (A-B)]}^2}{{[\cos (A+B)+\cos (A-B)]}^2}$$$$=\frac{2[\sin {}^2(A+B)+\sin {}^2(A-B)]}{{[\cos (A+B)+\cos (A-B)]}^2}$$$$ForA=\frac{\pi }{16},B=\frac{7\pi }{16};C=\frac{3\pi }{16},D=\frac{5\pi }{16}$$$$\sin (A+B)=\sin (C+D)=1;$$$$\cos (A+B)=\cos (C+D)=0,hence$$$$\mathrm{\Sigma }\tan {}^2\frac{(2n+1)\pi }{16}=$$$$\frac{2[1+\sin {}^2\frac{3\pi }{8}]}{\cos {}^2\frac{3\pi }{8}}+\frac{2[1+\sin {}^2\frac{\pi }{8}]}{\cos {}^2\frac{\pi }{8}}$$$$=\frac{4+4\sin {}^2\frac{3\pi }{8}}{2\cos {}^2\frac{3\pi }{8}}+\frac{4+4\sin {}^2\frac{\pi }{8}}{2\cos {}^2\frac{\pi }{8}}$$$$=\frac{4+2-2\cos \frac{3\pi }{4}}{1+\cos \frac{3\pi }{4}}+\frac{4+2-2\cos \frac{\pi }{4}}{1+\cos \frac{\pi }{4}}$$$$=\frac{6+\sqrt{2}}{(1-\frac{1}{\sqrt{2}})}+\frac{6-\sqrt{2}}{(1+\frac{1}{\sqrt{2}})}=\frac{6\sqrt{2}+2}{\sqrt{2}-1}+\frac{6\sqrt{2}-2}{\sqrt{2}+1}$$$$=(12+6\sqrt{2}+2\sqrt{2}+2)+(12-6\sqrt{2}-2\sqrt{2}+2)$$$$=28.$$

Commented by Tinkutara last updated on 21/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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