Prove-that-n-0-3-tan-2-2n-1-pi-16-28-

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

Prove that \underset{n = 0}{\sum^{3}} \tan^{2} \frac{(2 n + 1) π}{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}}

F o r A = \frac{π}{16}, B = \frac{7 π}{16}; C = \frac{3 π}{16}, D = \frac{5 π}{16}

\sin (A + B) = \sin (C + D) = 1;

\cos (A + B) = \cos (C + D) = 0, h e n c e

Σ \tan^{2} \frac{(2 n + 1) π}{16} =

\frac{2 [1 + \sin^{2} \frac{3 π}{8}]}{\cos^{2} \frac{3 π}{8}} + \frac{2 [1 + \sin^{2} \frac{π}{8}]}{\cos^{2} \frac{π}{8}}

= \frac{4 + 4 \sin^{2} \frac{3 π}{8}}{2 \cos^{2} \frac{3 π}{8}} + \frac{4 + 4 \sin^{2} \frac{π}{8}}{2 \cos^{2} \frac{π}{8}}

= \frac{4 + 2 - 2 \cos \frac{3 π}{4}}{1 + \cos \frac{3 π}{4}} + \frac{4 + 2 - 2 \cos \frac{π}{4}}{1 + \cos \frac{π}{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

Thank you very much Sir!