Question

${\sum }_{r=0}^{7}ta{n}^2\left(\frac{\pi r}{16}\right)=$___

• A. 28
• B. 35
• C. 1
• D. 3

Answer 1

The correct option is B

35

$I={\sum }_{r=0}^{7}ta{n}^2\frac{\pi r}{16})=ta{n}^2\frac{\pi }{16}+ta{n}^2\frac{2\pi }{16}+ta{n}^2\frac{3\pi }{16}+ta{n}^2\frac{4\pi }{16}+ta{n}^2\frac{5\pi }{16}+ta{n}^2\frac{6\pi }{16}+ta{n}^2\frac{7\pi }{16}=(ta{n}^2\frac{\pi }{16}+co{t}^2\frac{\pi }{16})+(ta{n}^2\frac{2\pi }{16}+co{t}^2\frac{2\pi }{16})+(ta{n}^2\frac{3\pi }{16}+co{t}^2\frac{3\pi }{16})+1={(tan\frac{\pi }{16}+cot\frac{\pi }{16})}^2-2+{(tan\frac{2\pi }{16}+cot\frac{2\pi }{16})}^2-2+{(tan\frac{3\pi }{16}+cot\frac{3\pi }{16})}^2-2+1=\frac{1}{si{n}^2\frac{\pi }{16}co{t}^2\frac{\pi }{16}}+\frac{1}{si{n}^2\frac{2\pi }{16}co{s}^2\frac{2\pi }{16}}+\frac{1}{si{n}^2\frac{3\pi }{16}co{t}^2\frac{3\pi }{16}}-5.$
$=4[\frac{1}{si{n}^2\frac{\pi }{8}}+\frac{1}{si{n}^2\frac{4\pi }{16}}+\frac{1}{si{n}^2\frac{6\pi }{16}}]-5=4[\frac{1}{si{n}^2\frac{\pi }{8}}+\frac{1}{co{s}^2\frac{\pi }{8}}]+3=\frac{4.4}{si{n}^2\frac{\pi }{4}}+3=35.$