∑_k=0^4 sin^2(2 k+1) (π)/20=

Correct answer: (B) 5/2. ∑_k=0^4 sin^2(2 k+1) (π)/20 =sin^2 (π)/20+sin^2 (3 π)/20+sin^2 (5 π)/20+sin^2 (7 π)/20+sin^2 (9 π)/20 =sin^2 ((π)/2-(9 π)/20 )+sin^2 ((π)/2-(7 π)/20 ) +sin^2 (5 π)/20+sin^2 (7 π)/20+sin^2 (9 π)/20 =cos^2 ((9 π)/20 )+cos^2 ((7 π)/20 )+sin^2 (5 π)/20+sin^2 (7 π)/20+sin^2 (5 π)/20 =1+1+sin^2 (π)/4=1+1+1/2=5/2

Q59 / 160 · AP EAMCET 2023 (18 May Shift 1) · Mathematics

$\sum_{k=0}^4 \sin ^2(2 k+1) \frac{\pi}{20}=$

A
$5$
Correct answer
B
$\frac{5}{2}$
C
$3$
D
$\frac{3}{2}$

Solution

$\sum_{k=0}^4 \sin ^2(2 k+1) \frac{\pi}{20}$ $\begin{aligned} & =\sin ^2 \frac{\pi}{20}+\sin ^2 \frac{3 \pi}{20}+\sin ^2 \frac{5 \pi}{20}+\sin ^2 \frac{7 \pi}{20}+\sin ^2 \frac{9 \pi}{20} \\ & =\sin ^2\left(\frac{\pi}{2}-\frac{9 \pi}{20}\right)+\sin ^2\left(\frac{\pi}{2}-\frac{7 \pi}{20}\right)\end{aligned}$ $+\sin ^2 \frac{5 \pi}{20}+\sin ^2 \frac{7 \pi}{20}+\sin ^2 \frac{9 \pi}{20}$ $=\cos ^2\left(\frac{9 \pi}{20}\right)+\cos ^2\left(\frac{7 \pi}{20}\right)+\sin ^2 \frac{5 \pi}{20}+\sin ^2 \frac{7 \pi}{20}+\sin ^2 \frac{5 \pi}{20}$ $=1+1+\sin ^2 \frac{\pi}{4}=1+1+\frac{1}{2}=\frac{5}{2}$

Hence, the correct answer is option: 'B'.

$\sum_{k=0}^4 \sin ^2(2 k+1) \frac{\pi}{20}=$

Solution

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