Trigonometric Identities Question 340
Question: The value of $ {{\sin }^{2}}5^{o}+{{\sin }^{2}}10^{o}+{{\sin }^{2}}15^{o}+…+ $ $ {{\sin }^{2}}85^{o}+{{\sin }^{2}}90^{o} $ is equal to
[Karnataka CET 1999]
Options:
A) 7
B) 8
C) 9
D) $ 9\frac{1}{2} $
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Answer:
Correct Answer: D
Solution:
Given expression is $ {{\sin }^{2}}5^{o}+{{\sin }^{2}}10^{o}+{{\sin }^{2}}15^{o}+…..+{{\sin }^{2}}85^{o}+{{\sin }^{2}}90^{o}. $ We know that $ \sin 90^{o}=1 $ or $ {{\sin }^{2}}90^{o}=1 $ . Similarly, $ \sin 45^{o}=\frac{1}{\sqrt{2}}or,si{n^{2}}45^{o}=\frac{1}{2} $ and the angles are in A.P. of 18 terms. We also know that $ {{\sin }^{2}}85^{o}={{[\sin (90^{o}-5^{o})]}^{2}} $ $ ={{\cos }^{2}}5^{o}. $ Therefore from the complementary rule, we find $ {{\sin }^{2}}5^{o}+{{\sin }^{2}}85^{o}={{\sin }^{2}}5^{o}+{{\cos }^{2}}5^{o}=1. $ Therefore, $ {{\sin }^{2}}5^{o}+{{\sin }^{2}}10^{o}+{{\sin }^{2}}15^{o}+…+{{\sin }^{2}}85^{o}+{{\sin }^{2}}90^{o} $ $ =(1+1+1+1+1+1+1+1)+1+\frac{1}{2}=9\frac{1}{2} $ .
BETA
