Trigonometric Identities Question 226
Question: The value of $ {{\sin }^{2}}5{}^\circ +{{\sin }^{2}}10{}^\circ +{{\sin }^{2}}15{}^\circ +{{\sin }^{2}} $ $ 20{}^\circ +…..+{{\sin }^{2}}90{}^\circ $ is
Options:
A) 7
B) 8
C) 9
D) $ \frac{19}{2} $
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Answer:
Correct Answer: D
Solution:
$ {{\sin }^{2}}5{}^\circ +{{\sin }^{2}}10{}^\circ +{{\sin }^{2}}15{}^\circ +…..+……+….. $ $ {{\sin }^{2}}75{}^\circ +{{\sin }^{2}}80{}^\circ +{{\sin }^{2}}85{}^\circ +{{\sin }^{2}}90{}^\circ $
$ \Rightarrow {{\sin }^{2}}5{}^\circ +{{\sin }^{2}}10{}^\circ +{{\sin }^{2}}15{}^\circ +…..{{\sin }^{2}} $ $ (90-15{}^\circ )+{{\sin }^{2}}(90-10)+sin^{2}(90-5)+1 $
$ \Rightarrow {{\sin }^{2}}5+{{\sin }^{2}}10+{{\sin }^{2}}15{}^\circ +……+{{\cos }^{2}}15{}^\circ $ $ +{{\cos }^{2}}10{}^\circ +{{\cos }^{2}}5{}^\circ +1 $
$ \Rightarrow ,(1+1+1+….8,times)+sin^{2}45{}^\circ +1 $
$ \Rightarrow 8+\frac{1}{2}+1=\frac{19}{2} $
BETA
