Trigonometric Identities Question 95
Question: $ {{\cos }^{2}}76^{o}+{{\cos }^{2}}16^{o}-\cos 76^{o}\cos 16^{o}= $
[EAMCET 2002]
Options:
A) -1/4
B) 1/2
C) 0
D) 3/4
Show Answer
Answer:
Correct Answer: D
Solution:
$ {{\cos }^{2}}76^{o}+{{\cos }^{2}}16^{o}-\cos 76^{o}\cos 16^{o} $ $ =\frac{1}{2}[ 1+\cos 152^{o}+1+\cos 32^{o}-\cos 92^{o}-\cos 60^{o} ] $ $ =\frac{1}{2}[ 2-\frac{1}{2}+\cos 152^{o}+\cos 32^{o}-\cos 92^{o} ] $ $ =\frac{1}{2}[ \frac{3}{2}+2\cos 92^{o}\cos 60^{o}-\cos 92^{o} ] $ $ =\frac{1}{2}[ \frac{3}{2}+\cos 92^{o}-\cos 92^{o} ] $ $ =\frac{3}{4} $ .
BETA
