Trigonometric Identities Question 98
Question: $ {{\cos }^{2}}\alpha +{{\cos }^{2}}(\alpha +120{}^\circ )+{{\cos }^{2}}(\alpha -120{}^\circ ) $ is equal to
[MP PET 1993]
Options:
A) 3/2
B) 1
C) 1/2
D) 0
Show Answer
Answer:
Correct Answer: A
Solution:
$ {{\cos }^{2}}\alpha +{{\cos }^{2}}(\alpha +120^{o})+{{\cos }^{2}}(\alpha -120^{o}) $ $ ={{\cos }^{2}}\alpha +{{{ \cos ,(\alpha +120^{o})+\cos ,(\alpha -120^{o}) }}^{2}} $ $ -2,\cos ,(\alpha +120^{o}),\cos ,(\alpha -120^{o}) $ $ ={{\cos }^{2}}\alpha +{{{ ,2,\cos \alpha ,\cos ,120^{o} }}^{2}}-2,{ {{\cos }^{2}}\alpha -{{\sin }^{2}},120^{o} } $ $ ={{\cos }^{2}}\alpha +{{\cos }^{2}}\alpha -2,{{\cos }^{2}}\alpha +2,{{\sin }^{2}},120^{o} $ $ =2{{\sin }^{2}}120^{o}=2\times \frac{3}{4}=\frac{3}{2} $ .
BETA
