Trigonometric Identities Question 93
Question: If $ a,{{\cos }^{3}}\alpha +3a,\cos \alpha ,{{\sin }^{2}}\alpha =m $ and $ a,{{\sin }^{3}}\alpha +3a,{{\cos }^{2}}\alpha \sin \alpha =n, $ then $ {{(m+n)}^{2/3}}+{{(m-n)}^{2/3}} $ is equal to
Options:
A) $ 2a^{2} $
B) $ 2{a^{1/3}} $
C) $ 2{a^{2/3}} $
D) $ 2a^{3} $
Show Answer
Answer:
Correct Answer: C
Solution:
Adding and subtracting the given relation, we get $ (m+n)=a{{\cos }^{3}}\alpha +3a\cos \alpha ,{{\sin }^{2}}\alpha $ $ +3a{{\cos }^{2}}\alpha .\sin \alpha +a{{\sin }^{3}}\alpha $ $ =a{{(\cos \alpha +\sin \alpha )}^{3}} $ and similarly $ (m-n)=a{{(\cos \alpha -\sin \alpha )}^{3}} $ Thus, $ {{(m+n)}^{2/3}}+{{(m-n)}^{2/3}} $ $ ={a^{2/3}}{{{\cos \alpha +\sin \alpha )}^{2}}+{{(\cos \alpha -\sin \alpha )}^{2}}} $ $ ={a^{2/3}}{2({{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha )}=2{a^{2/3}} $ .
BETA
