Question
For any $$\theta \in \left( {\frac{\pi }{4},\frac{\pi }{2}} \right)$$ the expression $$3{\left( {\sin \theta - \cos \theta } \right)^4} + 6{\left( {\sin \theta + \cos \theta } \right)^2} + 4\,{\sin ^6}\theta $$ equals:
A.
$$13 - 4\,{\cos ^2}\theta + 6\,{\sin ^2}\theta {\cos ^2}\theta $$
B.
$$13 - 4\,{\cos ^6}\theta $$
C.
$$13 - 4\,{\cos ^2}\theta + 6\,{\cos ^4}\theta $$
D.
$$13 - 4\,{\cos ^4}\theta + 2\,{\sin ^2}\theta {\cos ^2}\theta $$
Answer :
$$13 - 4\,{\cos ^6}\theta $$
Solution :
$$\eqalign{
& 3{\left( {\sin \theta - \cos \theta } \right)^4} + 6{\left( {\sin \theta + \cos \theta } \right)^2} + 4\,{\sin ^6}\theta \cr
& = 3{\left( {1 - 2\sin \theta \cos \theta } \right)^2} + 6\left( {1 + 2\sin \theta \cos \theta } \right) + 4\,{\sin ^6}\theta \cr
& = 3\left( {1 + 4{{\sin }^2}\theta\, {{\cos }^2}\theta - 4\sin \theta \cos \theta } \right) + 6 + 12\sin \theta \cos \theta + 4\,{\sin ^6}\theta \cr
& = 9 + 12\,{\sin ^2}\theta\, {\cos ^2}\theta + 4\,{\sin ^6}\theta \cr
& = 9 + 12\,{\cos ^2}\theta \left( {1 - {{\cos }^2}\theta } \right) + 4{\left( {1 - {{\cos }^2}\theta } \right)^3} \cr
& = 9 + 12\,{\cos ^2}\theta - 12\,{\cos ^4}\theta + 4\left( {1 - {{\cos }^6}\theta - 3\,{{\cos }^2}\theta + 3\,{{\cos }^4}\theta } \right) \cr
& = 9 + 4 - 4\,{\cos ^6}\theta \cr
& = 13 - 4\,{\cos ^6}\theta \cr} $$