Question

What is the value of $3{\left( {\sin x - \cos x} \right)^4} + 6{\left( {\sin x + \cos x} \right)^2} + 4\left( {{{\sin }^6}x + {{\cos }^6}x} \right)$?
a) $11$
b) $12$
c) $13$
d) $14$

Answer 1

Hint: The given problem requires us to simplify the given trigonometric expression. We will use the trigonometric identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$ to solve the given question. The question describes the wide-ranging applications of trigonometric identities and formulae. We must keep in mind the trigonometric identities while solving such questions.

Formula used: ${\sin ^2}\theta + {\cos ^2}\theta = 1$
${a^3} + {b^3} + 3ab\left( {a + b} \right) = {\left( {a + b} \right)^3}$

Complete step by step solution:
First, we simplify the given trigonometric expression using algebraic identity $\left( {{a^2} - 2ab + {b^2}} \right) = {\left( {a - b} \right)^2}$.
So, we have, $3{\left( {\sin x - \cos x} \right)^4} + 6{\left( {\sin x + \cos x} \right)^2} + 4\left( {{{\sin }^6}x + {{\cos }^6}x} \right)$.
$ \Rightarrow 3{\left[ {{{\left( {\sin x - \cos x} \right)}^2}} \right]^2} + 6\left( {{{\sin }^2}x + 2\sin x\cos x + {{\cos }^2}x} \right) + 4\left( {{{\sin }^6}x + {{\cos }^6}x} \right)$
$ \Rightarrow 3{\left( {{{\sin }^2}x - 2\sin x\cos x + {{\cos }^2}x} \right)^2} + 6\left( {{{\sin }^2}x + 2\sin x\cos x + {{\cos }^2}x} \right) + 4\left( {{{\sin }^6}x + {{\cos }^6}x} \right)$
Using the trigonometric identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$, we get,
$ \Rightarrow 3{\left( {1 - 2\sin x\cos x} \right)^2} + 6\left( {1 + 2\sin x\cos x} \right) + 4\left( {{{\sin }^6}x + {{\cos }^6}x} \right)$
$ \Rightarrow 3\left( {{1^2} - 2 \times 2\sin x\cos x + {{\left( {2\sin x\cos x} \right)}^2}} \right) + 6\left( {1 + 2\sin x\cos x} \right) + 4\left( {{{\sin }^6}x + {{\cos }^6}x} \right)$
$ \Rightarrow 3\left( {1 - 4\sin x\cos x + 4{{\sin }^2}x{{\cos }^2}x} \right) + 6 + 12\sin x\cos x + 4\left( {{{\sin }^6}x + {{\cos }^6}x} \right)$
Opening the brackets, we get,
$ \Rightarrow 3 - 12\sin x\cos x + 12{\sin ^2}x{\cos ^2}x + 6 + 12\sin x\cos x + 4{\sin ^6}x + 4{\cos ^6}x$
Cancelling the like terms with opposite terms, we get,
$ \Rightarrow 3 + 12{\sin ^2}x{\cos ^2}x + 6 + 4{\sin ^6}x + 4{\cos ^6}x - - - - \left( 1 \right)$
Now, we know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$. Cubing both sides of the equation, we get,
$ \Rightarrow {\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)^3} = {1^3}$
Using the algebraic identity ${a^3} + {b^3} + 3ab\left( {a + b} \right) = {\left( {a + b} \right)^3}$,
\[ \Rightarrow {\left( {{{\sin }^2}\theta } \right)^3} + {\left( {{{\cos }^2}\theta } \right)^3} + 3{\sin ^2}\theta {\cos ^2}\theta \left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) = {1^3}\]
\[ \Rightarrow {\sin ^6}\theta + {\cos ^6}\theta + 3{\sin ^2}\theta {\cos ^2}\theta = 1\]
Multiplying both sides by four, we get,
\[ \Rightarrow 4{\sin ^6}\theta + 4{\cos ^6}\theta + 12{\sin ^2}\theta {\cos ^2}\theta = 4 - - - - \left( 2 \right)\]
Substituting value from equation $\left( 2 \right)$ into expression $\left( 1 \right)$, we get,
\[ \Rightarrow 9 + 12{\sin ^2}x{\cos ^2}x + 4{\sin ^6}x + 4{\cos ^6}x\]
\[ \Rightarrow 9 + 4 = 13\]
So, we get the value of $3{\left( {\sin x - \cos x} \right)^4} + 6{\left( {\sin x + \cos x} \right)^2} + 4\left( {{{\sin }^6}x + {{\cos }^6}x} \right)$ as $13$.

Hence, option C is the correct answer.

Note:
Basic trigonometric identities include ${\sin ^2}\theta + {\cos ^2}\theta = 1$, ${\sec ^2}\theta = {\tan ^2}\theta + 1$ and $\cos e{c^2}\theta = {\cot ^2}\theta + 1$. These identities are of vital importance for solving any question involving trigonometric functions and identities. Algebraic identities such as ${a^3} + {b^3} + 3ab\left( {a + b} \right) = {\left( {a + b} \right)^3}$ and $\left( {{a^2} - 2ab + {b^2}} \right) = {\left( {a - b} \right)^2}$, and rules like transposition rule come into significant use while solving such problems.