Question

If $\sin x + {\sin ^2}x = 1$, then the value of expression ${\cos ^{12}}x + 3{\cos ^{10}}x + 3{\cos ^8}x + {\cos ^6}x - 1$ is equal to
(A) $0$
(B) $1$
(C) $ - 1$
(D) $2$

Answer 1

Hint: The given question deals with finding the value of trigonometric expression doing basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as ${\cos ^2}x + {\sin ^2}x = 1$. Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem. We must know the simplification rules to solve the problem with ease.

Complete step by step answer:
In the given problem, we have to find the value of trigonometric expression ${\cos ^{12}}x + 3{\cos ^{10}}x + 3{\cos ^8}x + {\cos ^6}x - 1$.
Given $\sin x + {\sin ^2}x = 1$.
Shifting the ${\sin ^2}x$ term to the right side of the equation. So, we get,
$ \Rightarrow \sin x = 1 - {\sin ^2}x$
Now, we know the trigonometric identity \[{\cos ^2}x = 1 - {\sin ^2}x\].
$ \Rightarrow \sin x = {\cos ^2}x$
Now, we will use the relation $\sin x = {\cos ^2}x$ for the evaluation of the value of expression given to us. So, we get,
${\cos ^{12}}x + 3{\cos ^{10}}x + 3{\cos ^8}x + {\cos ^6}x - 1$
Expressing some terms in form of product, we get,
$ \Rightarrow {\left( {{{\cos }^2}x} \right)^6} + 3{\left( {{{\cos }^2}x} \right)^4}{\left( {\cos x} \right)^2} + 3{\left( {{{\cos }^2}x} \right)^2}{\left( {\cos x} \right)^4} + {\left( {{{\cos }^2}x} \right)^3} - 1$
Now, using the relation $\sin x = {\cos ^2}x$ and substituting the value of ${\cos ^2}x$ as $\sin x$, we get,
$ \Rightarrow {\left( {\sin x} \right)^6} + 3{\left( {\sin x} \right)^4}{\left( {\cos x} \right)^2} + 3{\left( {\sin x} \right)^2}{\left( {\cos x} \right)^4} + {\left( {{{\cos }^2}x} \right)^3} - 1$
Simplifying the expression, we get,
$ \Rightarrow {\left( {{{\sin }^2}x} \right)^3} + 3{\left( {{{\sin }^2}x} \right)^2}\left( {{{\cos }^2}x} \right) + 3\left( {{{\sin }^2}x} \right){\left( {{{\cos }^2}x} \right)^2} + {\left( {{{\cos }^2}x} \right)^3} - 1$
Now, we use the algebraic identity ${\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}$.
$ \Rightarrow {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^3} - 1$
Again using the trigonometric identity $\left( {{{\sin }^2}x + {{\cos }^2}x} \right) = 1$, we get,
$ \Rightarrow {\left( 1 \right)^3} - 1$
Now, we know that any power of one is equal to one itself. So, we get,
$ \Rightarrow 1 - 1$
\[ \Rightarrow 0\]
Therefore, the value of trigonometric expression ${\cos ^{12}}x + 3{\cos ^{10}}x + 3{\cos ^8}x + {\cos ^6}x - 1$ is equal to zero. Hence, option (A) is the correct answer.

Note:
We must have a strong grip over the concepts of trigonometry, related formulae and rules to ace these types of questions. We must use the laws of exponents very carefully to get to the final answer. We also must keep in mind the trigonometric identities such as $\left( {{{\sin }^2}x + {{\cos }^2}x} \right) = 1$ to solve the given problem.