Question

If $\sin x + {\sin ^2}x + {\sin ^3}x = 1$ then what is the value of ${\cos ^6}x - 4{\cos ^4}x + 8{\cos ^2}x$ ?

Answer 1

Trigonometric identities and properties are used to solve this problem.We use transposing techniques to solve this problem. We use some identities like ${(a + b)^2} = {a^2} + 2ab + {b^2}$, ${(a - b)^2} = a - 2ab + {b^2}$ and ${\cos ^2}x + {\sin ^2}x = 1$ also. In this problem, transpositions play a key role.

Complete step by step answer:
Making transpositions play a crucial role in solving this problem.
First of all, let us try to spot out any identities present in the question or not.
So, make some transpositions to get a key to identities.
It is given that, $\sin x + {\sin ^2}x + {\sin ^3}x = 1$
Now, transpose the term ${\sin ^2}x$ to the right hand side.
$\Rightarrow \sin x + {\sin ^3}x = 1 - {\sin ^2}x$
We all know the identity ${\sin ^2}x + {\cos ^2}x = 1$

And from this identity, we can get to a conclusion that, $1 - {\sin ^2}x = {\cos ^2}x$
$\Rightarrow \sin x + {\sin ^3}x = {\cos ^2}x$
$\Rightarrow \sin x(1 + {\sin ^2}x) = {\cos ^2}x$
And, we know that ${\sin ^2}x = 1 - {\cos ^2}x$. So, substitute this in the above equation.
$\Rightarrow \sin x(1 + 1 - {\cos ^2}x) = {\cos ^2}x$
$\Rightarrow \sin x(2 - {\cos ^2}x) = {\cos ^2}x$
So, now squaring on both sides,
$\Rightarrow {(\sin x)^2}{(2 - {\cos ^2}x)^2} = {({\cos ^2}x)^2}$
$\Rightarrow ({\sin ^2}x)\left( {4 + {{\cos }^4}x - 2(2)({{\cos }^2}x)} \right) = {\cos ^4}x$

Now, substitute ${\sin ^2}x = 1 - {\cos ^2}x$ in the above equation.
$\Rightarrow \left( {1 - {{\cos }^2}x} \right)\left( {4 + {{\cos }^4}x - 4{{\cos }^2}x} \right) = {\cos ^4}x$
So, now, we have to expand this expression in order to get our desired expression.
$\Rightarrow \left( 1 \right)\left( {4 + {{\cos }^4}x - 4{{\cos }^2}x} \right) - {\cos ^2}x\left( {4 + {{\cos }^4}x - 4{{\cos }^2}x} \right) = {\cos ^4}x$
$\Rightarrow 4 - 8{\cos ^2}x - {\cos ^6}x + 4{\cos ^4}x = 0$
So, now grouping all constants on one side and all variables on another side, then,
$\therefore {\cos ^6}x - 4{\cos ^4}x + 8{\cos ^2}x = 4$
We got the desired expression.

Therefore, the value of expression ${\cos ^6}x - 4{\cos ^4}x + 8{\cos ^2}x$ is equal to four.

Note: Be careful while expanding the expressions, because if you replace one sign, the whole solution goes wrong. If you get this type of problem, try to identify the standard identities in the question. And always try to make your problem simpler.In transpositions, positive terms become negative and negative terms become positive when they are transposed to the other side of the symbol. And similarly, multiplication becomes division and vice-versa. Also remember some other identities like ${\sec ^2}x - {\tan ^2}x = 1$ and $\cos e{c^2}x - {\cot ^2}x = 1$ which may be much useful to you in your future problems.