Question

How do you use the power reducing formulas to rewrite the expression ${\sin ^4}x{\cos ^2}x$ in terms of the first power of cosine?

Answer 1

In this question, we want to reduce the given expression in terms of the first power of cosine. The first step is to use the law of indices for brackets to change the sine function into cosine using their relation identity. For that, apply the double angle formula for sine and cosine functions. Then apply the algebraic identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$to simplify the expression.
Here, we will use the trigonometric angle formulas as below.
$\cos 2x = 2{\cos ^2}x - 1$
$\cos 2x = 1 - 2{\sin ^2}x$

Complete step by step solution:
In this question, we will use the power reducing formulas to rewrite the given expression in terms of the first power of cosine.
The given expression is:
$\Rightarrow {\sin ^4}x{\cos ^2}x$
We can rewrite the above expression as
$\Rightarrow {\left( {{{\sin }^2}x} \right)^2}{\cos ^2}x$ ...(1)
As we know the trigonometric double angle formula for the cosine is $\cos 2x = 2{\cos ^2}x - 1$ and $\cos 2x = 1 - 2{\sin ^2}x$.
Now, from the formula $\cos 2x = 2{\cos ^2}x - 1$, we can write it as ${\cos ^2}x = \left( {\dfrac{{1 + \cos 2x}}{2}} \right)$ and ${\sin ^2}x = \left( {\dfrac{{1 - \cos 2x}}{2}} \right)$ .
Now, let us substitute the value of ${\cos ^2}x$ and ${\sin ^2}x$ in the equation (1).
$\Rightarrow {\left( {\dfrac{{1 - \cos 2x}}{2}} \right)^2}\left( {\dfrac{{1 + \cos 2x}}{2}} \right)$
That is equal to,
$\Rightarrow \dfrac{1}{4}{\left( {1 - \cos 2x} \right)^2} \times \dfrac{1}{2}\left( {1 + \cos 2x} \right)$
Let us apply the square to the first bracket. Apply the algebraic formula ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$.
$\Rightarrow \dfrac{1}{8}\left( {{1^2} - 2\left( 1 \right)\left( {\cos 2x} \right) + {{\cos }^2}2x} \right)\left( {1 + \cos 2x} \right)$
Let us simplify the above expression.
$\Rightarrow \dfrac{1}{8}\left( {1 - 2\cos 2x + {{\cos }^2}2x} \right)\left( {1 + \cos 2x} \right)$
Again apply the formula ${\cos ^2}x = \left( {\dfrac{{1 + \cos 2x}}{2}} \right)$ in the above expression.
$\Rightarrow \dfrac{1}{8}\left( {1 - 2\cos 2x + \left( {\dfrac{{1 + \cos 4x}}{2}} \right)} \right)\left( {1 + \cos 2x} \right)$
Now, let us take LCM.
$\Rightarrow \dfrac{1}{8}\left( {\dfrac{{2 - 4\cos 2x + 1 + \cos 4x}}{2}} \right)\left( {1 + \cos 2x} \right)$
Let us simplify the above expression.
$\Rightarrow \dfrac{1}{{16}}\left( {3 - 4\cos 2x + \cos 4x} \right)\left( {1 + \cos 2x} \right)$

Hence, the answer in terms of the first power of cosine is $\dfrac{1}{{16}}\left( {3 - 4\cos 2x + \cos 4x} \right)\left( {1 + \cos 2x} \right)$.

Note: We have to take care of the powers of the functions and signs. The double angle formulas are derived as below.
The double angle formula for cosine is:
$\cos 2x = {\cos ^2}x - {\sin ^2}x$
But we know that ${\sin ^2}x = 1 - {\cos ^2}x$
So,
$\cos 2x = {\cos ^2}x - \left( {1 - {{\cos }^2}x} \right)$
Let us open the bracket.
$\Rightarrow \cos 2x = {\cos ^2}x - 1 + {\cos ^2}x$
That is equal to,
$\Rightarrow \cos 2x = 2{\cos ^2}x - 1$
This is the cosine double angle formula in terms of cosine.
Similarly, we can derive the cosine double angle formula in terms of sine.
$\Rightarrow \cos 2x = 1 - 2{\sin ^2}x$