Question

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

Answer 1

In order to find the solution of the given question that is to power reducing formulas to rewrite the expression ${{\sin }^{4}}\left( x \right){{\cos }^{4}}\left( x \right)$ in terms of the first power of cosine apply the power reducing formulas that are $\sin \left( 2x \right)=2\sin \left( x \right)\cos \left( x \right)$, ${{\sin }^{2}}\left( x \right)=1-{{\cos }^{2}}\left( x \right)$, ${{\cos }^{2}}\left( 2x \right)=\dfrac{1}{2}\left( 1+\cos \left( 4x \right) \right)$ and ${{\cos }^{2}}\left( 4x \right)=\dfrac{1}{2}\left( 1+\cos \left( 8x \right) \right)$. Then simplify the given expression into the first power of cosine.

Complete step by step answer:
According to the question, given expression in the question is as follows:
${{\sin }^{4}}\left( x \right){{\cos }^{4}}\left( x \right)$
We can rewrite the above expression as follows:
$\Rightarrow {{\left( \sin \left( x \right)\cos \left( x \right) \right)}^{4}}$
Now multiply and divide $2$ with the bracket of the above expression, we get:
$\Rightarrow {{\left( \dfrac{2}{2}\sin \left( x \right)\cos \left( x \right) \right)}^{4}}$
$\Rightarrow \dfrac{1}{16}{{\left( 2\sin \left( x \right)\cos \left( x \right) \right)}^{4}}$
After this apply the identity $\sin \left( 2x \right)=2\sin \left( x \right)\cos \left( x \right)$ in the above expression, we get:
$\Rightarrow \dfrac{1}{16}{{\left( \sin \left( 2x \right) \right)}^{4}}$
We can rewrite the above expression as follows:
$\Rightarrow \dfrac{1}{16}{{\left( {{\sin }^{2}}\left( 2x \right) \right)}^{2}}$
After this apply the identity ${{\sin }^{2}}\left( x \right)=1-{{\cos }^{2}}\left( x \right)$ in the above expression, we get:
$\Rightarrow \dfrac{1}{16}{{\left( 1-{{\cos }^{2}}\left( 2x \right) \right)}^{2}}$
Simplify the above expression by opening the bracket, we get:
$\Rightarrow \dfrac{1}{16}\left( 1+{{\cos }^{4}}\left( 2x \right)-2{{\cos }^{2}}\left( 2x \right) \right)$
After this apply the identity ${{\cos }^{2}}\left( 2x \right)=\dfrac{1}{2}\left( 1+\cos \left( 4x \right) \right)$ in the above expression, we get:
$\Rightarrow \dfrac{1}{16}\left( 1+{{\left( {{\cos }^{2}}\left( 2x \right) \right)}^{2}}-2\cdot \dfrac{1}{2}\left( 1+\cos \left( 4x \right) \right) \right)$
$\Rightarrow \dfrac{1}{16}\left( 1+{{\left( {{\cos }^{2}}\left( 2x \right) \right)}^{2}}-1+\cos \left( 4x \right) \right)$
We can clearly see that addition of $1\text{ }\\!\\!\And\\!\\!\text{ }-1$ makes it zero, therefore we can rewrite the above expression as:
$\Rightarrow \dfrac{1}{16}\left( {{\left( {{\cos }^{2}}\left( 2x \right) \right)}^{2}}+\cos \left( 4x \right) \right)$
Now apply the identity ${{\cos }^{2}}\left( 2x \right)=\dfrac{1}{2}\left( 1+\cos \left( 4x \right) \right)$ in the above expression, we get:
$\Rightarrow \dfrac{1}{16}\left( {{\left( \dfrac{1}{2}\left( 1+\cos \left( 4x \right) \right) \right)}^{2}}+\cos \left( 4x \right) \right)$
Simplifying the above expression by opening the inner bracket we will get:
$\Rightarrow \dfrac{1}{16}\left( \dfrac{1}{4}\left( 1+2\cos \left( 4x \right)+{{\cos }^{2}}\left( 4x \right) \right)+\cos \left( 4x \right) \right)$
After this apply the identity ${{\cos }^{2}}\left( 4x \right)=\dfrac{1}{2}\left( 1+\cos \left( 8x \right) \right)$ in the above expression, we get:
$\Rightarrow \dfrac{1}{16}\left( \dfrac{1}{4}\left( 1+2\cos \left( 4x \right)+\dfrac{1}{2}\left( 1+\cos \left( 8x \right) \right) \right)+\cos \left( 4x \right) \right)$
Take $\dfrac{1}{2}$ common from the inner bracket of the above expression, we will have:
$\Rightarrow \dfrac{1}{16}\left( \dfrac{1}{8}\left( 2+4\cos \left( 4x \right)+1+\cos \left( 8x \right) \right)+\cos \left( 4x \right) \right)$
$\Rightarrow \dfrac{1}{16}\left( \dfrac{1}{8}\left( 3+4\cos \left( 4x \right)+\cos \left( 8x \right) \right)+\cos \left( 4x \right) \right)$
Now, take $\dfrac{1}{8}$ common from the inner bracket of the above expression, we will have:
$\Rightarrow \dfrac{1}{128}\left( 3+4\cos \left( 4x \right)+\cos \left( 8x \right)+8\cos \left( 4x \right) \right)$
Simplifying it further we will get:
$\Rightarrow \dfrac{1}{128}\left( 3+12\cos \left( 4x \right)+\cos \left( 8x \right) \right)$
Therefore, the reduced expression of ${{\sin }^{4}}\left( x \right){{\cos }^{4}}\left( x \right)$ in terms of first power of cosine is $\dfrac{1}{128}\left( 3+12\cos \left( 4x \right)+\cos \left( 8x \right) \right)$.

Note:
Students make a lot of mistakes in question like this. They tend to skip steps which increases the rate of miscalculation and getting the wrong answer. It’s important to remember that in such questions to try to solve without skipping each step and rechecking your answer again once solved.