# How do I simplify (sin^4x-2sin^2x+1)cosx?

## $\left({\sin}^{4} X - 2 {\sin}^{2} X + 1\right) \cos x$ I just cant see how to do this. Please give an easy explanation. I was doing so well in math until these types of problems come up and now I am stuck.

Feb 6, 2018

${\cos}^{5} x$

#### Explanation:

This type of problem is truly not that bad once you recognize that it involves a little algebra!

First, I'll rewrite the given expression to make the following steps easier to understand. We know that ${\sin}^{2} x$ is just a simpler way to write ${\left(\sin x\right)}^{2}$. Similarly, ${\sin}^{4} x = {\left(\sin x\right)}^{4}$.

We can now rewrite the original expression.

$\left({\sin}^{4} x - 2 {\sin}^{2} x + 1\right) \cos x$

$= \left[{\left(\sin x\right)}^{4} - 2 {\left(\sin x\right)}^{2} + 1\right] \cos x$

Now, here's the part involving algebra. Let $\sin x = a$. We can write ${\left(\sin x\right)}^{4} - 2 {\left(\sin x\right)}^{2} + 1$ as

${a}^{4} - 2 {a}^{2} + 1$

Does this look familiar? We just need to factor this! This is a perfect square trinomial. Since ${a}^{2} - 2 a b + {b}^{2} = {\left(a - b\right)}^{2}$, we can say

${a}^{4} - 2 {a}^{2} + 1 = {\left({a}^{2} - 1\right)}^{2}$

Now, switch back to the original situation. Re-substitute $\sin x$ for $a$.

$\left[{\left(\sin x\right)}^{4} - 2 {\left(\sin x\right)}^{2} + 1\right] \cos x$

$= {\left[{\left(\sin x\right)}^{2} - 1\right]}^{2} \cos x$

$= {\left(\textcolor{b l u e}{{\sin}^{2} x - 1}\right)}^{2} \cos x$

We can now use a trigonometric identity to simplify the terms in blue. Rearranging the identity ${\sin}^{2} x + {\cos}^{2} x = 1$, we get $\textcolor{b l u e}{{\sin}^{2} x - 1 = - {\cos}^{2} x}$.

$= {\left(\textcolor{b l u e}{- {\cos}^{2} x}\right)}^{2} \cos x$

Once we square this, the negative signs multiply to become positive.

$= \left({\cos}^{4} x\right) \cos x$

$= {\cos}^{5} x$

Thus, $\left({\sin}^{4} x - 2 {\sin}^{2} x + 1\right) \cos x = {\cos}^{5} x$.