# How do you verify sin x(1- 2 cos ^2 x + cos ^4 x) = sin ^5 x?

Aug 3, 2015

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

#### Explanation:

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

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

Aug 3, 2015

The key to the proof is to factor the left hand side

#### Explanation:

When I see a set of parenthesis I always want to see if it can be factored into a simpler form.

So it turns out that $\left(1 - 2 {\cos}^{2} \left(x\right) + {\cos}^{4} \left(x\right)\right)$ is a typical form and is factored by the the following form:

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

so our factored form is:

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

Now I see that we have ${\cos}^{2} \left(x\right)$ and that makes me want to check the pythagorean identity:

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

notice if we move the cosine to the other side then we have the
same expression.

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

so let us substitute the ${\sin}^{2} \left(x\right)$ into our equation:
$\sin \left(x\right) {\left({\sin}^{2} \left(x\right)\right)}^{2} = {\sin}^{5} \left(x\right)$

now it is just a matter of simplifying the left hand side.

$\sin \left(x\right) \cdot {\sin}^{4} \left(x\right) = {\sin}^{5} \left(x\right)$
${\sin}^{5} \left(x\right) = {\sin}^{5} \left(x\right)$

and thus it is demonstrated.
and that makes us happy.