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

#(sin^4X-2sin^2X+1)cosx# 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.

1 Answer
Feb 6, 2018

Answer:

#cos^5x#

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^2x# is just a simpler way to write #(sin x)^2#. Similarly, #sin^4x = (sin x)^4#.

We can now rewrite the original expression.

#(sin^4 x - 2 sin^2 x +1) cos x#

#=[ (sin x)^4 - 2 (sin x)^2 + 1] cos x#

Now, here's the part involving algebra. Let #sin x = a#. We can write # (sin x)^4 - 2 (sin x)^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 - 2ab + b^2 = (a-b)^2#, we can say

#a^4 - 2 a^2 + 1 = (a^2 - 1)^2#

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

#[ (sin x)^4 - 2 (sin x)^2 + 1] cos x#

#= [(sin x)^2 -1]^2 cos x#

#= (color(blue)(sin^2x - 1))^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 #color(blue)(sin^2 x -1 = -cos^2x)#.

#=(color(blue)(-cos^2x))^2 cos x#

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

#= (cos^4x) cos x#

#=cos^5x#

Thus, #(sin^4 x - 2 sin^2 x +1) cos x = cos^5x#.