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

Question

5323 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-03T02:35:08

Verify #sin x(1 - 2cos^2 x + cos^4 x) = sin^5 x#

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

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

Answer 2 · 2015-08-03T02:52:13

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

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 #(1-2cos^2(x)+cos^4(x))# is a typical form and is factored by the the following form:

#(a^2-2ab+b^2) = (a-b)^2#

so our factored form is:

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

Now I see that we have #cos^2(x)# and that makes me want to check the pythagorean identity:

#sin^2(x)+cos^2(x)=1#

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

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

so let us substitute the #sin^2(x)# into our equation:
#sin(x)(sin^2(x))^2 = sin^5(x)#

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

#sin(x)*sin^4(x) = sin^5(x)#
#sin^5(x) = sin^5(x)#

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