Proving Trig Ident.

Miike012

1. The problem statement, all variables and given/known data
sin(4s)/4 = cos^3(s)*Sin(s) - sin^3(s)*cos(s)

In the book they did....
2*sin(2s)*cos(2s)/4
= 2*2*sin(s)*cos(s)/4 *(cos^2(s) - sin^2(s))
(I understand everything up until they multiplyed the 2*2*sin(s)*cos(s)/4 expression by cos^2(s) - sin^2(s)...
where did cos^2(s) - sin^2(s) come from???

jhae2.718

[tex]\cos(2s) \equiv \cos^2(s) - \sin^2(s)[/tex]

Miike012

Yes that is true... but then why isnt the expression
2*sin(2s)*(cos^2 - Sin^2)4
?

jhae2.718

What they did was:
[tex]\frac{2\sin(2s)\cos(2s)}{4} = \frac{2\cdot \left(2\sin(s)\cos(s)\right)\left(\cos^2(s) - \sin^2(s)\right)}{4}[/tex]

which is simply substituting in [tex]2\sin(s)\cos(s)[/tex] for [tex]\sin(2s)[/tex], and [tex]\cos^2(s)-\sin^2(s)[/tex] for [tex]\cos(2s)[/tex].

Miike012

Thank you!

jhae2.718

Glad to help!