Proving Trig Ident: sin(4s)/4 = cos^3(s)*Sin(s) - sin^3(s)*cos(s)

[Miike012]

Homework Statement

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]

$$\cos(2s) \equiv \cos^2(s) - \sin^2(s)$$

 

[Miike012]

Yes that is true... but then why isn't the expression
2*sin(2s)*(cos^2 - Sin^2)4
?

 

[jhae2.718]

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

which is simply substituting in $$2\sin(s)\cos(s)$$ for $$\sin(2s)$$, and $$\cos^2(s)-\sin^2(s)$$ for $$\cos(2s)$$.

 

[Miike012]

Thank you!

 

[jhae2.718]

Glad to help!