
#1
Mar1911, 02:09 PM

P: 1,011

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??? 



#3
Mar1911, 02:18 PM

P: 1,011

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



#4
Mar1911, 02:23 PM

PF Gold
P: 1,153

Proving Trig Ident.
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]. 



#5
Mar1911, 02:25 PM

P: 1,011

Thank you!



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