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Proving Trig Ident.

by Miike012
Tags: ident, proving, trig
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Miike012
#1
Mar19-11, 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???
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jhae2.718
#2
Mar19-11, 02:13 PM
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[tex]\cos(2s) \equiv \cos^2(s) - \sin^2(s)[/tex]
Miike012
#3
Mar19-11, 02:18 PM
P: 1,011
Yes that is true... but then why isnt the expression
2*sin(2s)*(cos^2 - Sin^2)4
?

jhae2.718
#4
Mar19-11, 02:23 PM
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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].
Miike012
#5
Mar19-11, 02:25 PM
P: 1,011
Thank you!
jhae2.718
#6
Mar19-11, 02:27 PM
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Glad to help!


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