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0
1690
2
avatar+267 

This is where I've gotten so far:

 

 

\(sin(4x) = 2sin(2x)cos(2x) = 2Csin(2x)\)

 

I've then tried using the fact that sin(2x) = cos(2x-pi/2) to simplify but that didn't help me since I was left with a sin(2x) term anyway. I couldn't find an identity that would let me rewrite sin(2x) in terms of cos(2x)'s. 

 

Thank you.

 Dec 8, 2016

Best Answer 

 #1
avatar+128079 
+6

sin (4x)     and  cos (2x )   = C

 

sin (4x)  =  2sin(2x)cos(2x)

 

If cos (2x)  = C, then sin(2x)  = √ [1 - [cos(2x)]^2 ]  =  √ [1 - C^2]

 

So

 

sin(4x)  = 2 * √ [1 - C^2] * C

 

 

cool cool cool

 Dec 8, 2016
 #1
avatar+128079 
+6
Best Answer

sin (4x)     and  cos (2x )   = C

 

sin (4x)  =  2sin(2x)cos(2x)

 

If cos (2x)  = C, then sin(2x)  = √ [1 - [cos(2x)]^2 ]  =  √ [1 - C^2]

 

So

 

sin(4x)  = 2 * √ [1 - C^2] * C

 

 

cool cool cool

CPhill Dec 8, 2016
 #2
avatar+267 
+5

Thank you very much CPhil, always a life saver. I always tend to forget that identity >.<

Quazars  Dec 8, 2016

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