Processing math: 100%
 
+0  
 
0
1962
2
avatar

Can you prove sin4x=(4sinxcosx)(2cos^2x-1)? I just can't figure it out, thank you in advance!!!

 Mar 5, 2017

Best Answer 

 #1
avatar+130491 
+5

sin 4x   = (4sinxcosx) (2cos^2 (2x) - 1 )

 

4sinxcosx  =  2*2sinxcosx = 2sin2x

 

2cos^2(2x) - 1 =   cos(4x)

 

But.....as the graph below shows,

 

sin (4x)   is not equal to   2sin(2x)cos(4x)

 

 

 

So....this is not an identity.....

 

 

 

cool cool cool

 Mar 5, 2017
 #1
avatar+130491 
+5
Best Answer

sin 4x   = (4sinxcosx) (2cos^2 (2x) - 1 )

 

4sinxcosx  =  2*2sinxcosx = 2sin2x

 

2cos^2(2x) - 1 =   cos(4x)

 

But.....as the graph below shows,

 

sin (4x)   is not equal to   2sin(2x)cos(4x)

 

 

 

So....this is not an identity.....

 

 

 

cool cool cool

CPhill Mar 5, 2017
 #2
avatar+26397 
+5

Can you prove sin4x=(4sinxcosx)(2cos^2x-1)?

 

sin(4x)=[4sin(x)cos(x)][2cos2(x)1]cos(2x)=cos2(x)sin2(x)=cos2(x)[1cos2(x)]=2cos2(x)1sin(4x)=[4sin(x)cos(x)][cos(2x)]sin(2x)=2sin(x)cos(x)sin(4x)=2sin(2x)cos(2x)sin(4x)=sin(2x+2x)=sin(2x)cos(2x)+cos(2x)sin(2x)=2sin(2x)cos(2x)sin(4x)=sin(4x) 

 

laugh

 Mar 6, 2017

2 Online Users

avatar