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# Trig Identities

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

Guest Mar 5, 2017

#1
+77139
+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.....

CPhill  Mar 5, 2017
Sort:

#1
+77139
+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.....

CPhill  Mar 5, 2017
#2
+18626
+5

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

$$\begin{array}{|rcll|} \hline \sin(4x) &=& [4\sin(x)\cos(x)][2\cos^2(x)-1] \\ && \cos(2x) = \cos^2(x) - \sin^2(x) = \cos^2(x) - [1- \cos^2(x)] = 2\cos^2(x)-1 \\ \sin(4x) &=& [4\sin(x)\cos(x)][\cos(2x)] \\ && \sin(2x) = 2\cdot \sin(x)\cos(x) \\ \sin(4x) &=& 2\sin(2x)\cos(2x) \\ && \sin(4x)=\sin(2x+2x) = \sin(2x)\cos(2x) + \cos(2x)\sin(2x) = 2\sin(2x)\cos(2x) \\ \sin(4x) &=& \sin(4x) \ \checkmark \\ \hline \end{array}$$

heureka  Mar 6, 2017

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