How to verify identity of sin 4x tan 2x = 1 - cos 4x ?

Question

How to verify identity of sin 4x tan 2x = 1 - cos 4x  ?

Samantha Abate · Accepted Answer

The answer
#sin4x*tan2x =1-cos4x#

#L.H.S=sin4x*tan4x=sin4x*[(sin2x)/(cos2x)]#

#=sin2x*[(sin2(2x))/(cos2x)]=sin2x*[(2*sin2x*cos2x)/(cos2x)]#

#=sin2x*[2*sin2x]=2sin^2(2x)#

#2*[1/2(1-cos4x)]=1-cos4x=R.H.S# your name #sin4x*tan2x =1-cos4x# #L.H.S=sin4x*tan4x=sin4x*[(sin2x)/(cos2x)]# #=sin2x*[(sin2(2x))/(cos2x)]=sin2x*[(2*sin2x*cos2x)/(cos2x)]# #=sin2x*[2*sin2x]=2sin^2(2x)# #2*[1/2(1-cos4x)]=1-cos4x=R.H.S# Note that #sin2x=2*sinx*cosx# #cos2x=cos^2x-sin^2x# #sin^2x=1/2*[1-cos2x]# #cos^2x=1/2*[1+cos2x]#

Caroline Adams · Answer

undefined To verify the identity $\sin(4x) \tan(2x) = 1 - \cos(4x)$, we'll express both sides of the equation in terms of sine and cosine functions and then simplify.

Starting with the left side:

$\sin(4x) \tan(2x)$

Since $\tan(2x) = \frac{\sin(2x)}{\cos(2x)}$, we have:

$\sin(4x) \cdot \frac{\sin(2x)}{\cos(2x)}$

Expanding $\sin(4x)$ using the double-angle identity $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$, we get:

$2\sin(2x)\cos(2x) \cdot \frac{\sin(2x)}{\cos(2x)}$

This simplifies to:

$2\sin^2(2x)$

Now, let's simplify the right side of the equation:

$1 - \cos(4x)$

Using the double-angle identity $\cos(2\theta) = 1 - 2\sin^2(\theta)$, we can write $\cos(4x)$ as:

$\cos(2 \cdot 2x) = 1 - 2\sin^2(2x)$

Now, substitute this expression into the equation:

$1 - (1 - 2\sin^2(2x))$

Simplify this expression:

$1 - 1 + 2\sin^2(2x)$

$2\sin^2(2x)$

As we can see, both sides of the equation simplify to $2\sin^2(2x)$. Therefore, the identity $\sin(4x) \tan(2x) = 1 - \cos(4x)$ is verified.