Question

prove that cos4x=1-8sin square x cos square x

Answer 1

you can solve 
cos4x as
cos2(2x) =1-2sin2(2x)
             =1-2(2sinx . cosx)2                          { sin2x = 2sinx. cosx}
             =1-2(4sin2x.cos2x)
            =1-8sin2x.cos2x
LHS=RHS

Answer 2

As a result, the identity $cos 4x = 1 - 8sin2 x cos2 x$ is established.

As per the question given,

The identity $cos 4x = 1 - 8sin^2 x cos^2 x$can be proven using the double-angle formula for cosine:

$cos 2x = 1 - 2sin^2 x$

Now, let's square the above equation:

$cos^2 2x = 1 - 4sin^2 x + 4sin^4 x$

Using the identity $cos 2x = 1 - 2sin^2 x$ again:

$cos^2 2x = (cos 2x)^2 = (1 - 2sin^2 x)^2 = 1 - 4sin^2 x + 4sin^4 x$

Now, we can use the identity $cos 4x = cos 2(2x)$ to obtain:

$cos 4x = cos 2(2x) = cos^2 2x = 1 - 4sin^2 x + 4sin^4 x$

Expanding the term 4sin^4 x using the identity $sin^2 x = 1 - cos^2 x$:

$cos 4x = 1 - 4sin^2 x + 4(1 - cos^2 x)^2 = 1 - 8sin^2 x cos^2 x.$

So, the identity$cos 4x = 1 - 8sin^2 x cos^2 x$ is proven.

Also read,

https://brainly.in/question/4429731

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