Question

How do you solve that (2cosx+1)(2cosx-1)(2cos2x-1)=2cos4x+1Please help it's urgent!

Answer 1

(2cos x + 1)(2cos x - 1)(2cos 2x - 1 ) = 2cos 4x + 1 has proven and solved.

Further explanation

We will solve and proving an equation related to trigonometric identity.

$\boxed{ \ (2cos \ x + 1)(2cos \ x - 1)(2cos \ 2x - 1 ) = 2cos \ 4x + 1 \ }$

Let (2cos x + 1) as Part-A, (2cos x - 1) as Part-B, and (2cos 2x - 1) as Part-C.

We multiply Part-A with Part-B.

Recall that $\boxed{ \ (a + b)(a - b) = a^2 - b^2 \ }$

Do it carefully.

$\boxed{ \ (4cos^2x - 1)(2cos \ 2x - 1 ) = 2cos \ 4x + 1 \ }$ ... Equation-1

We use one property of trigonometric identities, i.e.,

$\boxed{ \ cos \ 2x = 2cos^2x - 1 \ } \rightarrow multiply \ by \ 2 \ \rightarrow \boxed{ \ 2cos \ 2x = 4cos^2x - 2 \ }$

Prepare it like this.

$\boxed{ \ 2cos \ 2x = 4cos^2x - 1 - 1 \ } \rightarrow \boxed{ \ 4cos^2x - 1 = 2cos \ 2x + 1 \ }$ ... Equation-2

Equation-2 is substituted into Equation-1.

$\boxed{ \ (2cos \ 2x + 1)(2cos \ 2x - 1 ) = 2cos \ 4x + 1 \ }$

$\boxed{ \ 4cos{^2}2x - 1 = 2cos \ 4x + 1 \ }$

On the left side, use the same property earlier with a little processed into, $\boxed{ \ 4cos{^2}2x - 1 = 2cos \ 4x + 1 \ }$

$\boxed{ \ 2cos \ 4x + 1 = 2cos \ 4x + 1 \ }$

So it has proven and solved.

Learn more

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Keywords: how do you solve that, (2cosx + 1)(2cosx - 1)(2cos2x - 1) = 2cos4x + 1, trigonometric identity, property, proving

Answer 2

On the left side, we can collapse some terms:

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

Recall the double angle identity for cosine:

$\cos^2x=\dfrac{1+\cos2x}2$

So we have

$4\cos^2x-1=\dfrac{4(1+\cos2x)}2-1=2\cos2x+1$

With the same reasoning, we can collapse that side further:

$(2\cos2x+1)(2\cos2x-1)=4\cos^22x-1=\dfrac{4(1+\cos4x)}2-1=2\cos4x+1$