(2cos x + 1)(2cos x - 1)(2cos 2x - 1 ) = 2cos 4x + 1 has proven and solved.
We will solve and proving an equation related to trigonometric identity.
[tex]\boxed{ \ (2cos \ x + 1)(2cos \ x - 1)(2cos \ 2x - 1 ) = 2cos \ 4x + 1 \ }[/tex]
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 [tex]\boxed{ \ (a + b)(a - b) = a^2 - b^2 \ }[/tex]
Do it carefully.
[tex]\boxed{ \ (4cos^2x - 1)(2cos \ 2x - 1 ) = 2cos \ 4x + 1 \ }[/tex] ... Equation-1
We use one property of trigonometric identities, i.e.,
[tex]\boxed{ \ cos \ 2x = 2cos^2x - 1 \ } \rightarrow multiply \ by \ 2 \ \rightarrow \boxed{ \ 2cos \ 2x = 4cos^2x - 2 \ }[/tex]
Prepare it like this.
[tex]\boxed{ \ 2cos \ 2x = 4cos^2x - 1 - 1 \ } \rightarrow \boxed{ \ 4cos^2x - 1 = 2cos \ 2x + 1 \ }[/tex] ... Equation-2
Equation-2 is substituted into Equation-1.
[tex]\boxed{ \ (2cos \ 2x + 1)(2cos \ 2x - 1 ) = 2cos \ 4x + 1 \ }[/tex]
[tex]\boxed{ \ 4cos{^2}2x - 1 = 2cos \ 4x + 1 \ }[/tex]
On the left side, use the same property earlier with a little processed into, [tex] \boxed{ \ 4cos{^2}2x - 1 = 2cos \ 4x + 1 \ }[/tex]
[tex]\boxed{ \ 2cos \ 4x + 1 = 2cos \ 4x + 1 \ }[/tex]
So it has proven and solved.
Keywords: how do you solve that, (2cosx + 1)(2cosx - 1)(2cos2x - 1) = 2cos4x + 1, trigonometric identity, property, proving
