$cos 4x=8cos^4(x)-8cos^2x + 1$ and $cos 4x=8sin^4(x)-8sin^2x + 1$ ,can u prove it?

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Answer 1 · 2017-07-08T08:07:57

$LHS=cos4x$

$=2cos^2 (2x)-1$

$=2(cos (2x))^2-1$

$=2(2cos ^2x-1)^2-1$

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

$=8cos ^4x-8cos^2x+2-1$

$=8cos ^4x-8cos^2x+1=RHS$

Again

$LHS=cos4x$

$=2cos^2 (2x)-1$

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

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

$=2-8sin^2x+8sin^4x-1$

$=8sin^4x-8sin^2x+1=RHS$

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

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

substitute in the equation as follows

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

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

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

$=8sin^4x-8sin^2x+1$

cos 4x=8cos^4(x)-8cos^2x + 1 cos 4x=8cos^4(x)-8cos^2x + 1 and cos 4x=8sin^4(x)-8sin^2x + 1cos 4x=8sin^4(x)-8sin^2x + 1,can u prove it?