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

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Answer 1 · 2017-07-08T08:07:57

$L H S = cos 4 x$ $LHS=cos4x$

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

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

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

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

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

$= 8 {cos}^{4} x - 8 {cos}^{2} x + 1 = R H S$ $=8cos ^4x-8cos^2x+1=RHS$

Again

$L H S = cos 4 x$ $LHS=cos4x$

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

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

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

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

$= 8 {sin}^{4} x - 8 {sin}^{2} x + 1 = R H S$ $=8sin^4x-8sin^2x+1=RHS$

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

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

substitute in the equation as follows

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

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

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

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

cos4x=8cos4(x)−8cos2x+1cos 4x=8cos^4(x)-8cos^2x + 1 and cos4x=8sin4(x)−8sin2x+1cos 4x=8sin^4(x)-8sin^2x + 1,can u prove it?