How can I rewrite 8 cos 4x in terms of cos x? If you could give a detailed explanation, that would be great!

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Answer 1 · 2017-11-17T17:31:55

$8 cos 4 x = 64 {cos}^{4} x - 64 {cos}^{2} x + 8 .$ $8cos4x=64cos^4x-64cos^2x+8.$

Recall that, $cos 2 x = 2 {cos}^{2} x - 1 .$ $cos2x=2cos^2x-1.$

$:. cos4x=2cos^2 2x-1,$

$=2(cos2x)^2-1,$

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

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

$:. cos4x=8cos^4x-8cos^2x+1.$

$rArr 8cos4x=64cos^4x-64cos^2x+8.$