How do you use the power reducing formulas to rewrite the expression $sin^8x$ in terms of the first power of cosine?

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Answer 1 · 2018-04-14T15:13:09

$sin^8x=1/128[35-56cos2x+28cos4x-8cos6x+cos8x]$

$rarrsin^8x$
$=[(2sin^2x)/2]^4$

$=1/16[{1-cos2x}^2]^2$

$=1/16[1-2cos2x+cos^2(2x)]^2$

$=1/16[(1-2cos2x)^2+2*(1-2cos2x)*cos^2(2x)+(cos^2(2x))^2]$

$=1/16[1-4cos2x+4cos^2(2x)+2cos^2(2x)-4cos^3(2x)+((2cos^2(2x))/2)^2]$

$=1/16[1-4cos2x+6cos^2(2x)-(3cos(2x)+cos6x)+((1+cos4x)/2)^2]$

$=1/16[1-4cos2x+3*{1+cos4x}-(3cos(2x)+cos6x)+((1+2cos4x+cos^2(4x))/4)]$

$=1/16[1-4cos2x+3+3cos4x-3cos(2x)-cos6x+((2+4cos4x+2cos^2(4x))/8)]$

$=1/16[4-7cos2x+3cos4x-cos6x+((2+4cos4x+1+cos8x)/8)]$

$=1/16[(4-7cos2x+3cos4x-cos6x+((3+4cos4x+cos8x)/8)]$

$=1/16[(8(4-7cos2x+3cos4x-cos6x)+3+4cos4x+cos8x)/8]$

$=1/16[(32-56cos2x+24cos4x-8cos6x+3+4cos4x+cos8x)/8]$

$=1/128[35-56cos2x+28cos4x-8cos6x+cos8x]$

How do you use the power reducing formulas to rewrite the expression sin^8xsin^8x in terms of the first power of cosine?