Asked by Serena
                Rewrite the expression in terms of the first power of the cosine 
sin^8(x)
how do i do this plz help !!
            
        sin^8(x)
how do i do this plz help !!
Answers
                    Answered by
            Damon
            
    I can get it in powers of cos but not first power
sin^8 = sin^2 * sin^2 *sin^2 *sin^2
but
sin^2 = 1 - cos^2
sin^2*sin^2 = 1-2cos^2+cos^4
(1-2cos^2+cos^4)^2
= 1-4cos^2x+6cos^4x-4cos^6x+cos^8x
    
sin^8 = sin^2 * sin^2 *sin^2 *sin^2
but
sin^2 = 1 - cos^2
sin^2*sin^2 = 1-2cos^2+cos^4
(1-2cos^2+cos^4)^2
= 1-4cos^2x+6cos^4x-4cos^6x+cos^8x
                    Answered by
            Steve
            
    recall that cos^2 x = (1+cos2x)/2
so, cos^4 x = (1+cos2x)^2/4 now cos^2 2x = (1+cos4x)/2
and so on. In the end, you get
sin^8(x)
= 1/128 (cos8x - 8cos6x + 28 cos4x - 56cos2x + 35)
    
so, cos^4 x = (1+cos2x)^2/4 now cos^2 2x = (1+cos4x)/2
and so on. In the end, you get
sin^8(x)
= 1/128 (cos8x - 8cos6x + 28 cos4x - 56cos2x + 35)
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