use the power-reducing formulas to rewrite the expression in terms of the first power of the cosine . . sin^6 x

1 answer:

andrey2020 [161]3 years ago

4 0

1/128 (3 - 4 Cos[4 x] + Cos[8 x])( sin^2(x) )^2 cos^4(x)

( (1 - cos^2(x) )^2 cos^4(x) 

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

cos^4(x) - 2cos^6(x) + cos^8(x) 

( cos^2(x) )^2 - 2( cos^2(x) )^3 + ( cos^2(x) )^4 <==> knowing the identity; cos^2(x) = (1/2) * (1 + cos(2x)) 

( (1/2) * (1 + cos(2x)) )^2 - 2( (1/2) * (1 + cos(2x)) )^3 + ( (1/2) * (1 + cos(2x)) )^4 

( (1/4) * (1 + 2cos(2x) + cos^2(2x) )) - 2( (1/8) * (cos^3(2x) + 3cos^2(2x) + 3cos(2x) + 1) + ( (1/16) * cos^4(2x) + 4cos^3(2x) + 6cos^2(2x) + 4cos(2x) + 1)

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Using binomial theorem:

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

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