题目

化简求值sin^4θ+sin^4(π/3-θ)+sin^4(π/3+θ)

提问时间：2021-01-03

答案

sin^4θ = [1/2(1-cos2θ)^2]^2 = 1/4{1-2cos2θ+(cos2θ)^2}
= 1/4{1-2cos2θ+(1+cos4θ)/2} = 3/8 - 1/4 cos2θ + 1/8cos4θ
∴sin^4θ+sin^4(π/3-θ)+sin^4(π/3+θ)
= sin^4θ+sin^4(θ-π/3)+sin^4(θ+π/3)
= 3/8-1/4cos2θ+1/8cos4θ + 3/8-1/4cos(2θ-2π/3)+1/8cos(4θ-4π/3) + 3/8-1/4cos(2θ+2π/3)+1/8cos(4θ+4π/3)
= 9/8 - 1/4{cos2θ+cos(2θ-2π/3)+cos(2θ+2π/3)} + 1/8{cos4θ+cos(4θ-4π/3)+cos(4θ+4π/3)}
= 9/8 - 1/4{cos2θ+cos2θcos2π/3+sincos2θsin2π/3 + cos2θcos2π/3-sincos2θsin2π/3}
+ 1/8{cos4θ+cos4θcos4π/3+sincos4θsin4π/3 + cos4θcos4π/3-sincos4θsin4π/3}
= 9/8 - 1/4{cos2θ+cos2θcos2π/3+cos2θcos2π/3} + 1/8{cos4θ+cos4θcos4π/3+cos4θcos4π/3}
= 9/8 - 1/4{cos2θ-1/2cos2θ-1/2cos2θ} + 1/8{cos4θ-1/2cos4θ-1/2cos4θ}
= 9/8 - 0 + 0
= 9/8