求证3+cos4α-4cos2α=(8sinα)^4证（1-cos2α)/(1+cos2α)=tan²α(1+sin2α)/(cosα+sinα)=cosα+sinα
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求证3+cos4α-4cos2α=(8sinα)^4证（1-cos2α)/(1+cos2α)=tan²α(1+sin2α)/(cosα+sinα)=cosα+sinα
求证3+cos4α-4cos2α=(8sinα)^4
证（1-cos2α)/(1+cos2α)=tan²α
(1+sin2α)/(cosα+sinα)=cosα+sinα
求证3+cos4α-4cos2α=(8sinα)^4证（1-cos2α)/(1+cos2α)=tan²α(1+sin2α)/(cosα+sinα)=cosα+sinα
证明：1）3+cos4α-4cos2α=2+(1+cos4α)-4cos2α
=2+2(cos2α)^2-4cos2α
=2[1+(cos2α)^2-2cos2α]
=2(1-cos2α)^2
=2*{1-[(cosα)^2-(sinα)^2}^2
=2*{1-(cosα)^2+(sinα)^2}^2
=2*{2(sinα)^2}^2
=8*(sinα)^4
2）（1-cos2α)/(1+cos2α)={1-[(cosα)^2-(sinα)^2]}/{1+[(cosα)^2-(sinα)^2]}
={1-(cosα)^2+(sinα)^2}/{1-(sinα)^2+(cosα)^2}
=2(sinα)^2/{2(cosα)^2}
=tan²α
3）(1+sin2α)/(cosα+sinα)=(sin²α+cos²α+2sinαcosα)/(cosα+sinα)
=(cosα+sinα)^2/(cosα+sinα)
=cosα+sinα
第一题
cos4α=1-2(sin2α)^2=1-8(sinα)^2*(cosα)^2=1-8(sinα)^2*[1-(sinα)^2]
=8(sinα)^4-8(sinα)^2+1
而4cos2α+8(sinα)^4-3
=4[1-2(sinα)^2]+8(sinα)^4-3=8(sinα)^4-8(sinα)^2+1
所以cos4α=4cos2α+8(sinα)^4-3
即3+cos4α-4cos2α=8sinα^4