求证3+cos4α-4cos2α=(8sinα)^4证（1-cos2α)/(1+cos2α)=tan²α(1+sin2α)/(cosα+sinα)=cosα+sinα
问题描述：
求证3+cos4α-4cos2α=(8sinα)^4
证（1-cos2α)/(1+cos2α)=tan²α
(1+sin2α)/(cosα+sinα)=cosα+sinα
答
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
答
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)^42）（1-c...