yqxs2010
春芽
共回答了18个问题采纳率:94.4% 举报
1.sin²αsin²β+cos²αcos²β-1/2cos2αcos2β
=(1-cos^2a)(1-cos^2b)+cos^2a*cos^2b-1/2cos2αcos2β
=1-cos^2a-cos^2b+2cos^2a*cos^2b-1/2cos2αcos2β
=1-(1+cos2a)/2-(1+cos2b)/2+2*(1+cos2a)/2*(1+cos2b)/2-1/2cos2αcos2β
=-cos2a/2-cos2b/2+1/2+cos2a/2+cos2b/2+1/2cos2αcos2β-1/2cos2αcos2β
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2:(sinθ+sin3θ+sin5θ)/(cosθ+cos3θ+cos5θ)
=(sin(3θ-2θ)+sin3θ+sin(3θ+2θ))/(cos(3θ-2θ)+cos3θ+cos(3θ+2θ))
=(2sin3θcos2θ+sin3θ)/(2cos3θcos2θ+cos3θ)
=sin3θ(2cos2θ+1)/cos3θ(2cos2+1)
=tan3θ
1年前
6