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求证 sin~4a-cos~4a=sin~2a-cos~2a tan~2asin~2a=tan~2a-sin~2a
人气:441 ℃ 时间:2020-03-20 05:30:24
解答
sin^4a-cos^4a
=(sin^2a-cos^2a)(sin^2a+cos^2a)
=(sin^2a-cos^2a)*1=sin^2a-cos^2a
tan^2asin^2a=(sce^2a-1)sin^2a
=tan^2a-sin^2a
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