搜搜做题作业网证明恒等式 三角比1. sin^2a+sin^2b-sin^2asin^2b+cos^2acos^2b=12. 2(1-
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证明恒等式 三角比
1. sin^2a+sin^2b-sin^2asin^2b+cos^2acos^2b=1
2. 2(1-sina)(1+cosa)=(1-sina+cosa)^2
3. (tan^2a-cot^2a)/(sin^2a-cos^2a)=sec^2a+csc^2a
过程答案
1. sin^2a+sin^2b-sin^2asin^2b+cos^2acos^2b=1
2. 2(1-sina)(1+cosa)=(1-sina+cosa)^2
3. (tan^2a-cot^2a)/(sin^2a-cos^2a)=sec^2a+csc^2a
过程答案
1.
(sina)^2+(sinb)^2-(sinasinb)^2+(cosacosb)^2
=(sina)^2-(sinasinb)^2 + 1-(cosb)^2+(cosacosb)^2
=(sina)^2[1-(sinb)^2] - (cosb)^2[1-(cosa)^2] +1
=(sinacosb)^2-(cosbsina)^2+1
=0+1
=1
2.
左边-右边
=2(1-sina+cosa-sinacosa)-[1+(sina)^2+(cosa)^2 -2sina+2cosa-2sinacosa]
=2(1-sina+cosa-sinacosa)-(2-2sina+2cosa-2sinacosa)
=0
3.
左边=[(sina/cosa)^2-(cosa/sina)^2]/[(sina)^2-(cosa)^2]
=[(sina)^2+(cosa)^2]/[(cosa)^2(sina)^2]
=(seca)^2+(csca)^2
(sina)^2+(sinb)^2-(sinasinb)^2+(cosacosb)^2
=(sina)^2-(sinasinb)^2 + 1-(cosb)^2+(cosacosb)^2
=(sina)^2[1-(sinb)^2] - (cosb)^2[1-(cosa)^2] +1
=(sinacosb)^2-(cosbsina)^2+1
=0+1
=1
2.
左边-右边
=2(1-sina+cosa-sinacosa)-[1+(sina)^2+(cosa)^2 -2sina+2cosa-2sinacosa]
=2(1-sina+cosa-sinacosa)-(2-2sina+2cosa-2sinacosa)
=0
3.
左边=[(sina/cosa)^2-(cosa/sina)^2]/[(sina)^2-(cosa)^2]
=[(sina)^2+(cosa)^2]/[(cosa)^2(sina)^2]
=(seca)^2+(csca)^2
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