三角函数证明
证明下列恒等式:(1)tan^2a×sin^2a=tan^2a-sin^2a(2)cosa+tana---------------=sinacosa/sina+1/cos...
证明下列恒等式:
(1)tan^2a×sin^2a=tan^2a-sin^2a (2)cosa+tana
--------------- =sina
cosa/sina+1/cosa
(3) cos^2a-sin^2a 1-tana
----------------------- =----------------
1+2sinacosa 1+tana 展开
(1)tan^2a×sin^2a=tan^2a-sin^2a (2)cosa+tana
--------------- =sina
cosa/sina+1/cosa
(3) cos^2a-sin^2a 1-tana
----------------------- =----------------
1+2sinacosa 1+tana 展开
1个回答
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(1)
tan^2a×sin^2a=tan^2a×(1-cos^2a)=tan^2a-sin^2a
(2)
(cosa+tana)/(cosa/sina+1/cosa)=sina(cosa+tana)/(cosa+tana)=sina
(3)
(cos^2a-sin^2a)/(1+2sinacosa)
=(cosa+sina)(cosa-sina)/(sina+cosa)^2
=(cosa-sina)/(sina+cosa)
=(1-tana)/(1+tana)
如果不懂,请Hi我,祝学习愉快!
tan^2a×sin^2a=tan^2a×(1-cos^2a)=tan^2a-sin^2a
(2)
(cosa+tana)/(cosa/sina+1/cosa)=sina(cosa+tana)/(cosa+tana)=sina
(3)
(cos^2a-sin^2a)/(1+2sinacosa)
=(cosa+sina)(cosa-sina)/(sina+cosa)^2
=(cosa-sina)/(sina+cosa)
=(1-tana)/(1+tana)
如果不懂,请Hi我,祝学习愉快!
来自:求助得到的回答
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