神马作文网证明下列恒等式：sin(a+b)*cos(a-b）=sina*cosa+sinb*cosb
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证明下列恒等式：
sin(a+b)*cos(a-b）=sina*cosa+sinb*cosb
sin(a+b)*cos(a-b）=sina*cosa+sinb*cosb
证明:
sin(a+b)*cos(a-b)
=[sinacosb+sinbcosa]*[cosacosb+sinasinb]
=[(sinacosa)*cos^2(b)+(sinbcosb)*cos^2(a)]
+[(sinbcosb)*sin^2(a)+(sinacosa)*sin^2(b)]
=[(sinacosa)*cos^2(b)+(sinacosa)*sin^2(b)]
+[(sinbcosb)*sin^2(a)+(sinbcosb)*cos^2(a)]
=(sinacosa)*[sin^2(b)+cos^2(b)]+
(sinbcosb)*[sin^2(a)+cos^2(a)]
=sina*cosa+sinb*cosb
即命题得证
sin(a+b)*cos(a-b)
=[sinacosb+sinbcosa]*[cosacosb+sinasinb]
=[(sinacosa)*cos^2(b)+(sinbcosb)*cos^2(a)]
+[(sinbcosb)*sin^2(a)+(sinacosa)*sin^2(b)]
=[(sinacosa)*cos^2(b)+(sinacosa)*sin^2(b)]
+[(sinbcosb)*sin^2(a)+(sinbcosb)*cos^2(a)]
=(sinacosa)*[sin^2(b)+cos^2(b)]+
(sinbcosb)*[sin^2(a)+cos^2(a)]
=sina*cosa+sinb*cosb
即命题得证
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