[3sin^2 (a/2)-2sin(a/2)cos(a/2)+cos^2 (a/2)]/[tana+1/tana]
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[3sin^2 (a/2)-2sin(a/2)cos(a/2)+cos^2 (a/2)]/[tana+1/tana]
[3sin^2 (a/2)-2sin(a/2)cos(a/2)+cos^2 (a/2)]/[tana+1/tana]
[3sin^2 (a/2)-2sin(a/2)cos(a/2)+cos^2 (a/2)]/[tana+1/tana]
分子=3(1-cosa)/2-sina+(1+cosa)/2
=2-cosa-sina
分母=sina/cosa+cosa/sina
=(sin²a+cos²a)/sinacosa
=1/sinacosa
所以原式=sinacosa(2-cosa-sina)
=2sinacosa-sin²acosa-sinacos²a
sin a+sin 2a +sin 3a +...+sin na怎么求和?
sin a+sin 2a +sin 3a +...+sin na怎么求和?
3 在三角形ABC中,已知(a2+b2)sin(A-B)=(a2-b2)sin(A+B) 求证:ABC是等腰或直角三角形(a^2+b^2)sin(A-B)=(a^2-b^2)sin(A+B),(sin^A+sin^B)sin(A-B)=(sin^A-sin^B)sin(A+B) sin^A*(sin(A+B)-sin(A-B))=sin^B*(sin(A-B)+sin(A+B)) sin^A*2c
若sin^2a
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求证:sin(a+b)sin(a-b)=sin^2a-sin^2b
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