求证1+sin a+cos a+2sin acos a/1+sin a+cos a=sin a+cos a
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求证1+sina+cosa+2sinacosa/1+sina+cosa=sina+cosa求证1+sina+cosa+2sinacosa/1+sina+cosa=sina+cosa求证1+sina+c
求证1+sin a+cos a+2sin acos a/1+sin a+cos a=sin a+cos a
求证1+sin a+cos a+2sin acos a/1+sin a+cos a=sin a+cos a
求证1+sin a+cos a+2sin acos a/1+sin a+cos a=sin a+cos a
sin a + cos a + (2sin acos a + 1) = sin a + cos a +(2sin acos a + (sin a)的平方 + ( cos a)的平方 )= (sin a + cos a)+ (sin a + cos a)的平方 = (sin a + cos a)(1+sin a+cos a)
上式除以 1+sin a+cos a=sin a+cos a
原式=sin^2 a+cos^2+sin a+cos a+2sin acos a/1+sin a+cos a
=(sin a+cos a)^2+(sin a+cos a)/1+sin a+cos a
=(sin a+cos a)(sin a+cos a+1)/1+sin a+cos a
=sin a+cos a
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