已知sina+sinb=siny,cosa+cosb=cosy,求证cos(a-y)=1/2
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已知sina+sinb=siny,cosa+cosb=cosy,求证cos(a-y)=1/2
原式=cosacosy+sinasiny=cosa(cosa+cosb)+sina(sina+sinb)
=1+cosacosb+sinasinb
sin^2(y)+cos^2(y)=1+1+2sinasinb+2cosacosb=1
故sinasinb+cosacosb=-1\2
1+cosacosb+sinasinb=1-1\2=1\2
证毕
将已知的两等式变形:a,y项在一起,b项一边,然后平方,发现什么?
情不自禁想将两个式子相加,那么结果就很明显了。
试试吧,加油~
证明:
cos(a-y)=cosa cosy+sina siny
=cos²a+cosa cosb+sin²a+sina sinb
=1+cosa cosb+sina sinb
而sin²y+cos²y=(sina+sinb)²+(cosa+cosb)²
=sin²a+sin²b...
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证明:
cos(a-y)=cosa cosy+sina siny
=cos²a+cosa cosb+sin²a+sina sinb
=1+cosa cosb+sina sinb
而sin²y+cos²y=(sina+sinb)²+(cosa+cosb)²
=sin²a+sin²b+2sina sinb+cos²a+cos²b+2cosa cosb
=2+2(sina sinb+cosa cosb)=1
∴sina sinb+cosa cosb=-½
带入原式,cos(a-y)=1/2得证。
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