Sinx+siny=根号2 Cosx+cosy=二分之根号三 求cos(x减y)和cos(x+y)
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Sinx+siny=根号2 Cosx+cosy=二分之根号三 求cos(x减y)和cos(x+y)
求好心人解答
求好心人解答
/>sinx+siny=√2
两边同时平方得:
sin²x+2sinxsiny+sin²y=2 ①
cosx+cosy=√3/2
两边同时平方得:
cos²x+2cosxcosy+cos²y=3/4 ②
由①+②得:
2+2(sinxsiny+cosxcosy)=2+3/4
2(sinxsiny+cosxcosy)=3/4
2cos(x-y)=3/4
cos(x-y)=3/8
sinx+siny=2sin[(x+y)/2]cos[(x-y)/2]
cosx+cosy=2cos[(x+y)/2]cos[(x-y)/2]
上面两式相除得:
tan[(x+y)/2]=2√6/3
cos(x+y)={1-tan²[(x+y)/2]}/{1+tan²[(x+y)/2]}
=[1-(2√6/3)²]/[1+(2√6/3)²]
=-5/11
答案:cos(x-y)=3/8,cos(x+y)=-5/11
两边同时平方得:
sin²x+2sinxsiny+sin²y=2 ①
cosx+cosy=√3/2
两边同时平方得:
cos²x+2cosxcosy+cos²y=3/4 ②
由①+②得:
2+2(sinxsiny+cosxcosy)=2+3/4
2(sinxsiny+cosxcosy)=3/4
2cos(x-y)=3/4
cos(x-y)=3/8
sinx+siny=2sin[(x+y)/2]cos[(x-y)/2]
cosx+cosy=2cos[(x+y)/2]cos[(x-y)/2]
上面两式相除得:
tan[(x+y)/2]=2√6/3
cos(x+y)={1-tan²[(x+y)/2]}/{1+tan²[(x+y)/2]}
=[1-(2√6/3)²]/[1+(2√6/3)²]
=-5/11
答案:cos(x-y)=3/8,cos(x+y)=-5/11
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