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第四题详解!
2个回答
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(asin2+bcos2)/(acos2-bsin2)=(a*tan2+b)/(a-b*tan2)=(tan2+(b/a))/(1-(b/a)tan2)
而:tanβ=tan(2+(6/π))=(tan2+tan(6/π))/(1-tan2*tan(6/π))
所以,显然:b/a=tan(6/π)
b/a=根号3/3
而:tanβ=tan(2+(6/π))=(tan2+tan(6/π))/(1-tan2*tan(6/π))
所以,显然:b/a=tan(6/π)
b/a=根号3/3
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