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4sinA+6cosA=?
人气:399 ℃ 时间:2020-05-19 13:10:53
解答
4sina+6cosa
=√52[(4/√52)sina+(6/√52)cosa]
设cosb=4/√52,且b是锐角
则(sinb)^2=1-(cosb)^2=36/52
sinb=6/√52
tanb=sinb/cosb=4/6
b=arctan(4/6)
所以4sina+6cosa
=√52(sinacosb+cosasinb)
=√52sin(a+b)
=√52*sin[a+arctan(4/6)]
总之asinx+bcosx=√(a^2+b^2)*sin[x+arctan(b/a)]
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