已知asinx+bcosx=0，Asin2x+Bcos2x=C，a²+b²≠0，

asinx+bcosx=√(a²+b²)sin(x+φ) φ是什么~

asinx+bcosx=√(a²+b²)sin(x+φ) φ是什么？

解：asinx+bcosx=[√(a²+b²)]{[a/√(a²+b²)]sinx+[b/√(a²+b²)]cosx}=[√(a²+b²)]sin(x+φ)
其中，a/√(a²+b²)=cosφ，b/√(a²+b²)=sinφ，即tanφ=sinφ/cosφ=b/a；
∴φ=arctan(b/a)，-π/2<φ<π/2；a，b同号时0<φ<π/2；a，b异号时-π/2<φ<0.

asinx+bcosx==√(a²+b²)sin(x+φ)
解释过程：
令y=asinx+bcosx
=√(a²+b²)[sinx*a/√(a²+b²)+cosx*b/√(a²+b²)]
令cosφ=a/√(a²+b²)
则sinφ=√(1-cos²φ)=b/√(a²+b²)
所以原式=√(a²+b²)(sinxcosφ+cosxsinφ)
=√(a²+b²)sin(x+φ)
考察的是辅助角公式的应用。

扩展资料辅助角公式是李善兰先生提出的一种高等三角函数公式，使用代数式表达为：
；该公式的主要作用是将多个三角函数的和化成单个函数，以此来求解有关最值问题。
见“求最值、值域”问题，启用有界性，或者辅助角公式：
1、|sinx|≤1，|cosx|≤1；
2、(asinx+bcosx)2=(a2+b2)sin2(x+φ)≤(a2+b2)；
3、asinx+bcosx=c有解的充要条件是a2+b2≥c2.
参考资料：百度百科-辅助角公式

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证明：设siny=-ba2+b2
，cosy=aa2+b2

则①可写成cosysinx-sinycosx=0，
∴sin（x-y）=0∴x-y=kπ（k为整数），
∴x=y+kπ
又sin2x=sin2（y+kπ）=sin2y=2sinycosy=-2aba2+b2

cos2x=cos2y=cos2y-sin2y=a2-b2a2+b2
代入②，得-2abAa2+b2+(a2-b2)Ba2+b2=C，
∴2abA+（b2-a2）B+（a2+b2）C=0．

不懂的，可以再问我哦~

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