三角函数中经常遇到求形如“y=asinx+bcosx+cdsinx+ecosx+f”型函数值域,对这一类分式型三角函数值域,从不同思维层次思考的求解方法不同,下面举一例说明其解法.题目:求函数f(x)=1+sinx2+cosx的值域.1.利用辅助角公式求解由y=1+sinx2+cosx变形为ycosx-sinx=1-2y可得y2+1cos(x+φ)=1-2y,其中φ由tanφ=-1y2+1确定.因为|cos(x+φ)|≤1,所以|1-2y|≤ Trigonometric functions are often encountered in the shape of such as “y = asinx + bcosx + cdsinx + ecosx + f ” type function range, for this type of trigonometric function range, from different levels of thinking of different methods of solving , The following gives an example of its solution. Title: Find the value of the function f (x) = 1 + sinx2 + cosx .1. Use the auxiliary angle formula to solve the y = 1 + sinx2 + cosx deformation ycosx-sinx = We have y2 + 1cos (x + φ) = 1-2y, where φ is determined by tanφ = -1y2 + 1. Since | cos (x + φ) | ≤ 1, | 1-2y | ≤