因为这里书写不便,故将我的答案做成图像贴于下方,谨供楼主参考(若图像显示过小,点击图片可放大) 感
解;1) 向量a=(cosx+sinx,√2cosx),b(cosx-sinx,√2sinx),
f(x)=4cosx(sinxcosπ/6+cosxsinπ/6)+1 =2√3sinxcosx+2
f(x)=cosxsin(x+π/6)-cos2x-1/4, =cosx(√3/2sinx+1/2
f(x)=(cosx)^4-2sinxcosx-(sinx)^4 =[(cosx)^2+
f(x)=cosxsin2x=2sinxcos2x=2sinx(1-sin2x)=2sinx-2si
f(x)=3sin^2x+2√3sinxcosx+5cos^2x =sin^2 x+2√3sin
解: (1) f(x)=2sinxcosx+2cos^2x =sin2x+cos2x +1
看来你是看到我以前的回答了。那就直接复制一下吧。 解答: (1)sinx≠0, x≠kπ,k∈
(1) f(x)=4cosxsin(x- π/3)+a =2[sin(x+x-π/3)-sin(