Question

急急急~！！！三角恒等变换的题目

（1）已知sin(x-0.75π)cos(x-0.25π)=-1/4, 则cos4x的值为多少

（2）求证1/sin2θ+1/tan2θ+1/sinθ=1/tan0.5θ

Answer 1

（1）sin(x-0.75π)cos(x-0.25π)=(-√2/2sinx-√2/2cosx)(√2/2sinx+√2/2cosx)=-sinxcosx-1/2=
-1/4，所以sinxcosx=1/2sin2x=-1/4,sin2x=-1/2,所以cos4x=1-2(sin2x)^2=1/2

（2）1/sin2θ+1/tan2θ+1/sinθ=(1+cos2θ)/sin2θ+1/sinθ=2(cosθ)^2/2sinθcosθ+1/sinθ=
1/tanθ+1/sinθ=(1+cosθ)/sinθ=2[cos(θ/2)]^2/2sin(θ/2)cos(θ/2)=1/tan0.5θ

回答完毕~望采纳哦