已知tanθ=1,求sinθcosθ\sin^2θ — 2cos^2θ,和 1\sin^2θ — sinθcosθ — cos^2θ

来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/26 21:42:00
已知tanθ=1,求sinθcosθ\sin^2θ—2cos^2θ,和1\sin^2θ—sinθcosθ—cos^2θ已知tanθ=1,求sinθcosθ\sin^2θ—2cos^2θ,和1\sin^

已知tanθ=1,求sinθcosθ\sin^2θ — 2cos^2θ,和 1\sin^2θ — sinθcosθ — cos^2θ
已知tanθ=1,求sinθcosθ\sin^2θ — 2cos^2θ,和 1\sin^2θ — sinθcosθ — cos^2θ

已知tanθ=1,求sinθcosθ\sin^2θ — 2cos^2θ,和 1\sin^2θ — sinθcosθ — cos^2θ
tanθ=1 --> θ=k*pi+pi/4,2θ=2k*pi+pi/2
1.sinθ * cosθ=sin(2θ)/2 = 1/2; sin^2θ = ( 1 - cos2θ ) /2 = 1/2; cos^2θ = ( 1 + cos2θ)/2 = 1/2
所以sinθcosθ\sin^2θ — 2cos^2θ=1/2 / ( 1/2 - 2 * 1/2) = -1
2.sinθ * cosθ=sin(2θ)/2 = 1/2; sin^2θ = ( 1 - cos2θ ) /2 = 1/2; cos^2θ = ( 1 + cos2θ)/2 = 1/2
所以分母=1/2 - 1/2 - 1/2; 分子为1,最后结果为-2