sin2θ+cos2θ=根号2cos(2θ-π/4)是如何得到的.要详细过程. 谢谢
3个回答
sin2@+cos2@
=根号2*(cos2@*根号2/2+sin2@*根号2/2)
=根号2*(cos2@*cosPai/4+sin2@*sinPai/4)
=根号2cos(2@-π/4)
相关问题
已知sinθ+cosθ=根号2/3,θ∈(π/2,π) 求cos4θ
已知cosθ =-(根号2)/3,θ 属于(π/2,π),求2/sin2θ -cosθ/sinθ
f(θ)=2cos^3 θ+sin^(2π-θ)+sin(π/2+θ)-3 / 2+2cos^(π+θ)+cos(-θ)
已知向量m=(cosθ,-sinθ),n=(根号2+sinθ,cosθ),θ∈(π,3π/2),且cos(θ/2+π/8
求证cos(π-θ)/cosθ[sin(3π/2-θ)-1]+cos(2π-θ)/cos(π+θ)sin(π/2+θ-s
已知sinθ+cosθ=根号2/3,θ∈(π/2,π),求sin^3θ+cos^3θ
已知tan2θ=-2根号2,π/4<θ<π/2.求(2cos²θ/2-sinθ-1)/[根号2sin(π/4+
急.设f(θ)=[2cos2θ+sin2(2π-θ)+sin(π/2+θ)-3]/[2+2cos2(π+θ)+cos(-
设f( θ)=2cos^3θ + sin^2 (2π-θ)+cos(-θ)-3 / 2+2cos^2(π+θ) + co
已知2cos 2 θ+5cosθ•sinθ-3sin 2 θ=0, θ∈( π 4 , π 2 ) ,则tanθ=___