若cos[sita+pai/4]={-7倍根号2}/26,sita属于{pai,3/2pai},则cos sita的值为

问题描述:

若cos[sita+pai/4]={-7倍根号2}/26,sita属于{pai,3/2pai},则cos sita的值为

cos(θ+π/4)=(-7√2)/26,θ∈(π,3π/2)
cosθcos(π/4)-sinθsin(π/4)=-7√2/26
(√2/2)cosθ-(√2/2)sinθ=-7√2/26
cosθ-sinθ=-7/13
又因为cos²θ+sin²θ=1
-2cosθsinθ=(cosθ-sinθ)²-(cos²θ+sin²θ)
(cosθ+sinθ)²=cos²θ+sin²θ+2cosθsinθ=2(cos²θ+sin²θ)-(cosθ-sinθ)²=2-49/169=289/169
又因为θ∈(π,3π/2)且cosθ-sinθ=-7/13