问题描述:
如sinθ+2cosθ=0,则有:tanθ=-2.
将(cos2θ-sinθ)/(1+cos²θ )变为tanθ的函数,
(cos2θ-sinθ)/(1+cos²θ )=(cos²θ-sin²θ-sinθ)/(1+cos²θ )
=(1-tan²θ-sinθ/cos²θ)/[(1/cos²θ)+1]
=(1-tan²θ-tanθsecθ)/[sec²θ+1]
=[1-tan²θ-tanθ√(sec²θ)]/[1+tan²θ+1]
=[1-(-2)²+2√(1+tan²θ)]/[2+(-2)²]
=(-3+2√5)/6=-1/2+√5/3
将(cos2θ-sinθ)/(1+cos²θ )变为tanθ的函数,
(cos2θ-sinθ)/(1+cos²θ )=(cos²θ-sin²θ-sinθ)/(1+cos²θ )
=(1-tan²θ-sinθ/cos²θ)/[(1/cos²θ)+1]
=(1-tan²θ-tanθsecθ)/[sec²θ+1]
=[1-tan²θ-tanθ√(sec²θ)]/[1+tan²θ+1]
=[1-(-2)²+2√(1+tan²θ)]/[2+(-2)²]
=(-3+2√5)/6=-1/2+√5/3
问题解答:
我来补答展开全文阅读