Question

TANθ、TAN（π/2-θ）是2X2-2KX=7-K2的两实根，且π〈θ〈5π/4，求COSθ-SINθ的值

Answer 1

解：TAN（π/2-θ）= COTθ
由韦达定理，有
TANθ + COTθ = K
TANθ * COTθ = (K2-7)/2
由判别式>0，有 K2 < 14
又TANθ * COTθ = 1
可得 (K2-7)/2 = 1
则 K = 3 或 K = -3
又因为π〈θ〈5π/4
所以 K = TANθ + COTθ > 0
K = 3
所以TANθ + COTθ = 3
TANθ * COTθ = 1
且π〈θ〈5π/4
解得 TANθ = (3-√5)/2
COTθ = (3+√5)/2
由万能公式 SIN2θ = 2/3
(COSθ-SINθ)^2 = 1-SIN2θ = 1/3
又π〈θ〈5π/4 有 COSθ-SINθ < 0
所以COSθ-SINθ = -√3/3

Answer 2

TAN（π/2-θ）=COTθ,
TANθ、COTθ是2X²‎-2KX+K²‎-7=0的两实根,
由根与系数关系得，TANθ+COTθ=2K/2=K, TANθCOTθ=TANθ(1/TANθ)=(K²‎-7)/2=1,
即K²‎=9，K=±3
π<θ<5π/4,TANθ>0,COTθ=1/TANθ>0,TANθ+COTθ>0,
由TANθ+COTθ=K,得,K>0,K=3,
TANθ+COTθ=SINθ/COSθ+COSθ/SINθ=(SIN²‎θ+COS²‎θ)/(COSθSINθ)=1/(COSθSINθ)=3,
COSθSINθ=1/3,
(COSθ-SINθ)²=COSθ²+SINθ²-2COSθSINθ=1-2/3=1/3
π<θ<5π/4,SINθ<0,COSθ<0,O<TANθ=SINθ/COSθ<1,COSθ<SINθ,COSθ-SINθ<0,
COSθ-SINθ=-√(1/3)=-√3/3

Answer 3

123