Question

已知tanα,1/tanα是关于x的方程3x^2-3kx+3k^2-13=0的两实根，且3π<α<7π/2.求cos(3π+α)+sin(π+α)

详细过程

Answer 1

【1】韦达定理
tanα + 1/tanα = k
tanα * 1/tanα = (3k^2-13)/3

【2】
tanα + 1/tanα = k = 4 / (3^1/2)
cos(3π+α)+sin(π+α) = -(sin α + cos α) = - 2^(1/2 ) sin ( α+ π/4)
由于3π<α<7π/2, 故sin α + cos α < 0

tanα + 1/tanα = [(sin x)^2 + (cos x)^2] / (sin x * cos x) = 1 / (sin x * cos x) = k
故 sinx * cosx = 1/k
sin α + cos α = - ((sin x)^2 + (cos x)^2 + 2 * sinx * cosx)^(1/2)
= - (1 + 2/k)^(1/2)