唔識依題咁難的數,可以詳解嗎

a)sinB + cosB = - a

a = - (sinB + cosB)

b = sinB cosB
b)
sinB + cosB= 1sinB + 1cosB= √(1²+1²) sin(B + ψ)= √2 sin(B + ψ)Since sin(B + ψ) ≤ 1 , When sin(B + ψ) = 1 , The max. value of sinB + cosB = √2a = - √2
b
= sinB cosB
= (sin²B + cos²B + 2sinB cosB - 1)/2
= ((sinB + cosB)² - 1)/2
= (√2² - 1)/2
= 1/2The required equation is x² + ax + b = 0
i.e.
x² - √2x + 1/2 = 0
Alternatively :
Let sinB + cosB = y ,sin²B + cos²B + 2sinB cosB = y² 1 + sin2B = y²y² - 1 = sin2B ≤ 1y² ≤ 2The Max. value of y = sinB + cosB = √2, soa = - √2remains are all same to the above .......


2011-06-05 17:59:24 補充：
點解會有√(1²+1²)

這是A.maths 課程有提及輔助角公式 :

a sinθ + b cosθ = √(a²+b²) sin( θ + ψ) , (0 < ψ < 2π)

ψ 是 sinψ = b / √(a²+b²) 及 cosψ = a / √(a²+b²) 的公解。

sinθ + cosθ 是 a = b = 1 的情況 ,
故有 sinθ + cosθ = √(1²+1²) sin(θ + ψ) = √2 sin(θ + ψ)

2011-06-05 18:00:47 補充：
ψ 是 sinψ = b / √(a²+b²) = 1 / √2 及 cosψ = a / √(a²+b²) = 1 / √2 的公解 ,

==>

ψ = (π/4 , 3π/4) 及 ψ = (π/4 , 7π/4) 的公解 , 即 ψ = π/4。
故sinθ + cosθ = √(1²+1²) sin(θ + π/4)

在本題中不必求出 ψ = π/4 , 所以沒求出。