F4 M2 Trigonometry

1. If Cos ß and Csc ßare the roots of x^2 + kx －3 = 0，where π/2 < ß <π，find the value of k . Express your answer in surd form.
SolCos ß*Csc ß=Cot ß=－3π/2 < ß < πCsc ß=√10，Cos ß=-3/√10Cos ß+Csc ß=－kk=－Cos ß－Csc ß=3/√10－√10=3√10/10－10√10/10=－7√10/10

2. If SinA +SinB = k/2，SinASinB=－k^2/ 2
a) Find SinA and SinB in terms of k.
Sol
SinB=k/2－SinA
SinASinB=SinA(k/2－SinA)=kSinA/2－Sin^ 2 A =－k^2/2
－2Sin^ 2 A +kSinA+k^2=0
2Sin^ 2 A－kSinA－k^2=0
(2SinA+k)(SinA－k)=0
SinA=－k/2 or SinA=k
(1) SinA=－k/2，SinB=k
(2) SinA=k，SinB=－k/2

b)Find the range of k so that both A and B are real.
Sina=-k/2
－1<=k/2<=1
－2<=k<=2---------------(1)
SinB=k
－1<=k<=1---------------(2)
綜合(10(2)
－1<=x<=1

3. f(ß) = (Cosß)^2 +Sinß－2.Find the ranges of values of f(ß)
Sol
(Cosß)^2 +Sinß－2
=1－Sin^2 ß+Sinß－2
=－(Sin^2 ß－Sinß)－1
=－(Sin^2 ß－Sinß+1/4)－1+1/4
=－(Sin ß－1/2)^2－3/4
－1<=Sin ß<=1
－3/2<=Sin ß－1/2<=1/2
0<=(Sin ß－1/2)^2<=9/4
－9/4<=－(Sin ß－1/2)^2<=0
－3<=－(Sin ß－1/2)^2－3/4<=－3/4
－3<=f( ß)<=－3/4

第一題Ans係咪-7√10 / /30 ?!