Expert: Ahmed Salami - 2/10/2010

Question
Cosa=1/2, -pi/2 less than 9 less than 0

a. sin(a+b)
b. cos(a+b)
c. tan(a+b)

Answer
Hi Rosa,
In the interval -π/2 < A < 0
cosA is positive and both sinA and tanA are negative.
Also, sinA = √[1 - cos²A]
and tanA = sinA/cosA
sin(A+B) = sinA.cosB + sinB.cosA
cos(A+B) = cosA.cosB - sinA.sinB
tan(A+B) = [tanA + tanB]/[1 - tanA.tanB]

Now,
cosA = 1/2
sinA = √[1 - (1/2)²]
    = √[1 - (1/4)]
    = √(3/4)
    = -(√3)/2
tanA = [-(√3)/2]/(1/2)
    = -√3

a) sin(A+B) = sinA.cosB + sinB.cosA
           = [-(√3)/2]cosB + (1/2)sinB

b) cos(A+B) = cosA.cosB - sinA.sinB
           = (1/2)cosB - [-(√3)/2]sinB
           = (1/2)cosB + [(√3)/2]sinB


c) tan(A+B) = [tanA + tanB]/[1 - tanA.tanB]
           = [-√3 + tanB]/[1 - (-√3)tanB]
           = [-√3 + tanB]/[1 + (√3)tanB]


Note that there's a '9' in your question. I've assumed you meant to write A and you also havent mentioned anything else about 'B'.

Regards