F.4 A-MTH Trigonometry (Quick)

1. If 2 tan^2 θ + 3 sec^2 θ =18 and π/2 < θ < π, find the value of cos θ.
Since π/2 < θ < π, cos θ < 0
2 tan^2 θ + 3 sec^2 θ =18
2 (1 + tan^2 θ) + 3 sec^2 θ = 20
2 sec^2 θ + 3 sec^2 θ = 20
sec^2 θ = 4
cos^2 θ = 1/4
cos θ = - 1/2 or 1/2 (rejected)
cos θ = - 1/2

2. If P (1+ √2, 1-√2) is a point on the terminal side of an angle a, find the values of sin a, cos a and tan a.
sin a = (√2 - 1)/√6
cos a = (1 + √2)/√6
tan a = (√2 - 1)/(1 + √2)

3.Given that cos a = -1/2, cot b =1 where π/2 < a < π and π < b < 3π/2, find the value of (2 sin b-sec a) / (2 cos a - csc b).
cos a = -1/2, cot b =1
sec a = -2, csc b = sqrt (1 + 1^2) = -√2
sin b = - 1 / √2
(2 sin b-sec a) / (2 cos a - csc b)
= ((2 * - 1 / √2) - (-2)) / ((2 * -1/2) - (-√2))
= (- √2 + 2) / (- 1 + √2)
= √2

4. If sin a = (t-1) / (t+1) and π/2 < a < π, find the values of cos a and cot a.
sin a = (t-1) / (t+1)
cos a = - sqrt (1 - sin^2 a)
cos a = - sqrt (1 - (t-1)^2 / (t+1)^2)
cos a = - sqrt (((t+1)^2 - (t-1)^2) / (t+1)^2))
cos a = - sqrt (4t / (t+1)^2))
cos a = - 2 √t / (t+1)
cot a = cos a / sin a
= (- 2 √t / (t+1)) / ((t-1) / (t+1))
cot a = - 2 √t / (t-1)
I hope this can help with your understanding. =)

Trigonometry