F.4 A-MTH Trigonometry (Quick)

2007-12-07 3:02 am
Simplify the following:
1. (sin (270+B) sin (90-B) cot (B-270)) / (sin (-B) cos (270-B) cot (180-B))
2. (sin (兀/2+a) cos (兀/2-a)) / (cos (兀-a)) + (sin(兀-a) cos (兀/2+a)) / (sin (兀+a))
3. Given that sin θ= -4/5 and θ is in the 4th quadrant, find the values of the following :
(a) cos θ (b) cot θ (c) cos (兀/2+θ) (d) tan (3兀/2-θ)

回答 (2)

2007-12-08 3:14 pm
✔ 最佳答案
1. (sin (270+B) sin (90-B) cot (B-270)) / (sin (-B) cos (270-B) cot (180-B))
= (sin (270+B) sin (90-B) tan (180-B)) / (sin (-B) cos (270-B) tan (B-270) )
= (sin (270+B) sin (90-B) tan (180-B)) / (sin (-B) cos (B-270) tan (B-270) )
= (sin (270+B) sin (90-B) tan (180-B)) / (sin (-B) sin (B-270))
= ((- cos B) * cos B * (- tan B)) / (- sin (B) cos (B))
= - 1
2. (sin (π/2+a) cos (π/2-a)) / (cos (π-a)) + (sin(π-a) cos (π/2+a)) / (sin (π+a))
= (cos a)(sin a) / (- cos a) + (sin a)(- sin a) / (-sin a)
= - sin a + sin a
= 0
3. Given that sin θ= -4/5 and θ is in the 4th quadrant, find the values of the following :
(a) cos θ (b) cot θ (c) cos (π/2+θ) (d) tan (3π/2-θ)
(a) θ is in the 4th quadrant, 0 < cos θ < 1
cos θ = sqrt (1 - sin^2 θ)
= sqrt (1 - 16/25)
= sqrt (9/25)
= 3/5
(b)
cot θ = cos θ / sin θ
= - 3/4
(c)
cos (π/2+θ)
= - sin θ
= 4/5
(d)
tan (3π/2-θ)
= sin (3π/2-θ) / cos (3π/2-θ)
= - cos θ / - sin θ
= cot θ
= - 3/4
I hope this can help with your understanding. =)
參考: 自己
2007-12-08 8:59 pm
Trigonometry


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