98.6k views
5 votes
If sin(A) = -7/9, where π < A < 3π/2, find: (a) cos(A) (b) tan(A) (c) csc(A) (d) sec(A) (e) cot(A)

User GenesRus
by
7.6k points

1 Answer

7 votes
To find these trigonometric values, we can use the fact that sin(A) is given as -7/9 and the given constraint on the angle A. Let's calculate each value:

(a) cos(A):
We know that sin^2(A) + cos^2(A) = 1.
So, cos^2(A) = 1 - sin^2(A) = 1 - (-7/9)^2 = 1 - 49/81 = 32/81.

Taking the square root of both sides, cos(A) = ±√(32/81).
Since A is in the third quadrant (π < A < 3π/2), cos(A) is negative.
Therefore, cos(A) = -√(32/81) = -4/9.

(b) tan(A):
We can use the relationship tan(A) = sin(A) / cos(A).
tan(A) = (-7/9) / (-4/9) = 7/4.

(c) csc(A):
csc(A) is the reciprocal of sin(A).
csc(A) = 1 / sin(A) = 1 / (-7/9) = -9/7.

(d) sec(A):
sec(A) is the reciprocal of cos(A).
sec(A) = 1 / cos(A) = 1 / (-4/9) = -9/4.

(e) cot(A):
cot(A) is the reciprocal of tan(A).
cot(A) = 1 / tan(A) = 1 / (7/4) = 4/7.

So, the values are:
(a) cos(A) = -4/9
(b) tan(A) = 7/4
(c) csc(A) = -9/7
(d) sec(A) = -9/4
(e) cot(A) = 4/7.
User Imckl
by
7.8k points