(-3/5, 2) is a point on the terminal side of theta, find the value of the six trig functions
r^2 = (-3/5)^2 + 2^2 = 109/25
r = √109/5
since (3/5,2) is in quadrant II
sinØ = 2/(√109/5) = 10/√109
cscØ = √109/10
cosØ = (-3/5) /(√109/5) = -3/√109
secØ = -√109/3
tanØ = 2/(-3/5) = -10/3
cotØ = -3/10
btw Ø = appr 106.7°
To find the values of the six trigonometric functions, we need to first determine the values of the opposite, adjacent, and hypotenuse sides of the right triangle associated with the given point (-3/5, 2) on the terminal side of theta.
Let's label the coordinates of the given point as (x, y) = (-3/5, 2). The x-coordinate represents the adjacent side, and the y-coordinate represents the opposite side of the right triangle.
We can calculate the values of the adjacent side (a) and the opposite side (b) using the Pythagorean Theorem:
a^2 + b^2 = c^2
Where c is the hypotenuse.
In this case, we have:
(-3/5)^2 + 2^2 = c^2
9/25 + 4 = c^2
9/25 + 100/25 = c^2
109/25 = c^2
Taking the square root of both sides, we get:
c = √(109/25)
c = √109/5
Now that we know the values of the opposite (b = 2), adjacent (a = -3/5), and hypotenuse (c = √109/5) sides of the right triangle, we can calculate the trigonometric functions.
The six trigonometric functions are:
1. Sine (sin) = opposite/hypotenuse = b/c
sin(theta) = 2/(√109/5) = 10√109/109
2. Cosine (cos) = adjacent/hypotenuse = a/c
cos(theta) = (-3/5)/(√109/5) = -3√109/109
3. Tangent (tan) = opposite/adjacent = b/a
tan(theta) = 2/(-3/5) = -10/3
4. Cosecant (csc) = 1/sine = 1/(opposite/hypotenuse) = c/b
csc(theta) = (√109/5)/2 = √109/10
5. Secant (sec) = 1/cosine = 1/(adjacent/hypotenuse) = c/a
sec(theta) = (√109/5)/(-3/5) = -√109/3
6. Cotangent (cot) = 1/tangent = 1/(opposite/adjacent) = a/b
cot(theta) = (-3/5)/2 = -3/10
Therefore, the values of the six trigonometric functions for the point (-3/5, 2) on the terminal side of theta are:
sin(theta) = 10√109/109
cos(theta) = -3√109/109
tan(theta) = -10/3
csc(theta) = √109/10
sec(theta) = -√109/3
cot(theta) = -3/10