The point, in on the terminal side of an angle in standard position find the exact value of the six trigonometric functions of the angle
a. (8,15)
b.(-9,-40)
Yes Pythagorean theorem to find the hypotenuse, then solve.
To find the exact values of the six trigonometric functions of an angle in standard position, we need to determine the values of the three sides of the right triangle related to that angle: the hypotenuse, adjacent side, and opposite side.
Let's solve for the values for each given point:
a. (8,15)
To find the hypotenuse (r) of the right triangle, use the Pythagorean theorem:
r = sqrt(8^2 + 15^2) = sqrt(64 + 225) = sqrt(289) = 17
Next, to find the angle, we can use the ratio of the sides. Since we have the x-coordinate and y-coordinate of the point:
sin(angle) = opposite/hypotenuse = 15/17
cos(angle) = adjacent/hypotenuse = 8/17
tan(angle) = opposite/adjacent = 15/8
Now, let's find the reciprocal functions:
cosec(angle) = 1/sin(angle) = 17/15
sec(angle) = 1/cos(angle) = 17/8
cot(angle) = 1/tan(angle) = 8/15
Therefore, for point (8,15):
sin(angle) = 15/17
cos(angle) = 8/17
tan(angle) = 15/8
cosec(angle) = 17/15
sec(angle) = 17/8
cot(angle) = 8/15
b. (-9,-40)
Similarly, let's find the values for the given point:
To find the hypotenuse (r) of the right triangle:
r = sqrt((-9)^2 + (-40)^2) = sqrt(81 + 1600) = sqrt(1681) = 41
Next, find the angle:
sin(angle) = opposite/hypotenuse = -40/41
cos(angle) = adjacent/hypotenuse = -9/41
tan(angle) = opposite/adjacent = -40/-9 = 40/9
Reciprocal functions:
cosec(angle) = 1/sin(angle) = -41/40
sec(angle) = 1/cos(angle) = -41/9
cot(angle) = 1/tan(angle) = 9/40
Hence, for point (-9,-40):
sin(angle) = -40/41
cos(angle) = -9/41
tan(angle) = 40/9
cosec(angle) = -41/40
sec(angle) = -41/9
cot(angle) = 9/40