Question

A point on the terminal side of an angle 0 in standard position is (-15, 8). Find the exact value of each of the six trigonometric functions of 0.

Answer 1

!<Answer>!

To find the exact value of each of the six trigonometric functions of angle θ, where a point on the terminal side of the angle in standard position is (-15, 8), we can use the given coordinates to calculate the necessary values.

First, we need to determine the distance r from the origin to the point (-15, 8) using the Pythagorean theorem:

r = √((-15)^2 + 8^2)

r = √(225 + 64)

r = √289

r = 17

Next, we can determine the values of the trigonometric functions using the given coordinates (-15, 8) and the calculated value of r.

1. The sine function (sin θ) is equal to the y-coordinate divided by r:

sin θ = 8/17

2. The cosine function (cos θ) is equal to the x-coordinate divided by r:

cos θ = -15/17

3. The tangent function (tan θ) is equal to the y-coordinate divided by the x-coordinate:

tan θ = 8/-15

4. The cosecant function (csc θ) is equal to the reciprocal of the sine function:

csc θ = 1/(8/17) = 17/8

5. The secant function (sec θ) is equal to the reciprocal of the cosine function:

sec θ = 1/(-15/17) = -17/15

6. The cotangent function (cot θ) is equal to the reciprocal of the tangent function:

cot θ = 1/(8/-15) = -15/8

So, the exact values of the six trigonometric functions of angle θ are:

sin θ = 8/17

cos θ = -15/17

tan θ = 8/-15 = -8/15

csc θ = 17/8

sec θ = -17/15

cot θ = -15/8

These values represent the ratios of the sides of a right triangle formed by the given angle in standard position.

Answer 2

Answer:

$\sin(\theta) = \dfrac{8}{17}$

$\cos(\theta) =- \dfrac{15}{17}$

$\tan(\theta) = -\dfrac{8}{15}$

$\csc(\theta) = \dfrac{17}{8}$

$\sec(\theta) = -\dfrac{17}{15}$

$\cot(\theta) = -\dfrac{15}{8}$

Step-by-step explanation:

To find the trigonometric functions of the angle $\theta$ in standard position given the point $(-15, 8)$, we can use the properties of right triangles and the definitions of trigonometric functions.

Given that the point $(-15, 8)$ is in the second quadrant, we can construct a right triangle with legs of length 15 (horizontal distance) and 8 (vertical distance). The hypotenuse of this triangle can be found using the Pythagorean theorem:

$c^2 = a^2 + b^2$

$c^2 = (-15)^2 + 8^2$

$c^2 = 225 + 64$

$c^2 = 289$

$c = 17$

Now, let's find the trigonometric functions:

Sine ($\sin$):

$\sin(\theta) = \dfrac{\textsf{opposite}}{\textsf{hypotenuse}} \\\\ = \dfrac{8}{17}$

Cosine ($\cos$):

$\cos(\theta) = \dfrac{\textsf{adjacent}}{\textsf{hypotenuse}} \\\\= \dfrac{-15}{17} \\\\ = -\dfrac{15}{17}$

Tangent ($\tan$):

$\tan(\theta) = \dfrac{\textsf{opposite}}{\textsf{adjacent}}\\\\ = \dfrac{8}{-15}\\\\ = -\dfrac{8}{15}$

Cosecant ($\csc$):

$\csc(\theta) = \dfrac{1}{\sin(\theta)} \\\\= \dfrac{1}{\dfrac{8}{17}} \\\\ = \dfrac{17}{8}$

Secant ($\sec$):

$\sec(\theta) = \dfrac{1}{\cos(\theta)} \\\\ = \dfrac{1}{\dfrac{-15}{17}}\\\\ = -\dfrac{17}{15}$

Cotangent ($\cot$):

$\cot(\theta) = \dfrac{1}{\tan(\theta)} \\\\= \dfrac{1}{-\dfrac{8}{15}} \\\\= -\dfrac{15}{8}$