Math Problem Statement

The point given below is on the terminal side of an angle θ. Find the exact value of each of the six trigonometric functions of θ. (-10, 24)

Solution

Let's solve the problem by determining the six trigonometric functions of the given angle θ\theta, based on the point (−10,24)(-10, 24) on the terminal side of the angle.

Step 1: Determine the radius (hypotenuse) rr

The radius rr is given by the Pythagorean theorem: r=x2+y2r = \sqrt{x^2 + y^2} Substituting x=−10x = -10 and y=24y = 24: r=(−10)2+242=100+576=676=26r = \sqrt{(-10)^2 + 24^2} = \sqrt{100 + 576} = \sqrt{676} = 26 So, r=26r = 26.

Step 2: Find the six trigonometric functions

The six trigonometric functions are based on the definitions:

  1. sin⁡θ=yr\sin \theta = \frac{y}{r}
  2. cos⁡θ=xr\cos \theta = \frac{x}{r}
  3. tan⁡θ=yx\tan \theta = \frac{y}{x}
  4. csc⁡θ=ry\csc \theta = \frac{r}{y} (reciprocal of sine)
  5. sec⁡θ=rx\sec \theta = \frac{r}{x} (reciprocal of cosine)
  6. cot⁡θ=xy\cot \theta = \frac{x}{y} (reciprocal of tangent)

1. sin⁡θ\sin \theta:

sin⁡θ=2426=1213\sin \theta = \frac{24}{26} = \frac{12}{13}

2. cos⁡θ\cos \theta:

cos⁡θ=−1026=−513\cos \theta = \frac{-10}{26} = \frac{-5}{13}

3. tan⁡θ\tan \theta:

tan⁡θ=24−10=−125\tan \theta = \frac{24}{-10} = -\frac{12}{5}

4. csc⁡θ\csc \theta (reciprocal of sin⁡θ\sin \theta):

csc⁡θ=2624=1312\csc \theta = \frac{26}{24} = \frac{13}{12}

5. sec⁡θ\sec \theta (reciprocal of cos⁡θ\cos \theta):

sec⁡θ=26−10=−135\sec \theta = \frac{26}{-10} = -\frac{13}{5}

6. cot⁡θ\cot \theta (reciprocal of tan⁡θ\tan \theta):

cot⁡θ=−1024=−512\cot \theta = \frac{-10}{24} = -\frac{5}{12}

Final Answers:

  1. sin⁡θ=1213\sin \theta = \frac{12}{13}
  2. cos⁡θ=−513\cos \theta = \frac{-5}{13}
  3. tan⁡θ=−125\tan \theta = -\frac{12}{5}
  4. csc⁡θ=1312\csc \theta = \frac{13}{12}
  5. sec⁡θ=−135\sec \theta = -\frac{13}{5}
  6. cot⁡θ=−512\cot \theta = -\frac{5}{12}

Would you like further details on any of these steps or have any questions?

Related Questions:

  1. What quadrant is the angle θ\theta located in, based on the coordinates (−10,24)(-10, 24)?
  2. How would the values of the trigonometric functions change if the point were in a different quadrant?
  3. Can you use the same approach for an angle in standard position with different coordinates?
  4. What are the reference angles for the given values of sin⁡θ\sin \theta and cos⁡θ\cos \theta?
  5. How do the signs of trigonometric functions differ in different quadrants?

Tip:

Always remember that the signs of the trigonometric functions depend on the quadrant in which the terminal side of the angle lies.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Pythagorean Theorem

Formulas

sin(θ) = y/r
cos(θ) = x/r
tan(θ) = y/x
csc(θ) = r/y
sec(θ) = r/x
cot(θ) = x/y

Theorems

Trigonometric identities

Suitable Grade Level

Grades 10-12

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