Math Problem Statement

Given cot θ = 7/24 in QIII, find the value of the remaining trigonometric functions.

Solution

The image shows a trigonometric problem where you're given cotθ=724\cot \theta = \frac{7}{24} in the third quadrant (QIII), and you're asked to find the values of the other trigonometric functions.

Step-by-step solution:

1. Use the identity for cotangent:

cotθ=adjacentopposite\cot \theta = \frac{\text{adjacent}}{\text{opposite}} Thus, the adjacent side is 7 and the opposite side is 24.

2. Find the hypotenuse using the Pythagorean theorem:

hypotenuse=adjacent2+opposite2=72+242=49+576=625=25\text{hypotenuse} = \sqrt{\text{adjacent}^2 + \text{opposite}^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25

3. Determine the signs of the trigonometric functions:

In quadrant III:

  • Sine (sinθ\sin \theta) and cosine (cosθ\cos \theta) are both negative.
  • Tangent (tanθ\tan \theta) and cotangent (cotθ\cot \theta) are positive.

4. Find the trigonometric functions:

  • Sine: sinθ=oppositehypotenuse=2425\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{24}{25} Since sinθ\sin \theta is negative in QIII: sinθ=2425\sin \theta = -\frac{24}{25}

  • Cosine: cosθ=adjacenthypotenuse=725\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{7}{25} Since cosθ\cos \theta is negative in QIII: cosθ=725\cos \theta = -\frac{7}{25}

  • Tangent: tanθ=oppositeadjacent=247\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{24}{7} Since tanθ\tan \theta is positive in QIII: tanθ=247\tan \theta = \frac{24}{7}

  • Secant (secθ\sec \theta): secθ=1cosθ=1725=257\sec \theta = \frac{1}{\cos \theta} = \frac{1}{-\frac{7}{25}} = -\frac{25}{7}

  • Cosecant (cscθ\csc \theta): cscθ=1sinθ=12425=2524\csc \theta = \frac{1}{\sin \theta} = \frac{1}{-\frac{24}{25}} = -\frac{25}{24}

  • Cotangent: cotθ=adjacentopposite=724\cot \theta = \frac{\text{adjacent}}{\text{opposite}} = \frac{7}{24}

Summary:

  • sinθ=2425\sin \theta = -\frac{24}{25}
  • cosθ=725\cos \theta = -\frac{7}{25}
  • tanθ=247\tan \theta = \frac{24}{7}
  • secθ=257\sec \theta = -\frac{25}{7}
  • cscθ=2524\csc \theta = -\frac{25}{24}
  • cotθ=724\cot \theta = \frac{7}{24} (already given)

Would you like further details on any step or clarification?

Related Questions:

  1. What is the significance of the signs of trigonometric functions in different quadrants?
  2. How does the Pythagorean theorem apply to trigonometric identities?
  3. Can you find the values of trigonometric functions if only sine or cosine is given?
  4. How would these trigonometric functions change if the angle was in the first quadrant instead of the third?
  5. How do you determine the signs of secant and cosecant based on cosine and sine?

Tip:

Remember the acronym ASTC to recall the signs of trigonometric functions in each quadrant:

  • A (All positive in QI)
  • S (Sine positive in QII)
  • T (Tangent positive in QIII)
  • C (Cosine positive in QIV).

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Trigonometric Functions
Quadrants
Pythagorean Theorem

Formulas

cot θ = adjacent / opposite
Pythagorean Theorem: hypotenuse = √(adjacent^2 + opposite^2)
sin θ = opposite / hypotenuse
cos θ = adjacent / hypotenuse
tan θ = opposite / adjacent
sec θ = 1 / cos θ
csc θ = 1 / sin θ
cot θ = adjacent / opposite

Theorems

Pythagorean Theorem

Suitable Grade Level

Grades 10-12