Problem 105 - Precalculus with Limits

Question

In Exercises \(105-108,\) use the trigonometric substitution to write the algebraic equation as a trigonometric equation of \(\theta\) where \(-\pi / 2<\theta<\pi / 2 .\) Then find \(\sin \theta\) and \(\cos \theta\). $$ 3=\sqrt{9-x^{2}}, \quad x=3 \sin \theta $$

Step-by-Step Solution

Answer

\(\sin \theta = 1\) and \(\cos \theta = 0\)

Step 1: Substitution

Substitute \( x = 3 \sin \theta \) into the equation to get \( 3 = \sqrt{9 - (3 \sin \theta)^2} \). After this, simplify the equation.

Step 2: Simplify the Equation

Applying the square rule to the right side of the equation, it becomes \( 3 = \sqrt{9 - 9 \sin^2 \theta} \). Then we can rewrite the right side of the equation using the Pythagorean Trigonometric Identity \( \cos^2 \theta = 1 - \sin^2 \theta \) to get \( 3 = \sqrt{9 - 9 \cos^2 \theta} \). Dividing throughout by 3, we obtain \( 1 = \sqrt{1 - \cos^2 \theta} \), since \( 1 = \sin \theta \) in this case, we can now find \( \cos \theta \) by using the Pythagorean Trigonometric Identity \( \sin^2 \theta + \cos^2 \theta = 1 \).

Step 3: Find the Value of Cosine

Since by Pythagorean trigonometric identity, we know that \( \sin^2 \theta + \cos^2 \theta = 1 \), we can rearrange it to \( \cos^2 \theta = 1 - \sin^2 \theta \). Hence, \( \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 -1} = 0 \).