Question

Given that \(\sin \theta = a/b\), then \(\cos \theta\) is equal to :

• A. \(\frac{\sqrt{b^2 - a^2}}{b}\)
• B. \(b/a\)
• C. \(a/b\)
• D. \(\frac{\sqrt{b^2 - a^2}}{a}\)

Hint:

Using the "SOH CAH TOA" mnemonic: \(\sin\) is Opposite/Hypotenuse (\(a/b\)). By Pythagoras, the Adjacent side is \(\sqrt{b^2 - a^2}\). \(\cos\) is Adjacent/Hypotenuse.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Trigonometric ratios are related via the Pythagorean identity. In a right-angled triangle, if \(\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}\), we can find the adjacent side using Pythagoras' theorem.
Step 2: Key Formula or Approach:
Identity: \( \sin^2 \theta + \cos^2 \theta = 1 \implies \cos \theta = \sqrt{1 - \sin^2 \theta} \)
Step 3: Detailed Explanation:
1. Substitute \(\sin \theta = \frac{a}{b}\) into the identity: \[ \cos \theta = \sqrt{1 - \left(\frac{a}{b}\right)^2} \] 2. Simplify the expression: \[ \cos \theta = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{\frac{b^2 - a^2}{b^2}} \] 3. Extract the denominator from the square root: \[ \cos \theta = \frac{\sqrt{b^2 - a^2}}{b} \]
Step 4: Final Answer:
\(\cos \theta = \frac{\sqrt{b^2 - a^2}}{b}\).