Question

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

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

Hint:

Alternatively, think of a right triangle where opposite side = \(a\) and hypotenuse = \(b\).
By Pythagoras theorem, adjacent side = \(\sqrt{\text{hypotenuse}^{2} - \text{opposite}^{2}} = \sqrt{b^{2} - a^{2}}\).
Thus, \(\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{\sqrt{b^{2} - a^{2}}}{b}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Trigonometric ratios are related by the identity \(\sin^{2} \theta + \cos^{2} \theta = 1\).
Step 2: Key Formula or Approach:
\[ \cos \theta = \sqrt{1 - \sin^{2} \theta} \]
Step 3: Detailed Explanation:
Given \(\sin \theta = \frac{a}{b}\).
Substitute this into the identity:
\[ \cos^{2} \theta = 1 - \left( \frac{a}{b} \right)^{2} \]
\[ \cos^{2} \theta = 1 - \frac{a^{2}}{b^{2}} \]
Taking the LCM:
\[ \cos^{2} \theta = \frac{b^{2} - a^{2}}{b^{2}} \]
Taking the square root:
\[ \cos \theta = \sqrt{\frac{b^{2} - a^{2}}{b^{2}}} = \frac{\sqrt{b^{2} - a^{2}}}{b} \]
Step 4: Final Answer:
\(\cos \theta = \frac{\sqrt{b^{2} - a^{2}}}{b}\).