Question

If $\sin \theta = \frac{3}{5}$ and $\theta$ lies in the first quadrant, find $\cos \theta$.

• A. \(\frac{4}{5}\)
• B. \(\frac{3}{4}\)
• C. \(\frac{5}{3}\)
• D. \(\frac{5}{4}\)

Hint:

Tip: Always consider the quadrant of the angle to decide the sign of trigonometric values.

Answer 1

The Correct Option is A

Given that \(\sin \theta = \frac{3}{5}\) and \(\theta\) is in the first quadrant, we need to find \(\cos \theta\). In the first quadrant, both sine and cosine values are positive. We know the identity \(\sin^2 \theta + \cos^2 \theta = 1\).

Substitute \(\sin \theta = \frac{3}{5}\) into the identity:

\(\left(\frac{3}{5}\right)^2 + \cos^2 \theta = 1\)

\(\frac{9}{25} + \cos^2 \theta = 1\)

Solve for \(\cos^2 \theta\):

\(\cos^2 \theta = 1 - \frac{9}{25}\)

\(\cos^2 \theta = \frac{25}{25} - \frac{9}{25}\)

\(\cos^2 \theta = \frac{16}{25}\)

Take the positive square root (since \(\theta\) is in the first quadrant):

\(\cos \theta = \sqrt{\frac{16}{25}} = \frac{4}{5}\)

Thus, the correct answer is \(\frac{4}{5}\).

Answer 2

Step 1: Use Pythagorean identity 
Since \(\sin^2 \theta + \cos^2 \theta = 1\), \[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \]

Step 2: Find \(\cos \theta\) 
Since \(\theta\) is in the first quadrant, \(\cos \theta>0\), \[ \cos \theta = \sqrt{\frac{16}{25}} = \frac{4}{5} \]