Question:medium

If \[ \sin\theta=\frac{3}{5}, \] then \(\cos\theta=\)

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If quadrant is not given, both positive and negative values may be possible after taking square root.
  • \(\frac{4}{5}\) but not \(-\frac{4}{5}\)
  • \(-\frac{4}{5}\) or \(\frac{4}{5}\)
  • \(-\frac{4}{5}\) but not \(\frac{4}{5}\)
  • \(\frac{3}{5}\) but not \(-\frac{3}{5}\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
This question uses the fundamental Pythagorean identity of trigonometry to find the value of cosine when sine is known. It's crucial to consider all possible quadrants where the given condition is true.
Step 2: Key Formula or Approach:
The Pythagorean identity is: \[ \sin^2\theta + \cos^2\theta = 1 \] We can rearrange this to solve for $\cos\theta$: \[ \cos^2\theta = 1 - \sin^2\theta \] \[ \cos\theta = \pm \sqrt{1 - \sin^2\theta} \] Step 3: Detailed Explanation:
We are given $\sin\theta = \frac{3}{5}$. Substitute this value into the formula: \[ \cos^2\theta = 1 - \left(\frac{3}{5}\right)^2 \] \[ \cos^2\theta = 1 - \frac{9}{25} \] \[ \cos^2\theta = \frac{25}{25} - \frac{9}{25} = \frac{16}{25} \] Now, take the square root of both sides to find $\cos\theta$: \[ \cos\theta = \pm \sqrt{\frac{16}{25}} \] \[ \cos\theta = \pm \frac{4}{5} \] The problem does not specify the quadrant in which the angle $\theta$ lies.
$\sin\theta = \frac{3}{5}$ is positive, which occurs in Quadrant I and Quadrant II.
In Quadrant I, $\cos\theta$ is positive, so $\cos\theta = \frac{4}{5}$.
In Quadrant II, $\cos\theta$ is negative, so $\cos\theta = -\frac{4}{5}$.
Since both scenarios are possible, $\cos\theta$ can be either $\frac{4}{5}$ or $-\frac{4}{5}$.
Step 4: Final Answer:
Both positive and negative values for cosine are possible. Therefore, option (B) is the correct answer.
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