Question

macy is told that an angle has a sine of checkmark5/5 and is in quadrant 2 find cosine and tangent using identities and the quadrant

Answer 1

To find the cosine and tangent of an angle with a sine of \(\frac{5}{5}\) (which simplifies to \(1\)) in quadrant II, we can start by applying the Pythagorean identity and considering the properties of trigonometric functions in different quadrants.

1. **Finding the value of cosine:**
We know that \(\sin^2 \theta + \cos^2 \theta = 1\).

Here, we have:
\[
\sin \theta = 1
\]
So,
\(\sin^2 \theta = 1^2 = 1\).

Using the Pythagorean identity:
\[
1 + \cos^2 \theta = 1
\]
Subtracting \(1\) from both sides, we have:
\[
\cos^2 \theta = 1 - 1 = 0
\]

Taking the square root gives:
\[
\cos \theta = 0
\]

In quadrant II, the cosine function is negative. However, since \(\cos \theta = 0\), it is exactly \(0\).

2. **Finding the value of tangent:**
The tangent function is defined as:
\[
\tan \theta = \frac{\sin \theta}{\cos \theta}
\]

Substituting the values we have:
\[
\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{1}{0}
\]

Since dividing by zero is undefined, we conclude:
\[
\tan \theta \text{ is undefined.}
\]

**Summary:**
- \(\cos \theta = 0\)
- \(\tan \theta\) is undefined.

This corresponds to the angle being \(90^\circ\) (or \(\frac{\pi}{2}\) radians), which is indeed located in the second quadrant.