Question
macy is told that an angle has a sine of checkmark symbol5/5 and is in quadrant 2 find cosine and tangent using identities and the quadrant
Answers
GPT-4o mini
To find the cosine and tangent of an angle given that the sine of the angle is \( \frac{5}{5} = 1 \) and that the angle is in the second quadrant, we first need to note that there’s an inconsistency.
1. **Understanding Sine Value**: The sine function obtains values in the range of \(-1\) to \(1\). If the sine of an angle is \(1\), this occurs at \(90^\circ\) or \(\frac{\pi}{2}\) radians, which is not within the options of quadrants since \(90^\circ\) is not part of any quadrant (it is a boundary).
2. **Clarifying Given Information**: If we consider that the sine value is meant to be interpreted differently (for example, if you would like to represent the sine value as \( \frac{5}{\sqrt{25}} = \frac{1}{5} \) which actually makes more sense), we need to use that value for further trigonometric calculations.
Assuming you meant:
- Sine, \( \sin(\theta) = \frac{1}{5} \)
Now we can find the cosine and tangent using trigonometric identities.
### Step 1: Find Cosine using Pythagorean Identity
The Pythagorean Identity states:
\[
\sin^2(\theta) + \cos^2(\theta) = 1
\]
Thus,
\[
\left(\frac{1}{5}\right)^2 + \cos^2(\theta) = 1
\]
Calculating \( \left(\frac{1}{5}\right)^2 \):
\[
\frac{1}{25} + \cos^2(\theta) = 1
\]
Subtracting \( \frac{1}{25} \) from both sides gives:
\[
\cos^2(\theta) = 1 - \frac{1}{25} = \frac{25}{25} - \frac{1}{25} = \frac{24}{25}
\]
To find \(\cos(\theta)\):
\[
\cos(\theta) = \pm \sqrt{\frac{24}{25}} = \pm \frac{\sqrt{24}}{5} = \pm \frac{2\sqrt{6}}{5}
\]
Since \( \theta \) is in the second quadrant, where cosine is negative:
\[
\cos(\theta) = -\frac{2\sqrt{6}}{5}
\]
### Step 2: Find Tangent
Using the identity for tangent:
\[
\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}
\]
Substituting:
\[
\tan(\theta) = \frac{\frac{1}{5}}{-\frac{2\sqrt{6}}{5}} = -\frac{1}{2\sqrt{6}}
\]
To rationalize the denominator:
\[
\tan(\theta) = -\frac{1}{2\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = -\frac{\sqrt{6}}{12}
\]
### Summary
So, in conclusion we have:
- \( \cos(\theta) = -\frac{2\sqrt{6}}{5} \)
- \( \tan(\theta) = -\frac{\sqrt{6}}{12} \)
Please let me know if the sine value you provided should be interpreted differently!
1. **Understanding Sine Value**: The sine function obtains values in the range of \(-1\) to \(1\). If the sine of an angle is \(1\), this occurs at \(90^\circ\) or \(\frac{\pi}{2}\) radians, which is not within the options of quadrants since \(90^\circ\) is not part of any quadrant (it is a boundary).
2. **Clarifying Given Information**: If we consider that the sine value is meant to be interpreted differently (for example, if you would like to represent the sine value as \( \frac{5}{\sqrt{25}} = \frac{1}{5} \) which actually makes more sense), we need to use that value for further trigonometric calculations.
Assuming you meant:
- Sine, \( \sin(\theta) = \frac{1}{5} \)
Now we can find the cosine and tangent using trigonometric identities.
### Step 1: Find Cosine using Pythagorean Identity
The Pythagorean Identity states:
\[
\sin^2(\theta) + \cos^2(\theta) = 1
\]
Thus,
\[
\left(\frac{1}{5}\right)^2 + \cos^2(\theta) = 1
\]
Calculating \( \left(\frac{1}{5}\right)^2 \):
\[
\frac{1}{25} + \cos^2(\theta) = 1
\]
Subtracting \( \frac{1}{25} \) from both sides gives:
\[
\cos^2(\theta) = 1 - \frac{1}{25} = \frac{25}{25} - \frac{1}{25} = \frac{24}{25}
\]
To find \(\cos(\theta)\):
\[
\cos(\theta) = \pm \sqrt{\frac{24}{25}} = \pm \frac{\sqrt{24}}{5} = \pm \frac{2\sqrt{6}}{5}
\]
Since \( \theta \) is in the second quadrant, where cosine is negative:
\[
\cos(\theta) = -\frac{2\sqrt{6}}{5}
\]
### Step 2: Find Tangent
Using the identity for tangent:
\[
\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}
\]
Substituting:
\[
\tan(\theta) = \frac{\frac{1}{5}}{-\frac{2\sqrt{6}}{5}} = -\frac{1}{2\sqrt{6}}
\]
To rationalize the denominator:
\[
\tan(\theta) = -\frac{1}{2\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = -\frac{\sqrt{6}}{12}
\]
### Summary
So, in conclusion we have:
- \( \cos(\theta) = -\frac{2\sqrt{6}}{5} \)
- \( \tan(\theta) = -\frac{\sqrt{6}}{12} \)
Please let me know if the sine value you provided should be interpreted differently!