Find the exact function value.\
The problem is to find the exact value oftan(5π6)\tan\left(\frac{5\pi}{6}\right)tan(65π)
✅ Correct Answer:
tan(5π6)=−13\tan\left(\frac{5\pi}{6}\right) = -\frac{1}{\sqrt{3}}tan(65π)=−31
✍️ Explanation
To find the exact value of tan(5π6)\tan\left(\frac{5\pi}{6}\right)tan(65π), we begin by identifying which quadrant this angle lies in and using a reference angle.
Step 1: Understand the angle
5π6 radians=150∘\frac{5\pi}{6} \text{ radians} = 150^\circ65π radians=150∘
This angle lies in Quadrant II (between 90∘90^\circ90∘ and 180∘180^\circ180∘).
Step 2: Find the reference angle
To find the reference angle in Quadrant II, subtract the angle from π\piπ (or 180°):Reference angle=π−5π6=π6\text{Reference angle} = \pi – \frac{5\pi}{6} = \frac{\pi}{6}Reference angle=π−65π=6π
So, the reference angle is π6\frac{\pi}{6}6π.
Step 3: Use known values for tan(π6)\tan(\frac{\pi}{6})tan(6π)
From the unit circle, we know:tan(π6)=13\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}tan(6π)=31
Step 4: Determine the sign in Quadrant II
In Quadrant II:
- Sine is positive
- Cosine is negative
- Therefore, Tangent (which is sin/cos\sin/\cossin/cos) is negative
So:tan(5π6)=−tan(π6)=−13\tan\left(\frac{5\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}tan(65π)=−tan(6π)=−31
🔁 Final Answer Box:
tan(5π6)=−13\boxed{\tan\left(\frac{5\pi}{6}\right) = -\frac{1}{\sqrt{3}}}tan(65π)=−31
This value is exact and comes from understanding the unit circle and reference angles, which is a key part of trigonometry.
