To solve the expression \( 6 \tan 300^\circ - 5 \cos 240^\circ \), we'll use the tangent and cosine values of the given angles:

1. \( \tan 300^\circ \)
2. \( \cos 240^\circ \)

### Step-by-Step Solution

1. **Find \(\tan 300^\circ\):**
   - The angle \(300^\circ\) is in the fourth quadrant.
   - The reference angle is \(360^\circ - 300^\circ = 60^\circ\).
   - \(\tan 300^\circ = \tan 60^\circ = \sqrt{3}\), but since it's in the fourth quadrant, where tangent is negative, \(\tan 300^\circ = -\sqrt{3}\).

2. **Find \(\cos 240^\circ\):**
   - The angle \(240^\circ\) is in the third quadrant.
   - The reference angle is \(240^\circ - 180^\circ = 60^\circ\).
   - \(\cos 240^\circ = \cos 60^\circ = \frac{1}{2}\), but since it's in the third quadrant, where cosine is negative, \(\cos 240^\circ = -\frac{1}{2}\).

### Substitute and Simplify:

\[ 6 \tan 300^\circ - 5 \cos 240^\circ \]

\[ 6 \left(-\sqrt{3}\right) - 5 \left(-\frac{1}{2}\right) \]

\[ -6\sqrt{3} - (-\frac{5}{2}) \]

\[ -6\sqrt{3} + \frac{5}{2} \]

### Final Answer:
\[ 6 \tan 300^\circ - 5 \cos 240^\circ = -6\sqrt{3} + \frac{5}{2} \]

Would you like more details or have any questions?

Here are 5 questions you might want to ask next:
1. How do you find the tangent values of angles in other quadrants?
2. What are the sine and cosine values for other key angles like 45°, 135°, etc.?
3. Can you explain how the signs of trigonometric functions change in each quadrant?
4. What are the fundamental identities in trigonometry and how are they used?
5. How do you convert between degrees and radians in trigonometry?

**Tip:** Understanding the unit circle can help you quickly determine the values of trigonometric functions for any angle.

Learn how to solve the expression 6 tan 300° - 5 cos 240° using trigonometric identities. This problem involves calculating tangent and cosine values for specific angles and applying them to a trigonometric expression. Suitable for high school students studying trigonometry.

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