Math Problem Statement

6 tan 300° - 5 cos 240°

Solution

To solve the expression 6tan3005cos2406 \tan 300^\circ - 5 \cos 240^\circ, we'll use the tangent and cosine values of the given angles:

  1. tan300\tan 300^\circ
  2. cos240\cos 240^\circ

Step-by-Step Solution

  1. Find tan300\tan 300^\circ:

    • The angle 300300^\circ is in the fourth quadrant.
    • The reference angle is 360300=60360^\circ - 300^\circ = 60^\circ.
    • tan300=tan60=3\tan 300^\circ = \tan 60^\circ = \sqrt{3}, but since it's in the fourth quadrant, where tangent is negative, tan300=3\tan 300^\circ = -\sqrt{3}.
  2. Find cos240\cos 240^\circ:

    • The angle 240240^\circ is in the third quadrant.
    • The reference angle is 240180=60240^\circ - 180^\circ = 60^\circ.
    • cos240=cos60=12\cos 240^\circ = \cos 60^\circ = \frac{1}{2}, but since it's in the third quadrant, where cosine is negative, cos240=12\cos 240^\circ = -\frac{1}{2}.

Substitute and Simplify:

6tan3005cos2406 \tan 300^\circ - 5 \cos 240^\circ

6(3)5(12)6 \left(-\sqrt{3}\right) - 5 \left(-\frac{1}{2}\right)

63(52)-6\sqrt{3} - (-\frac{5}{2})

63+52-6\sqrt{3} + \frac{5}{2}

Final Answer:

6tan3005cos240=63+526 \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.

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Math Problem Analysis

Mathematical Concepts

Trigonometry

Formulas

Tangent and cosine formulas

Theorems

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Suitable Grade Level

High School