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Trigonometric identities are equations involving trigonometric functions that are true for all values of the included variables. Some of the most important ones are the Pythagorean identities, such as \( \sin^2(\theta) + \cos^2(\theta) = 1 \), and angle sum and difference identities, like \( \sin(a + b) = \sin a \cos b + \cos a \sin b \).

These identities are fundamental because they allow us to simplify and manipulate trigonometric expressions. In the given solution, we use the identity for \( \sin \left( 180^{\circ} - \theta \right) = \sin \theta \) to find that \( \sin(5 \pi / 6) = \sin(150^{\circ}) = \sin(30^{\circ}) = \frac{1}{2} \). Similarly, for cosine, we have \( \cos(180^{\circ} - \theta) = -\cos(\theta) \), so \( \cos(5 \pi / 6) = \cos(150^{\circ}) = -\cos(30^{\circ}) = -\frac{\sqrt{3}}{2} \).

Mastering these identities helps in solving various trigonometric problems effectively.

The unit circle is a circle with a radius of one, centered at the origin of the coordinate system. It is a powerful tool in trigonometry, helping us find the exact values of trigonometric functions for key angles.

The angles on the unit circle are often measured in radians. For instance, \( \frac{\pi}{6} \), \( \frac{\pi}{4} \), and \( \frac{\pi}{3} \) correspond to 30°, 45°, and 60° respectively.

Use the unit circle to locate \(5 \pi / 6\), which is 150°. It tells us that \( \sin(5 \pi / 6) = \frac{1}{2} \) and \( \cos(5 \pi / 6) = - \frac{\sqrt{3}}{2} \). The signs of sine and cosine depend on the quadrant where the angle lies.

Recognizing these values swiftly can make problem-solving smoother and quicker.

Rationalizing the denominator is a technique used to eliminate radicals from the denominator of a fraction. This often involves multiplying both the numerator and the denominator by the conjugate of the denominator.

In our expression \( \frac{1}{2 - \sqrt{3}} \), multiplying by the conjugate \(2 + \sqrt{3} \) gives us: \frac{1}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{2 + \sqrt{3}}{(2)^2 - (\sqrt{3})^2} = \frac{2 + \sqrt{3}}{4 - 3} = 2 + \sqrt{3} \.

Rationalizing is a crucial step in simplifying trigonometric expressions, ensuring they have neat and manageable forms. Ensuring we understand this process can make seemingly complex fractions much easier to handle.

Trigonometric functions behave differently depending on the quadrant of the angle.

The unit circle is divided into four quadrants:

• Quadrant I: Both sine and cosine are positive.
• Quadrant II: Sine is positive and cosine is negative.
• Quadrant III: Both sine and cosine are negative.
• Quadrant IV: Sine is negative and cosine is positive.

In our example, \(5 \pi / 6\) lies in the second quadrant. Hence, \( \sin(5 \pi / 6) = \frac{1}{2} \) is positive and \( \cos(5 \pi / 6) = - \frac{\sqrt{3}}{2} \) is negative.

Understanding these sign changes is essential because it directly impacts the simplification of trigonometric expressions. Knowing which function is positive or negative helps avoid errors when solving these types of problems.