Problem 66 - Calculus for Biology and Medicine

Question

Evaluate the following expressions without using a calculator: (a) \(\sin \left(\frac{3 \pi}{4}\right)\) (b) \(\cos \left(-\frac{13 \pi}{6}\right)\) (c) \(\tan \left(\frac{4 \pi}{3}\right)\)

Step-by-Step Solution

Answer

(a) \(\frac{\sqrt{2}}{2}\), (b) \(\frac{\sqrt{3}}{2}\), (c) \(\sqrt{3}\).

Step 1: Determine the Reference Angle of Part (a)

The angle \( \frac{3\pi}{4} \) is in the second quadrant. The reference angle is \( \pi - \frac{3\pi}{4} = \frac{\pi}{4} \).

Step 2: Use Known Values for Part (a)

In the second quadrant, sine is positive. Therefore, \( \sin \left( \frac{3\pi}{4} \right) = \sin \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \).

Step 3: Simplify Angle for Part (b)

The angle \( -\frac{13\pi}{6} \) can be rewritten by adding \( 2\pi \) multiples: \( -\frac{13\pi}{6} + 2\pi = \frac{-13\pi + 12\pi}{6} = -\frac{\pi}{6} \).

Step 4: Use Known Values for Part (b)

\( \cos (-\theta) = \cos \theta \), so \( \cos \left( -\frac{\pi}{6} \right) = \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} \).

Step 5: Determine the Reference Angle of Part (c)

The angle \( \frac{4\pi}{3} \) is in the third quadrant. The reference angle is \( \frac{4\pi}{3} - \pi = \frac{\pi}{3} \).

Step 6: Use Known Values for Part (c)

In the third quadrant, tangent is positive. Therefore, \( \tan \left( \frac{4\pi}{3} \right) = \tan \left( \frac{\pi}{3} \right) = \sqrt{3} \).

Key Concepts

Sine

The sine function is an important trigonometric function used to relate the angles of a right-angled triangle to the lengths of its sides. Sine of an angle in a right triangle is defined as the ratio of the length of the side opposite to the angle, to the length of the hypotenuse, which is the triangle's longest side. Mathematically, this is expressed as:\[\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}\]When dealing with angles on the unit circle, the sine function measures the y-coordinate of a point on the circle. The unit circle has a radius of 1, which makes it easy to visualize these coordinates as trigonometric values.

• In the first and second quadrants, sine values are positive.
• In the third and fourth quadrants, sine values are negative.

In the exercise example, \(\sin \left( \frac{3\pi}{4} \right)\) takes a second quadrant angle. Here, the sine value is positive as mentioned, derived from the reference angle \(\frac{\pi}{4}\), which is known to be \(\frac{\sqrt{2}}{2}\).This results in:\[\sin \left( \frac{3\pi}{4} \right) = \frac{\sqrt{2}}{2}\]

Cosine

Cosine is another fundamental trigonometric function often used alongside sine in trigonometric problems. For a given angle in a right triangle, the cosine is the ratio of the adjacent side's length to the hypotenuse:\[\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}\]On the unit circle, the cosine function represents the x-coordinate of a point corresponding to an angle. This makes it quite helpful for visualizing trigonometric ratios in circular motion.

• Cosine is positive in the first and fourth quadrants.
• Cosine is negative in the second and third quadrants.

In the exercise problem, for \(\cos \left( -\frac{13\pi}{6} \right)\), you first simplify the angle by adding \(2\pi\) to move within a standard circle rotation. Simplifying gives\(-\frac{\pi}{6}\).Since cosine is an even function, \(\cos(-\theta) = \cos(\theta)\).Thus, \(\cos \left( -\frac{\pi}{6} \right) = \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}\).

Tangent

The tangent function is derived from the sine and cosine functions. It is a critical component in trigonometry, particularly for finding angles and distances. Tangent of an angle in a right triangle is the ratio of the sine to the cosine of that angle:\[\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\text{Opposite}}{\text{Adjacent}}\]Tangent values are also significant in the unit circle. They can become quite large as they approach vertical lines (or undefined), aligning with the concept of angles of inclination.

• Tangent is positive in the first and third quadrants.
• Tangent is negative in the second and fourth quadrants.

In the exercise case with \(\tan \left( \frac{4\pi}{3} \right)\), you identify this angle in the third quadrant where tangent is positive. The reference angle \(\frac{\pi}{3}\) has a known tangent value of \(\sqrt{3}\). Thus, \[\tan \left( \frac{4\pi}{3} \right) = \sqrt{3}\].