Problem 2 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Compute the following exactly. Do not use calculator approximations. (a) \(\cos \left(\frac{\pi}{12}\right)=\cos \left(\frac{\pi}{4}-\frac{\pi}{6}\right)\) (b) \(\sin \left(\frac{-\pi}{12}\right)=\sin \left(\frac{\pi}{6}-\frac{\pi}{4}\right)\) (c) \(\tan \left(\frac{\pi}{12}\right)\)

Step-by-Step Solution

Answer

\(\cos \left(\frac{\pi}{12}\right) = \frac{\sqrt{6}+\sqrt{2}}{4}\), \(\sin \left(-\frac{\pi}{12}\right) = \frac{\sqrt{2}-\sqrt{6}}{4}\), \(\tan \left(\frac{\pi}{12}\right) = \frac{\sqrt{2}-\sqrt{6}}{\sqrt{6}+\sqrt{2}}\)

Step 1: Compute cos(π/12)

Express \(\frac{\pi}{12}\) as \(\frac{\pi}{4}-\frac{\pi}{6}\) and apply the following identity for cosine of difference of two angles: \(\cos(A - B) = \cos A * \cos B + \sin A * \sin B\). Here, \(A=\frac{\pi}{4}\) and \(B=\frac{\pi}{6}\). So \(\cos \left(\frac{\pi}{12}\right)=\cos \left(\frac{\pi}{4}\right) * \cos \left(\frac{\pi}{6}\right) + \sin \left(\frac{\pi}{4}\right) * \sin \left(\frac{\pi}{6}\right) = \frac{\sqrt{2}}{2} * \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} * \frac{1}{2} = \frac{\sqrt{6}+\sqrt{2}}{4}\)

Step 2: Compute sin(-π/12)

Express \(-\frac{\pi}{12}\) as \(\frac{\pi}{6}-\frac{\pi}{4}\) and apply the following identity for sine of difference of two angles: \(\sin(A - B) = \sin A * \cos B - \cos A * \sin B\). Here, \(A=\frac{\pi}{6}\) and \(B=\frac{\pi}{4}\). So \(\sin \left(-\frac{\pi}{12}\right) = \sin \left(\frac{\pi}{6}\right) * \cos \left(\frac{\pi}{4}\right) - \cos \left(\frac{\pi}{6}\right) * \sin \left(\frac{\pi}{4}\right) = \frac{1}{2} * \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} * \frac{\sqrt{2}}{2} = \frac{\sqrt{2} - \sqrt{6}}{4}\)

Step 3: Compute tan(π/12)

We apply the following identity: \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\). Thus, \(\tan \left(\frac{\pi}{12}\right) = \frac{\sin \left(\frac{\pi}{12}\right)}{\cos \left(\frac{\pi}{12}\right)} = \frac{\frac{\sqrt{2} - \sqrt{6}}{4}}{\frac{\sqrt{6}+\sqrt{2}}{4}} = \frac{\sqrt{2} - \sqrt{6}}{\sqrt{6}+\sqrt{2}}\)

Key Concepts

Understanding the Cosine Function

The cosine function is a fundamental trigonometric function closely related to angles and sides of right triangles. It describes the horizontal coordinate of a point on the unit circle at any given angle. To compute the cosine of an angle, we use various identities, one of which is the cosine of a difference of two angles.

An identity used in trigonometry is \[\cos(A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B\]This formula allows us to find the cosine of an angle that is the difference of two known angles.

In the exercise, we expressed \(\frac{\pi}{12}\) as \(\frac{\pi}{4} - \frac{\pi}{6}\). Then plugging these into our identity gives us a calculated value of the cosine function:

• \(\cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)
• \(\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\)
• \(\sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)
• \(\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}\)

By applying these values, we end up with:\[\cos \left(\frac{\pi}{12}\right) = \frac{\sqrt{6} + \sqrt{2}}{4}\] Understanding this process helps grasp not just cosine values, but also how to manipulate angles through identities.

Exploring the Sine Function

Sine is another key trigonometric function that measures the vertical coordinate of the unit circle at any angle. It's essential in understanding oscillations and waves. One useful formula is the sine of a difference of angles:

\[\sin(A - B) = \sin A \cdot \cos B - \cos A \cdot \sin B\]In the exercise, we took advantage of this identity by expressing \(-\frac{\pi}{12}\) as \(\frac{\pi}{6} - \frac{\pi}{4}\). Inputting these into the identity helps us derive the exact sine value:

• \(\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}\)
• \(\cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)
• \(\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\)
• \(\sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)

By computing this, we find:\[\sin \left(-\frac{\pi}{12}\right) = \frac{\sqrt{2} - \sqrt{6}}{4}\] This calculation reflects the negative angle's effects and the intricacies of the sine function.

The Tangent Function Explained

Tangent, a third primary trigonometric function, is defined as the ratio of sine to cosine for any given angle. This function is vital in analyzing slope and finding the steepness of curves. The main identity we use is:

\[\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\]When computing the tangent of an angle, like in our exercise with \(\tan\left(\frac{\pi}{12}\right)\), we first determine the sine and cosine for that angle through previously calculated identities.

Given that:- \(\sin \left(\frac{\pi}{12}\right) = \frac{\sqrt{2} - \sqrt{6}}{4}\)- \(\cos \left(\frac{\pi}{12}\right) = \frac{\sqrt{6} + \sqrt{2}}{4}\)

We find the tangent by dividing the sine by the cosine:\[\tan \left(\frac{\pi}{12}\right) = \frac{\sqrt{2} - \sqrt{6}}{\sqrt{6} + \sqrt{2}}\] This highlights how tangent combines elements of both sine and cosine. Through tangent, we see different perspectives on angle relationships and how slopes can be interpreted in trigonometry.