Problem 18 - Precalculus: Functions and Graphs

Question

If \(\cos \alpha=\frac{24}{25}\) and \(\sin \alpha<0,\) find the exact value of \(\cos \left(\alpha+\frac{\pi}{6}\right)\)

Step-by-Step Solution

Answer

\(\cos \left(\alpha + \frac{\pi}{6}\right) = \frac{24\sqrt{3} + 7}{50}.\)

Step 1: Recognize the Quadrant

Since \(\cos \alpha = \frac{24}{25}\) and \(\sin \alpha<0\), \(\alpha\) is in the fourth quadrant where cosine is positive and sine is negative.

Step 2: Use Pythagorean Identity

Use the identity \(\sin^2 \alpha + \cos^2 \alpha = 1\). We have \(\cos^2 \alpha = \left(\frac{24}{25}\right)^2 = \frac{576}{625}\). \(\sin^2 \alpha = 1 - \frac{576}{625} = \frac{49}{625}\). Since \(\sin \alpha < 0\), \(\sin \alpha = -\frac{7}{25}\).

Step 3: Use Angle Addition Formula for Cosine

The formula for the cosine of the sum of two angles is \(\cos(a + b) = \cos a \cos b - \sin a \sin b\). Apply it here: \(\cos \left(\alpha + \frac{\pi}{6}\right) = \cos \alpha \cos \frac{\pi}{6} - \sin \alpha \sin \frac{\pi}{6}\).

Step 4: Calculate \(\cos(\frac{\pi}{6})\) and \(\sin(\frac{\pi}{6})\)

From trigonometric values, \(\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\) and \(\sin\frac{\pi}{6} = \frac{1}{2}\).

Step 5: Substitute and Simplify

Substitute the known values into the formula: \(\cos \left(\alpha + \frac{\pi}{6}\right) = \frac{24}{25}\times \frac{\sqrt{3}}{2} - \left(-\frac{7}{25}\right)\times \frac{1}{2}\). Simplify to get: \[\frac{24\sqrt{3}}{50} + \frac{7}{50} = \frac{24\sqrt{3} + 7}{50}.\]

Key Concepts

Pythagorean Identity

The Pythagorean Identity is a fundamental concept in trigonometry that relates the squares of sine and cosine functions. This identity states: \[ \sin^2 \alpha + \cos^2 \alpha = 1 \] It comes from the Pythagorean theorem in a right triangle, where the hypotenuse is always 1 in the unit circle. This makes it very useful in finding unknown trigonometric values when one value is known. For example, in our problem, we knew that \(\cos \alpha = \frac{24}{25}\). We can use the Pythagorean Identity to find \(\sin \alpha\).

• Square the known \(\cos \alpha\): \[ \cos^2 \alpha = \left(\frac{24}{25}\right)^2 = \frac{576}{625} \]
• Substitute this into the identity formula and solve for \(\sin^2 \alpha\): \[ \sin^2 \alpha = 1 - \frac{576}{625} = \frac{49}{625} \]

Since we know \(\sin \alpha < 0\) due to the quadrant location, we find that \(\sin \alpha = -\frac{7}{25}\). This identity is crucial for solving problems involving trigonometric expressions.

Angle Addition Formula

The Angle Addition Formula is another essential tool in trigonometry, which allows you to find the trigonometric function of two added angles. The formula for cosine is: \[ \cos(a + b) = \cos a \cos b - \sin a \sin b \] This is used when we want to find the cosine of the sum of two angles, such as \(\alpha + \frac{\pi}{6}\) in our exercise. By knowing the trigonometric functions of individual angles, we can find the cosine of their sum.

• We start with \(\cos \alpha = \frac{24}{25}\) and \(\sin \alpha = -\frac{7}{25}\) from previous calculations.
• Use known values: \(\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}\) and \(\sin \frac{\pi}{6} = \frac{1}{2}\).
• Substitute these into the Angle Addition Formula: \[ \cos\left(\alpha + \frac{\pi}{6}\right) = \frac{24}{25} \cdot \frac{\sqrt{3}}{2} + \frac{7}{25} \cdot \frac{1}{2} \]
• Simplify the expression to find: \[ \frac{24\sqrt{3}}{50} + \frac{7}{50} = \frac{24\sqrt{3} + 7}{50} \]

By using this formula, you can easily handle problems involving the addition of angles.

Quadrants

Understanding quadrants is crucial in trigonometry to determine the sign of trigonometric functions based on angle positions. The four quadrants of the coordinate plane are divided by the x and y-axis. Each quadrant gives a different sign to trigonometric functions:

• First Quadrant: (0 to \(\frac{\pi}{2}\)) where all trigonometric values are positive.
• Second Quadrant: (\(\frac{\pi}{2}\) to \(\pi\)) where sine is positive.
• Third Quadrant: (\(\pi\) to \(\frac{3\pi}{2}\)) where tangent is positive.
• Fourth Quadrant: (\(\frac{3\pi}{2}\) to \(2\pi\)) where cosine is positive.

In our exercise, \(\alpha\) is in the fourth quadrant where cosine is positive, but \(\sin \alpha < 0\). This is because the angle return value is on the x-axis side, making cosine positive and placing sine on the negative y-axis below it. Understanding which quadrant the angle is in helps correctly apply signs to \(\sin \alpha\) and \(\cos \alpha\), essential for accurately solving trigonometry problems.