Chapter 6: Problem 18
If \(\cos \alpha=\frac{24}{25}\) and \(\sin \alpha<0,\) find the exact value of \(\cos \left(\alpha+\frac{\pi}{6}\right)\)
Short Answer
Expert verified
\(\cos \left(\alpha + \frac{\pi}{6}\right) = \frac{24\sqrt{3} + 7}{50}.\)
Step by step solution
01
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.
02
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 \(\sin \alpha < 0\), \(\sin \alpha = -\frac{7}{25}\).
03
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: \(\cos \left(\alpha + \frac{\pi}{6}\right) = \cos \alpha \cos \frac{\pi}{6} - \sin \alpha \sin \frac{\pi}{6}\).
04
Calculate \(\cos(\frac{\pi}{6})\) and \(\sin(\frac{\pi}{6})\)
From trigonometric values, \(\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\) and \(\sin\frac{\pi}{6} = \frac{1}{2}\).
05
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: \[\frac{24\sqrt{3}}{50} + \frac{7}{50} = \frac{24\sqrt{3} + 7}{50}.\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \(\sin^2 \alpha\): \[ \sin^2 \alpha = 1 - \frac{576}{625} = \frac{49}{625} \]
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: \[ \frac{24\sqrt{3}}{50} + \frac{7}{50} = \frac{24\sqrt{3} + 7}{50} \]
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.