Problem 25
Question
Find the exact value of each trigonometric function. \(\cot 510^{\circ}\)
Step-by-Step Solution
Verified Answer
\(\cot 510^\circ = -\sqrt{3}\).
1Step 1: Convert Angle to Standard Position
Given the angle of \(510^\circ\), we first need to convert it to a standard position by finding its coterminal angle between \(0^\circ\) and \(360^\circ\). We do this by subtracting \(360^\circ\) from \(510^\circ\) until the result is within the standard range. \(510^\circ - 360^\circ = 150^\circ\) which is in the desired range.
2Step 2: Understand the Cotangent Function
The cotangent function, \(\cot\theta\), is defined as \(\frac{1}{\tan\theta}\), where \(\tan\theta\) is the tangent of the angle. Thus, \(\cot\theta = \frac{\cos\theta}{\sin\theta}\). Now, we need to find the sine and cosine of \(150^\circ\).
3Step 3: Evaluate Trigonometric Values at 150°
At \(150^\circ\), the angle is in the second quadrant where sine is positive and cosine is negative. The reference angle for \(150^\circ\) is \(30^\circ\). Thus, \(\sin 150^\circ = \frac{1}{2}\) and \(\cos 150^\circ = -\frac{\sqrt{3}}{2}\).
4Step 4: Calculate Cotangent Value
Using the sine and cosine values, calculate \(\cot 150^\circ = \frac{\cos 150^\circ}{\sin 150^\circ} = \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}} = -\sqrt{3}\).
5Step 5: Conclusion
The cotangent of \(510^\circ\), using the coterminal angle of \(150^\circ\), is \(-\sqrt{3}\).
Key Concepts
Coterminal AnglesCotangent FunctionSine and Cosine Values
Coterminal Angles
Coterminal angles are angles which share the same initial and terminal sides but may differ by full rotations. This means if you keep adding or subtracting full rotations of \(360^\circ\) to an angle, the resulting angles are coterminal with each other.
In the current exercise, you start with \(510^\circ\). To find a coterminal angle within the standard position, which ranges from \(0^\circ\) to \(360^\circ\), you subtract \(360^\circ\). In this case:
In the current exercise, you start with \(510^\circ\). To find a coterminal angle within the standard position, which ranges from \(0^\circ\) to \(360^\circ\), you subtract \(360^\circ\). In this case:
- \(510^\circ - 360^\circ = 150^\circ\)
Cotangent Function
The cotangent function, denoted by \(\cot\theta\), is one of the six basic trigonometric functions. It is specifically defined as the reciprocal of the tangent function:
\[ \cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta} \]
This means you need both the cosine and sine values of an angle to find its cotangent value. For \(150^\circ\) from our exercise, you will use these sine and cosine values:
\[ \cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta} \]
This means you need both the cosine and sine values of an angle to find its cotangent value. For \(150^\circ\) from our exercise, you will use these sine and cosine values:
- \(\sin 150^\circ\) and \(\cos 150^\circ\)
Sine and Cosine Values
Sine and cosine values for angles are fundamental in trigonometry. They determine the behavior of the waveforms we commonly encounter. For an angle such as \(150^\circ\), you use its reference angle to find its sine and cosine values.
The reference angle for \(150^\circ\) is \(30^\circ\), and since \(150^\circ\) is in the second quadrant, the following rules apply:
The reference angle for \(150^\circ\) is \(30^\circ\), and since \(150^\circ\) is in the second quadrant, the following rules apply:
- Sine is positive: \(\sin 150^\circ = \frac{1}{2}\)
- Cosine is negative: \(\cos 150^\circ = -\frac{\sqrt{3}}{2}\)
Other exercises in this chapter
Problem 25
Find the exact value of each function. $$ \frac{\cos 60^{\circ}+\sin 30^{\circ}}{4} $$
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Determine whether each triangle should be solved by beginning with the Law of Sines or Law of Cosines. Then solve each triangle. Round measures of sides to the
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Rewrite each degree measure in radians and each radian measure in degrees. \(\frac{11 \pi}{4}\)
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Determine whether each triangle has no solution, one solution, or two solutions. Then solve each triangle. Round measures of sides to the nearest tenth and meas
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