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Chapter 4: Problem 78

In Exercises \(61-86,\) use reference angles to find the exact value of each expression. Do not use a calculator. $$\sec 510^{\circ}$$

Short Answer

Expert verified
The value of \(\sec 510^{\circ}\) is \(- \frac{2}{\sqrt{3}}\)

Step by step solution

01

Convert the given angle to a reference angle

The given angle, \(510^{\circ}\), exceeds \(360^{\circ}\). To convert this angle to a reference angle, subtract multiples of \(360^{\circ}\) from the angle until it falls within \(0^{\circ}\) and \(360^{\circ}\). This gives: \(510^{\circ} - 360^{\circ} = 150^{\circ}\). Therefore, the reference angle is \(150^{\circ}\). This angle is situated in the second quadrant.
02

Find the value of \(\cos 150^{\circ}\)

In the second quadrant, cosine is negative. It is important to remember that \(cos(180^{\circ} - \theta) = - cos \theta \), hence \(\cos 150^{\circ} = - \cos (180^{\circ} - 150^{\circ} ) = - \cos 30^{\circ}\). The value of \(\cos 30^{\circ}\) is known as \(\frac{\sqrt{3}}{2}\). Therefore, \(\cos 150^{\circ} = - \frac{\sqrt{3}}{2}\).
03

Find the value of \(\sec 150^{\circ}\)

Since \(\sec \theta = \frac{1}{\cos \theta}\), using this formula the value of \(\sec 150^{\circ}\) can be obtained. This gives \(\sec 150^{\circ} = \frac{1}{\cos 150^{\circ}} = \frac{1}{- \frac{\sqrt{3}}{2}} = - \frac{2}{\sqrt{3}}\).

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Most popular questions from this chapter

A clock with an hour hand that is 15 inches long is hanging on a wall. At noon, the distance between the tip of the hour hand and the ceiling is 23 inches. At 3 P.M., the distance is 38 inches; at 6 P.M., 53 inches; at 9 P.M., 38 inches; and at midnight the distance is again 23 inches. If \(y\) represents the distance between the tip of the hour hand and the ceiling \(x\) hours after noon, make a graph that displays the information for \(0 \leq x \leq 24\)

Use a vertical shift to graph one period of the function. $$y=2 \cos \frac{1}{2} x+1$$

Determine the range of the following functions. Then give a viewing rectangle, or window, that shows two periods of the function's graph. a. \(f(x)=\sec \left(3 x+\frac{\pi}{2}\right)\) b. \(g(x)=3 \sec \pi\left(x+\frac{1}{2}\right)\)

Use a graphing utility to graph two periods of the function. $$y=-2 \cos \left(2 \pi x-\frac{\pi}{2}\right)$$

Write the equation for a cosecant function satisfying the given conditions. $$\text { period: } 3 \pi ; \text { range: }(-\infty,-2] \cup[2, \infty)$$
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