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Chapter 4: Problem 33

Find the lengths of both circular arcs on the unit circle connecting the points (1,0) and \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\).

Short Answer

Expert verified
The lengths of the two circular arcs on the unit circle connecting the points (1, 0) and \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\) are \(\frac{\pi}{4}\) (shorter arc) and \(\frac{7\pi}{4}\) (longer arc).

Step by step solution

01

Find the angle between the points

First, we need to find the angle between the given points and the positive x-axis. We can use the inverse tangent function for this purpose: For point (1, 0): $$\theta_1 = \tan^{-1}\left(\frac{0}{1}\right) = 0$$ For point \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\): $$\theta_2 = \tan^{-1}\left(\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}\right) = \tan^{-1}(1) = \frac{\pi}{4}$$
02

Find the smaller angle between the two points

Now that we have the angles between each point and the positive x-axis, we can find the smaller angle between the points by subtracting the smaller angle from the larger angle: $$\theta = \theta_2 - \theta_1 = \frac{\pi}{4} - 0 = \frac{\pi}{4}$$
03

Calculate the shorter arc length

Now, we can find the length of the shorter arc connecting the points using the formula for arc length: $$\text{arc length} = \text{radius} \times \text{angle (in radians)}$$ As our circle is a unit circle, the radius is 1. So the shorter arc length is: $$\text{shorter arc length} = 1 \times \frac{\pi}{4} = \frac{\pi}{4}$$
04

Calculate the longer arc length

To find the length of the longer arc, we need to calculate the remaining portion of the circle. Since a full circle has an angle of \(2\pi\) radians, we can find the angle of the longer arc by subtracting the smaller angle we found earlier: $$\theta_{\text{longer arc}} = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$$ Now we can find the longer arc length using the same formula as before: $$\text{longer arc length} = 1 \times \frac{7\pi}{4} = \frac{7\pi}{4}$$ So, the lengths of the two circular arcs on the unit circle connecting the points (1, 0) and \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\) are \(\frac{\pi}{4}\) and \(\frac{7\pi}{4}\).

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Most popular questions from this chapter

For Exercises \(29-34,\) assume the surface of the earth is a sphere with radius 3963 miles. The latitude of \(a\) point \(P\) on the earth's surface is the angle between the line from the center of the earth to \(P\) and the line from the center of the earth to the point on the equator closest to \(P\), as shown below for latitude \(40^{\circ} .\) Dallas has latitude \(32.8^{\circ}\) north. Find the radius of the circle formed by the points with the same latitude as Dallas.

In Exercises 5-38, find exact expressions for the indicated quantities, given that $$\cos \frac{\pi}{12}=\frac{\sqrt{2+\sqrt{3}}}{2} \text { and } \sin \frac{\pi}{8}=\frac{\sqrt{2-\sqrt{2}}}{2}$$ [These values for \(\cos \frac{\pi}{12}\) and \(\sin \frac{\pi}{8}\) will be derived in Examples 3 and 4 in Section 5.5.] \(\sin \frac{3 \pi}{8}\)

Show that $$\sin (\pi-\theta)=\sin \theta$$ for every angle \(\theta\).

In Exercises 5-38, find exact expressions for the indicated quantities, given that $$\cos \frac{\pi}{12}=\frac{\sqrt{2+\sqrt{3}}}{2} \text { and } \sin \frac{\pi}{8}=\frac{\sqrt{2-\sqrt{2}}}{2}$$ [These values for \(\cos \frac{\pi}{12}\) and \(\sin \frac{\pi}{8}\) will be derived in Examples 3 and 4 in Section 5.5.] \(\sin \frac{13 \pi}{12}\)

In Exercises 5-38, find exact expressions for the indicated quantities, given that $$\cos \frac{\pi}{12}=\frac{\sqrt{2+\sqrt{3}}}{2} \text { and } \sin \frac{\pi}{8}=\frac{\sqrt{2-\sqrt{2}}}{2}$$ [These values for \(\cos \frac{\pi}{12}\) and \(\sin \frac{\pi}{8}\) will be derived in Examples 3 and 4 in Section 5.5.] \(\cos \frac{5 \pi}{12}\)
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