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Chapter 4: Problem 30

Find the lengths of both circular arcs of the unit circle connecting the point \(\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\) and the point whose radius corresponds to 2 radians.

Short Answer

Expert verified
The lengths of both circular arcs of the unit circle connecting the given points are \(Length_{smaller} = \frac{\pi}{4}\) and \(Length_{larger} = \frac{7\pi}{4}\).

Step by step solution

01

1. Determine the angles of the given points

The first point is \((-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})\). Since it's a point on the unit circle, we know the angle will be either in the first or second quadrant. As its x-coordinate is negative and y-coordinate is positive, it's in the second quadrant. The angle can be found using the tangent function and its inverse: \(tan(\theta)=-\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}=-1\) Take the inverse tangent or arctangent to find the angle: \(\theta=arctan(-1)=\frac{3\pi}{4}\) Now let's find the angle for the second point, whose radius corresponds to 2 radians. As the radius is given in radians, the angle is: \(\phi=2\)
02

2. Calculate the difference in angles to find out the smaller arc

Now that we have both angles, we can find the difference between them to determine the smaller arc. We know that the angle difference is positive and less than \(\pi\) ( or 180掳). So, we'll calculate the absolute value of the difference between the angles: \(Arc_{smaller} = |\phi - \theta| = |2-\frac{3\pi}{4}| = \left|-\frac{\pi}{4}\right| = \frac{\pi}{4}\)
03

3. Find the lengths of both arcs

To find the lengths, multiply the angle measures by the radius of the unit circle, which is 1. The smaller arc's length is: \(Length_{smaller} = Radius \cdot Arc_{smaller} = 1\cdot \frac{\pi}{4} = \frac{\pi}{4}\) The larger arc is the remaining part of the circumference of the unit circle. The circumference of the unit circle is: \(Circumference = 2\pi \cdot Radius = 2\pi\) So, the larger arc's length is: \(Length_{larger} = Circumference - Length_{smaller} = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}\) Thus, the lengths of both circular arcs of the unit circle connecting the given points are \(Length_{smaller} = \frac{\pi}{4}\) and \(Length_{larger} = \frac{7\pi}{4}\).

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Most popular questions from this chapter

Find exact expressions for the indicated quantities. \(\cos \left(\frac{\pi}{2}-u\right)\)

Show that $$\sin ^{2} \theta=\frac{\tan ^{2} \theta}{1+\tan ^{2} \theta}$$ for all \(\theta\) except odd multiples of \(\frac{\pi}{2}\).

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.] \(\tan \frac{\pi}{8}\)

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.] \(\sin \frac{9 \pi}{8}\)

Find the smallest positive number \(x\) such that $$ \tan x=3 \tan \left(\frac{\pi}{2}-x\right) $$
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