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In each case, determine the arc \(t\) that satisfies the given conditions or explain why no such arc exists. (a) \(\tan (t)=1, \cos (t)=-\frac{1}{\sqrt{2}},\) and \(0

Short Answer

Expert verified
(a) \(t=\frac{3\pi}{4}\) (b) \(t=\frac{\pi}{2}\) (c) \(t=\frac{3\pi}{4}\) (d) No such arc exists. (e) \(t=\frac{7\pi}{4}\)

Step by step solution

01

Analyze the given conditions

The given conditions are \(\tan(t)=1\) and \(\cos(t)=-\frac{1}{\sqrt{2}}\). We need to find angle t such that both conditions are satisfied and t lies in between \(0\) and \(2\pi\).
02

Calculate angle t from the tangent condition

From \(\tan(t)=1\), we know that angle t corresponds to \(\frac{\pi}{4}\) and an additional \(\pi\)k, where k is any integer. So, \(t= \frac{\pi}{4} + \pi k\).
03

Check if the cosine condition is satisfied by angle t

We need to check if \(\cos(t)=-\frac{1}{\sqrt{2}}\). By plugging \(t= \frac{\pi}{4} + \pi k\), we find \(\cos(t)=\cos(\frac{\pi}{4}+\pi k)\). It may happen that \(t = \frac{3\pi}{4}\) satisfies \(\cos(t) = -\frac{1}{\sqrt{2}}\), since \(\frac{3\pi}{4} = \frac{\pi}{4} + \pi\) and falls between \(0\) and \(2\pi\). So, the arc \(t\) that satisfies the given conditions is \(t = \frac{3\pi}{4}\). (b) \(\sin (t)=1, \sec (t)\) is undefined, and \(0<t<\pi\).
04

Analyze the given conditions

The given conditions are \(\sin(t)=1\) and \(\sec(t)\) is undefined. We need to find angle t such that both conditions are satisfied and t lies in between \(0\) and \(\pi\).
05

Calculate angle t from the sine condition

From \(\sin(t)=1\), we know that angle t corresponds to \(\frac{\pi}{2}\) and additional \(2\pi\)k, where k is any integer. So, \(t = \frac{\pi}{2} + 2\pi k\).
06

Check if the secant condition is satisfied by angle t

We need to check if sec(t) is undefined. Secant is undefined for \(\cos(t)=0\). The angle \(\frac{\pi}{2}\) satisfies \(\cos(t)=0\). So, the arc \(t\) that satisfies the given conditions is \(t = \frac{\pi}{2}\). (c) \(\sin (t)=\frac{\sqrt{2}}{2}, \sec (t)=-\sqrt{2},\) and \(0<t<\pi\).
07

Analyze the given conditions

The given conditions are \(\sin(t)=\frac{\sqrt{2}}{2}\) and \(\sec(t) = -\sqrt{2}\). We will find the angle t such that both conditions are satisfied and t lies in between \(0\) and \(\pi\).
08

Calculate angle t from the sine condition

From \(\sin(t)=\frac{\sqrt{2}}{2}\), we find out that the corresponding angle t is \(\frac{\pi}{4}\) and \(\frac{3\pi}{4}\).
09

Check if the secant condition is satisfied by angle t

We need to check if \(\sec(t)=-\sqrt{2}\). We know the secant is a reciprocal function of cosine function. We need to check if \(\cos(t) = -\frac{1}{\sqrt{2}}\). The angle \(\frac{3\pi}{4}\) satisfies \(\cos(t) = -\frac{1}{\sqrt{2}}\). So, the arc \(t\) that satisfies the given conditions is \(t = \frac{3\pi}{4}\). (d) \(\sec (t)=-\frac{2}{\sqrt{3}}, \tan (t)=\sqrt{3},\) and \(0<t<2 \pi\).
10

Analyze the given conditions

The given conditions are \(\sec(t) = -\frac{2}{\sqrt{3}}\) and \(\tan(t) = \sqrt{3}\). We will find the angle t such that both conditions are satisfied and t lies in between \(0\) and \(2\pi\).
11

Calculate angle t from the tangent condition

From \(\tan(t)=\sqrt{3}\), we find out that the corresponding angle t is \(\frac{\pi}{3}\) and an additional \(\pi\)k for integer k. So, \(t=\frac{\pi}{3}+\pi k\).
12

Check if the secant condition is satisfied by angle t

We need to check if \(\sec(t) = -\frac{2}{\sqrt{3}}\). We know the secant is a reciprocal function of cosine function. We need to check if \(\cos(t) = -\frac{\sqrt{3}}{2}\). Upon calculation, none of the angles satisfy both conditions. Therefore, there is no arc \(t\), such that both \(\sec(t) = -\frac{2}{\sqrt{3}}\) and \(\tan(t) = \sqrt{3}\). (e) \(\csc (t)=\sqrt{2}, \tan (t)=-1,\) and \(0<t<2 \pi\).
13

Analyze the given conditions

The given conditions are \(\csc(t)=\sqrt{2}\) and \(\tan(t)=-1\). We will find the angle t such that both conditions are satisfied and t lies in between \(0\) and \(2\pi\).
14

Calculate angle t from the tangent condition

From \(\tan(t)=-1\), we find out that the corresponding angle t is \(\frac{3\pi}{4}\) and \(\frac{7\pi}{4}\).
15

Check if the cosecant condition is satisfied by angle t

We need to check if \(\csc(t)=\sqrt{2}\). We know the cosecant is a reciprocal function of sine function. We need to check if \(\sin(t) = \frac{1}{\sqrt{2}}\). The angle \(\frac{7\pi}{4}\) satisfies \(\sin(t) = \frac{1}{\sqrt{2}}\). So, the arc \(t\) that satisfies the given conditions is \(t = \frac{7\pi}{4}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tangent Function
The tangent function, often denoted as \( \tan(\theta) \), is one of the primary trigonometric functions used to relate an angle with the ratio of two sides in a right-angled triangle. Specifically, for an angle \( \theta \), the tangent is the ratio of the opposite side to the adjacent side. This function is periodic, meaning it repeats its values at regular intervals, with a period of \( \pi \) radians or 180 degrees.

When solving trigonometric equations, it is vital to understand that the tangent function can give insights into the angle, especially when we're given the condition that \( \tan(t)=1 \), as in the textbook exercise. This condition implies that the angle has to be \( \frac{\pi}{4} \) plus any multiple of \( \pi \), as tangent exhibits symmetry over its period. This can be used to find specific angles that fall within a desired range, such as \( 0 < t < 2\pi \) in our example.

However, the tangent function isn't defined when the cosine of the angle is zero, as it would involve division by zero. When the function 'jumps' from negative to positive infinity or vice versa, we say it has a vertical asymptote at those points where the cosine equals zero.
Sine Function
The sine function, represented as \( \sin(\theta) \), is fundamental in trigonometry and corresponds to the ratio of the length of the side of the triangle opposite the angle \( \theta \) to the length of the hypotenuse. Just like other trigonometric functions, sine has a periodic behavior, with a period of \( 2\pi \) radians or 360 degrees.

In our exercise, the condition \( \sin(t)=1 \) implies that \( t \) must be \( \frac{\pi}{2} \) plus any integer multiple of \( 2\pi \). However, as we're constrained by the interval \( 0 < t < \pi \), the only angle that meets this condition in the given interval is \( \frac{\pi}{2} \). Additionally, when \( \sin(t)= \frac{\sqrt{2}}{2} \), there are typically two angles within one period that satisfy this condition, specifically \( \frac{\pi}{4} \) and \( \frac{3\pi}{4} \) in the first period. Learning where these values lie on the unit circle is very helpful for quickly identifying correct angles.
Secant Function
The secant function, written as \( \sec(\theta) \), is the reciprocal of the cosine function. For any angle \( \theta \), it is defined as the ratio of the hypotenuse to the adjacent side of a triangle. The secant function has a period of \( 2\pi \) radians or 360 degrees, akin to the sine function. Crucially, \( \sec(\theta) \) is not defined when \( \cos(\theta) \) is zero, as it also represents division by zero.

One of the textbook exercise conditions presents the secant function being undefined, which hints to us that \( \cos(t) \) must be zero within the specified interval, leading to the solution of \( \frac{\pi}{2} \). Another part of the exercise sets \( \sec(t) = -\sqrt{2} \), inferring that \( \cos(t) \) must be \( -\frac{1}{\sqrt{2}} \). Understanding the relationship between secant and cosine and the idea of reciprocal functions is crucial in analyzing and solving such trigonometric conditions.

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Most popular questions from this chapter

This exercise provides an alternate method for determining the exact values of \(\cos \left(\frac{\pi}{3}\right)\) and \(\sin \left(\frac{\pi}{3}\right)\). The diagram to the right shows the terminal point \(P(x, y)\) for an arc of length \(t=\frac{\pi}{3}\) on the unit circle. The points \(A(1,0)\), \(B(0,1),\) and \(S\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\) are also shown. Notice that \(B\) is the terminal point of the \(\operatorname{arc} t=\frac{\pi}{2}\). From Exercise (10), we know that \(S\) is the terminal point of an arc of length \(\frac{\pi}{6}\). We now notice that the length of the arc from \(A\) to \(P\) is \(\frac{\pi}{3} .\) In addition, since the length of the arc from \(A\) to \(B\) is \(\frac{\pi}{2}\) and the and the length of the arc from \(B\) to \(P\) is \(\frac{\pi}{3} .\) This means that the distance from \(P\) to \(B\) is equal $$ \frac{\pi}{2}-\frac{\pi}{3}=\frac{\pi}{6} $$ Since both of the arcs have length \(\frac{\pi}{6},\) the distance from \(A\) to \(S\) is equal to the distance from \(P\) to \(B\). (a) Use the distance formula to determine the distance from \(A\) to \(S\). (b) Use the distance formula to write a formula (in terms of \(x\) and \(y\) ) for the distance from \(P\) to \(B\). (c) Set the distances from (a) and (b) equal to each other and solve the resulting equation for \(y\). To do this, begin by squaring both sides of the equation. In order to solve for \(y,\) it may be necessary to use the fact that \(x^{2}+y^{2}=1\) (d) Use the value for \(y\) in (c) and the fact that \(x^{2}+y^{2}=1\) to determine the value for \(x\). Explain why this proves that $$\cos \left(\frac{\pi}{3}\right)=\frac{1}{2} \text { and } \sin \left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}$$.

Use a process similar to the one in exercise (5) to complete each of the following: (a) Suppose it is known that \(-\frac{1}{4}<\sin (t)<0\) and that \(\cos (t)>0 .\) What can be concluded about \(\cos (t) ?\) (b) Suppose it is known that \(0 \leq \sin (t) \leq \frac{3}{7}\) and that \(\cos (t)<0 .\) What can be concluded about \(\cos (t) ?\)

A compact disc (CD) has a diameter of 12 centimeters (cm). Suppose that the CD is in a CD-player and is rotating at 225 revolutions per minute. What is the angular velocity of the CD (in radians per second) and what is the linear velocity of a point on the edge of the CD?

Convert each of the following radian measurements for angles into degree measures for the angles. When necessary, write each result as a 4 decimal place approximation. (a) \(\frac{3}{8} \pi\) radians (b) \(\frac{9}{7} \pi\) radians (c) \(-\frac{7}{15} \pi\) radians (d) 1 radian (e) 2.4 radians (f) 3 radians

Determine the quadrant that contains the terminal point of each given arc with initial point (1,0) on the unit circle. (a) \(\frac{7 \pi}{4}\) (b) \(-\frac{7 \pi}{4}\) (c) \(\frac{3 \pi}{5}\) (d) \(\frac{-3 \pi}{5}\) (e) \(\frac{7 \pi}{3}\) (f) \(\frac{-7 \pi}{3}\) (g) \(\frac{5 \pi}{8}\) (h) \(\frac{-5 \pi}{8}\) (i) 2.5 (j) -2.5 (k) 3 (1) \(3+2 \pi\) (m) \(3-\pi\) (n) \(3-2 \pi\)

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