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Chapter 4: Problem 17

Suppose \(-\frac{\pi}{2}<\theta<0\) and \(\tan \theta=-3\). Evaluate: (a) \(\cos \theta\) (b) \(\sin \theta\)

Short Answer

Expert verified
(a) \(\cos \theta = \frac{-1}{\sqrt{10}}\) (b) \(\sin \theta = \frac{3}{\sqrt{10}}\)

Step by step solution

01

Form a right triangle with angle \(\theta\)

Since \(\tan \theta = \frac{\sin \theta}{\cos \theta} = -3\), we can see that the opposite and adjacent sides of angle \(\theta\) in a right triangle must have a ratio of \(-3\). In the given range of \(\theta (-\frac{\pi}{2}<\theta<0)\), we know that both sine and cosine are negative, so let the opposite side be 3 units and the adjacent side be -1 unit. This forms a right triangle with angle \(\theta\) having these sides ratio.
02

Calculate the hypotenuse

Using the Pythagorean theorem, we can calculate the hypotenuse of the triangle. Let the hypotenuse be denoted as \(h\). We have: \[h^2 = (-1)^2 + 3^2 = 1 + 9 = 10\Rightarrow h= \sqrt{10}\]
03

Calculate \(\cos \theta\)

Now that we have the hypotenuse, we can calculate the cosine of \(\theta\) using the ratio of the adjacent side to the hypotenuse: \[\cos \theta= \frac{adjacent}{hypotenuse} = \frac{-1}{\sqrt{10}}\] Thus, the value of \(\cos \theta\) is \(\frac{-1}{\sqrt{10}}\).
04

Calculate \(\sin \theta\)

Similarly, we can calculate the sine of \(\theta\) using the ratio of the opposite side to the hypotenuse: \[\sin \theta = \frac{opposite}{hypotenuse} = \frac{3}{\sqrt{10}}\] Thus, the value of \(\sin \theta\) is \(\frac{3}{\sqrt{10}}\). To summarize, the values of \(\cos \theta\) and \(\sin \theta\) are as follows: (a) \(\cos \theta = \frac{-1}{\sqrt{10}}\) (b) \(\sin \theta = \frac{3}{\sqrt{10}}\)

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Most popular questions from this chapter

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.] \(\cos \frac{5 \pi}{12}\)

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.] \(\sin \frac{25 \pi}{12}\)

Given that $$\cos 15^{\circ}=\frac{\sqrt{2+\sqrt{3}}}{2} \text { and } \sin 22.5^{\circ}=\frac{\sqrt{2-\sqrt{2}}}{2}$$ Find exact expressions for the indicated quantities. [These values for \(\cos 15^{\circ}\) and \(\sin 22.5^{\circ}\) will be derived in Examples 3 and 4 in Section 5.5.] \(\sec 15^{\circ}\)

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

Show that $$\cos \left(x+\frac{\pi}{2}\right)=-\sin x$$ for every number \(x\).
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