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Chapter 4: Problem 9

Suppose \(0<\theta<\frac{\pi}{2}\) and \(\cos \theta=\frac{1}{5} .\) Evaluate: (a) \(\sin \theta\) (b) \(\tan \theta\)

Short Answer

Expert verified
(a) \(\sin \theta = \sqrt{\frac{24}{25}}\) (b) \(\tan \theta = \sqrt{24}\)

Step by step solution

01

Use the Pythagorean Identity for sine and cosine

Recall the Pythagorean identity relating sine and cosine, which states that \(\sin^2\theta + \cos^2\theta = 1\). We are given \(\cos \theta = \frac{1}{5}\) and we need to find \(\sin \theta\). Therefore, substitute the value of \(\cos \theta\) into the identity: \[\sin^2\theta + \left(\frac{1}{5}\right)^2 = 1\]
02

Solve for \(\sin^2 \theta\)

Now we need to find the value of \(\sin^2 \theta\). From the equation in Step 1, solve for \(\sin^2\theta\): \[\sin^2\theta = 1 - \left(\frac{1}{5}\right)^2 = 1 - \frac{1}{25} = \frac{24}{25}\]
03

Solve for \(\sin \theta\)

Finally, we need to find the value of \(\sin \theta\), given \(\sin^2 \theta = \frac{24}{25}\). Since \(\theta\) is in the range \(0<\theta<\frac{\pi}{2}\), \(\sin \theta\) is positive, and we can take the positive square root: \[\sin\theta = \sqrt{\frac{24}{25}}\] So, \(\sin \theta = \sqrt{\frac{24}{25}}\). b) Finding \(\tan \theta\)
04

Use the identity for tangent in terms of sine and cosine

Recall the definition of the tangent function, which states that \(\tan\theta = \frac{\sin\theta}{\cos\theta}\). We have already found the values of \(\sin \theta\) and \(\cos \theta\). Therefore, we can use these values to find \(\tan \theta\).
05

Substitute values of sine and cosine into the tangent identity

We know that \(\sin\theta = \sqrt{\frac{24}{25}}\) and \(\cos\theta = \frac{1}{5}\). Substituting these values into the tangent identity, we get: \[\tan\theta = \frac{\sqrt{\frac{24}{25}}}{\frac{1}{5}}\]
06

Solve for \(\tan \theta\)

Simplify the fraction to find the value of \(\tan\theta\): \[\tan\theta = \frac{\sqrt{\frac{24}{25}}}{\frac{1}{5}} = 5\sqrt{\frac{24}{25}} = \sqrt{24}\] So, \(\tan \theta = \sqrt{24}\).

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

Suppose \(n\) is an integer. Find formulas for \(\sec (\theta+n \pi), \csc (\theta+n \pi)\), and \(\cot (\theta+n \pi)\) in terms of \(\sec \theta, \csc \theta,\) and \(\cot \theta\).

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.] \(\sin \left(-\frac{\pi}{12}\right)\)

Simplify the expression $$ (\tan \theta)\left(\frac{1}{1-\cos \theta}-\frac{1}{1+\cos \theta}\right) . $$

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