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Chapter 5: Problem 8

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

Short Answer

Expert verified
(a) \(\sin \theta = -\frac{2\sqrt{6}}{5}\) (b) \(\tan \theta = -2\sqrt{6}\)

Step by step solution

01

Recall the Pythagorean identity

Recall the Pythagorean identity for trigonometric functions: \(\sin^2 \theta + \cos^2 \theta = 1\). Since we know the value of \(\cos \theta\), we can substitute this into the equation and then solve for \(\sin \theta\).
02

Determine the sine value of 胃

Given \(\cos \theta =\frac{1}{5}\), let's substitute this value into the Pythagorean identity: \(\sin^2 \theta + \left(\frac{1}{5}\right)^2 = 1\) Now, solve for \(\sin^2 \theta\): \(\sin^2 \theta = 1 - \left(\frac{1}{5}\right)^2\) \(\sin^2 \theta = 1 - \frac{1}{25}\) \(\sin^2 \theta = \frac{24}{25}\) Take the square root of both sides, keeping in mind that since \(-\frac{\pi}{2}<\theta<0\), the sine value will be negative: \(\sin \theta = -\sqrt{\frac{24}{25}}\) So, \(\sin \theta = -\frac{2\sqrt{6}}{5}\).
03

Determine the tangent value of 胃

Recall the definition of the tangent function: \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). We have the values for \(\sin \theta\) and \(\cos \theta\), so let's substitute them into the equation: \(\tan \theta = \frac{-\frac{2\sqrt{6}}{5}}{\frac{1}{5}}\) Now, simplify the expression by multiplying the numerator and denominator by 5: \(\tan \theta = \frac{-2\sqrt{6}}{1}\) So, \(\tan \theta = -2\sqrt{6}\). In conclusion, the solutions to the exercise are: (a) \(\sin \theta = -\frac{2\sqrt{6}}{5}\) (b) \(\tan \theta = -2\sqrt{6}\)

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Most popular questions from this chapter

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 4 and 5 in Section 6.3.] $$ \sin \frac{5 \pi}{12} $$

Suppose \(u\) and \(v\) are in the interval \(\left(\frac{\pi}{2}, \pi\right),\) with \(\tan u=-2\) and \(\tan v=-3\) Find exact expressions for the indicated quantities. $$ \cos (v-6 \pi) $$

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

Show that $$ (\cos \theta+\sin \theta)^{2}=1+2 \cos \theta \sin \theta $$ for every number \(\theta\). [Expressions such as \(\cos \theta \sin \theta\) mean $$ (\cos \theta)(\sin \theta), \operatorname{not} \cos (\theta \sin \theta) .] $$

Suppose \(u\) and \(v\) are in the interval \(\left(\frac{\pi}{2}, \pi\right),\) with \(\tan u=-2\) and \(\tan v=-3\) Find exact expressions for the indicated quantities. $$ \tan \left(\frac{\pi}{2}-u\right) $$
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