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Chapter 5: Problem 10

Use the given values to find the values (if possible) of all six trigonometric functions. $$\sin (-x)=-\frac{1}{3}, \quad \tan x=-\frac{\sqrt{2}}{4}$$

Short Answer

Expert verified
Based on the given information, the values of the six trigonometric functions are: \(\sin x = \frac{1}{3}\), \(\cos x = -\frac{2\sqrt{2}}{3}\), \(\tan x = -\frac{\sqrt{2}}{4}\), \(\csc x = 3\), \(\sec x = -\frac{3\sqrt{2}}{4}\), \(\cot x = -2\sqrt{2}\)

Step by step solution

01

Determine sine and cosine of the angle using given values

The sine of the negative angle is given as \(-\frac{1}{3}\). However, we know that \(\sin(-x) = -\sin x\). Hence, \(\sin x = -\sin(-x) = -(-\frac{1}{3}) = \frac{1}{3}\). The tangent of the angle is given as \(-\frac{\sqrt{2}}{4}\). We know that tangent is defined as the ratio of sine to cosine, i.e., \(\tan x = \frac{\sin x}{\cos x}\). Solving for \(\cos x\), we get \(\cos x = \frac{\sin x}{\tan x} = \frac{\frac{1}{3}}{-\frac{\sqrt{2}}{4}} = -\frac{4}{3\sqrt{2}}\)
02

Simplify the cosine of the angle

Simplify \(\cos x\) = -\(\frac{4}{3\sqrt{2}}\) by rationalizing the denominator: \(\cos x = -\frac{4\sqrt{2}}{6} = -\frac{2\sqrt{2}}{3}\)
03

Calculate the remaining trigonometric functions

The other trigonometric functions can be obtained from sine and cosine. The cosecant is the reciprocal of sine, so \(\csc x = \frac{1}{\sin x} = 3\). The secant is the reciprocal of cosine, so \(\sec x = \frac{1}{\cos x} = -\frac{3}{2\sqrt{2}} = -\frac{3\sqrt{2}}{4}\) and the tangent is the ratio of sine to cosine, which was given as \(\tan x = -\frac{\sqrt{2}}{4}\), so the cotangent is the reciprocal of the tangent, thus \(\cot x = -\frac{4}{\sqrt{2}} = -2\sqrt{2}\)

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