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Chapter 4: Problem 15

evaluate (if possible) the sine, cosine, and tangent at the real number. $$t=-\frac{\pi}{6}$$

Short Answer

Expert verified
The values of the functions at \( t=-\frac{\pi}{6} \) are: \(\sin(-\frac{\pi}{6}) = -\frac{1}{2}\), \(\cos(-\frac{\pi}{6}) = \frac{\sqrt{3}}{2}\), and \(\tan(-\frac{\pi}{6}) = -\frac{\sqrt{3}}{3}\).

Step by step solution

01

Identify the Quadrant

The real number \( t=-\frac{\pi}{6} \) corresponds to an angle of -30 degrees in standard position (counterclockwise from positive x-axis). So, the angle \( t \) is in the fourth quadrant.
02

Evaluate the Functions

In the fourth quadrant, cosine is positive and sine is negative. The reference angle (the acute angle made with the x-axis) is 30 degrees which is \(\frac{\pi}{6}\). From the special triangles, or the unit circle, we know that \(\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}\) and \(\sin(\frac{\pi}{6}) = \frac{1}{2}\). Thus, \(\cos(-\frac{\pi}{6}) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}\) and \(\sin(-\frac{\pi}{6}) = -\sin(\frac{\pi}{6}) = -\frac{1}{2}\)
03

Calculate Tangent

The tangent function is the ratio of sine to cosine. That is, \(\tan(-\frac{\pi}{6}) = \frac{\sin(-\frac{\pi}{6})}{\cos(-\frac{\pi}{6})} = \frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}} = -\frac{\sqrt{3}}{3}\).

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