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Chapter 6: Problem 19

Find the exact values of the sine, cosine, and tangent of the angle. $$285^{\circ}$$

Short Answer

Expert verified
The exact sine, cosine, and tangent values for the angle \(285^{\circ}\) are: \(\sin(285^{\circ}) = -(\sqrt{6} + \sqrt{2}) / 4\), \(\cos(285^{\circ}) = (\sqrt{6} - \sqrt{2}) / 4\), and \(\tan(285^{\circ}) = - (2 + \sqrt{3})\).

Step by step solution

01

Find the Reference Angle

The angle of \(285^{\circ}\) lies in the IV quadrant. To find the reference angle, subtract the full \(360^{\circ}\) until you get an angle less than \(90^{\circ}\). So: \(360^{\circ} - 285^{\circ} = 75^{\circ}\)
02

Find the Trigonometric Values

Now you need to find the sine, cosine, and tangent of the reference angle \(75^{\circ}\). We know that: \(\sin(75^{\circ}) = \sqrt{6} + \sqrt{2} \) over \(4\), \(\cos(75^{\circ}) = \sqrt{6} - \sqrt{2} \) over \(4\), and \(\tan(75^{\circ}) = 2 + \sqrt{3}\). But these are not the final values. They are the values for the reference angle only.
03

Conclusion

In the 4th quadrant, sine is negative, cosine is positive and tangent is negative. So, for the original angle \(285^{\circ}\), the trigonometric values will be: \(\sin(285^{\circ}) = - (\sqrt{6} + \sqrt{2}) \) over \(4\), \(\cos(285^{\circ}) = \sqrt{6} - \sqrt{2} \) over \(4\), and \(\tan(285^{\circ}) = - (2 + \sqrt{3})\).

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