Problem 19 - Algebra and Trigonometry Real Mathematics, Real People

Question

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

Step-by-Step Solution

Answer

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 1: 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: \(360^{\circ} - 285^{\circ} = 75^{\circ}\)

Step 2: 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.

Step 3: 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 \(\tan(285^{\circ}) = - (2 + \sqrt{3})\).