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Chapter 6: Problem 20

Find the exact values of the sine, cosine, and tangent of the angle. $$15^{\circ}$$

Short Answer

Expert verified
The exact values of the sine, cosine and tangent of 15 degrees are respectively: \(\frac{\sqrt{6}-\sqrt{2}}{4}\), \(\frac{\sqrt{6}+\sqrt{2}}{4}\), and \(2 - \sqrt{3}\).

Step by step solution

01

Find the exact value of sine

First, use the formula for the sine of a difference of two angles. \(sin (a - b) = sin a cos b - cos a sin b\). Using a=45 degrees and b=30 degrees, we get that \(sin(45 - 30) = sin(45)cos(30) - cos(45)sin(30)\) = \(\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}\cdot \frac{1}{2}\) = \(\frac{\sqrt{6}-\sqrt{2}}{4}\).
02

Find the exact value of cosine

Second, use the formula for the cosine of a difference of two angles. \(cos(a - b) = cos a cos b + sin a sin b\). Using a=45 degrees and b=30 degrees, we get that \(cos(45 - 30) = cos(45)cos(30) + sin(45)sin(30)\) = \(\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot \frac{1}{2}\) = \(\frac{\sqrt{6}+\sqrt{2}}{4}\).
03

Find the exact value of tangent

Third, apply the formula \(tan a = \frac{sin a}{cos a}\). We know that \(sin 15 = \frac{\sqrt{6}-\sqrt{2}}{4}\) and \(cos 15 = \frac{\sqrt{6}+\sqrt{2}}{4}\). Substituting these values we get \(tan 15 = \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}\). This, when simplified further becomes \(2 - \sqrt{3}\).

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