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tags:
Cosine Fundamental Trigonometric Identity Sine Tangent Trigonometry Trigonometry Series

Using Identities to Find Exact Trig Values Exercise 4

Author: Hannah Solomon

The following problem is solved in this video. It is recommended that you try to solve the problem before watching the video. You can click "Reveal Answer" to see the answer to the problem.

Problem: Use half-angle identities to find the exact value of \(\sin(15^\circ)\), \(\cos(15^\circ)\), and \(\tan(15^\circ).\)

Solutions: \(\sin 15^\circ = \dfrac{1}{2}\sqrt{2-\sqrt{3}},\qquad \) \(\cos 15^\circ = \dfrac{1}{2}\sqrt{2+\sqrt{3}},\qquad\) \(\tan 15^\circ = 2-\sqrt{3}\)

 

Solution Method:

Half-Angle Identities

\[\begin{aligned} \sin{\frac{x}{2}} &= \pm \sqrt{\frac{1-\cos{x}}{2}} \\ \cos{\frac{x}{2}} &= \pm \sqrt{\frac{1+\cos{x}}{2}} \\ \hspace{.32in} \tan{\frac{x}{2}} &= \frac{1-\cos{x}}{\sin{x}} = \frac{\sin{x}}{1+\cos{x}} \end{aligned}\]

\(15^\circ\) is not a special angle, but it is half of \(30^\circ\), which is. So we can use the half-angle identities to find the trig values at \(15^\circ.\) Notice for \(\sin{\frac{x}{2}}\) and \(\cos{\frac{x}{2}}\), we can have a positive or negative sign. So we have to determine what quadrant our angle is in. \(15^\circ\) is in Q1, so (ASTC), all three ratios will have positive sign.

\[\begin{aligned} \sin{\frac{30^\circ}{2}} &= \sqrt{\frac{1-\cos{30^\circ}}{2}} \\ &= \sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}} \\ &= \sqrt{\frac{2-\sqrt{3}}{4}} \\ &= \frac{1}{2}\sqrt{2-\sqrt{3}} \end{aligned}\]

\[\begin{aligned} \cos{\frac{30^\circ}{2}} &= \sqrt{\frac{1-\cos{30^\circ}}{2}} \\ &= \sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}} \\ &= \sqrt{\frac{2+\sqrt{3}}{4}} \\ &= \frac{1}{2}\sqrt{2+\sqrt{3}} \end{aligned}\] \[\begin{aligned} \text{ } \\ \tan{\frac{30^\circ}{2}} &= \frac{1-\cos{30^\circ}}{\sin{30^\circ}} \\ &= \frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}} \\ &= 2\left( \frac{2-\sqrt{3}}{2} \right) \\ &= 2-\sqrt{3} \end{aligned}\]

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