Using Identities to Find Exact Trig Values Exercise 4
Exercises
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.
\[\begin{aligned} \sin{\frac{30^\circ}{2}} &= \sqrt{\frac{1-\cos{30^\circ}}{2}} \\ &= \sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}} \\ &= \sqrt{\frac{2-\sqrt{3}}{4}} \\ &= \frac{1}{2}\sqrt{2-\sqrt{3}} \end{aligned}\]
\[\begin{aligned} \cos{\frac{30^\circ}{2}} &= \sqrt{\frac{1-\cos{30^\circ}}{2}} \\ &= \sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}} \\ &= \sqrt{\frac{2+\sqrt{3}}{4}} \\ &= \frac{1}{2}\sqrt{2+\sqrt{3}} \end{aligned}\] \[\begin{aligned} \text{ } \\ \tan{\frac{30^\circ}{2}} &= \frac{1-\cos{30^\circ}}{\sin{30^\circ}} \\ &= \frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}} \\ &= 2\left( \frac{2-\sqrt{3}}{2} \right) \\ &= 2-\sqrt{3} \end{aligned}\]
