9.3 Double-angle, half-angle, and reduction formulas  (Page 3/8)

We will work on simplifying the left side of the equation:

Since $\mathrm{15°}=\frac{\mathrm{30°}}{2},$ we use the half-angle formula for sine:

Remember that we can check the answer with a graphing calculator.

Using the given information, we can draw the triangle shown in [link] . Using the Pythagorean Theorem, we find the hypotenuse to be 17. Therefore, we can calculate $\sin \alpha =-\frac{8}{17}$ and $\cos \alpha =-\frac{15}{17}.$

1. Before we start, we must remember that if $\alpha$ is in quadrant III, then $\mathrm{180°}<\alpha <\mathrm{270°},$ so $\frac{\mathrm{180°}}{2}<\frac{\alpha }{2}<\frac{\mathrm{270°}}{2}.$ This means that the terminal side of $\frac{\alpha }{2}$ is in quadrant II, since $\mathrm{90°}<\frac{\alpha }{2}<\mathrm{135°}.$

   To find $\sin \frac{\alpha }{2},$ we begin by writing the half-angle formula for sine. Then we substitute the value of the cosine we found from the triangle in [link] and simplify.

   $$\begin{array}{ccc}\hfill \sin \frac{\alpha }{2}& =& \pm \sqrt{\frac{1-\cos \alpha }{2}}\hfill \\ & =& \pm \sqrt{\frac{1-\left(-\frac{15}{17}\right)}{2}}\hfill \\ & =& \pm \sqrt{\frac{\frac{32}{17}}{2}}\hfill \\ & =& \pm \sqrt{\frac{32}{17}\cdot \frac{1}{2}}\hfill \\ & =& \pm \sqrt{\frac{16}{17}}\hfill \\ & =& \pm \frac{4}{\sqrt{17}}\hfill \\ & =& \frac{4\sqrt{17}}{17}\hfill \end{array}$$

   We choose the positive value of $\sin \frac{\alpha }{2}$ because the angle terminates in quadrant II and sine is positive in quadrant II.

2. To find $\cos \frac{\alpha }{2},$ we will write the half-angle formula for cosine, substitute the value of the cosine we found from the triangle in [link] , and simplify.
   $$\begin{array}{ccc}\hfill \cos \frac{\alpha }{2}& =& \pm \sqrt{\frac{1+\cos \alpha }{2}}\hfill \\ & =& \pm \sqrt{\frac{1+\left(-\frac{15}{17}\right)}{2}}\hfill \\ & =& \pm \sqrt{\frac{\frac{2}{17}}{2}}\hfill \\ & =& \pm \sqrt{\frac{2}{17}\cdot \frac{1}{2}}\hfill \\ & =& \pm \sqrt{\frac{1}{17}}\hfill \\ & =& -\frac{\sqrt{17}}{17}\hfill \end{array}$$

   We choose the negative value of $\cos \frac{\alpha }{2}$ because the angle is in quadrant II because cosine is negative in quadrant II.

3. To find $\tan \frac{\alpha }{2},$ we write the half-angle formula for tangent. Again, we substitute the value of the cosine we found from the triangle in [link] and simplify.
   $$\begin{array}{ccc}\hfill \tan \frac{\alpha }{2}& =& \pm \sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }}\hfill \\ & =& \pm \sqrt{\frac{1-(-\frac{15}{17})}{1+(-\frac{15}{17})}}\hfill \\ & =& \pm \sqrt{\frac{\frac{32}{17}}{\frac{2}{17}}}\hfill \\ & =& \pm \sqrt{\frac{32}{2}}\hfill \\ & =& -\sqrt{16}\hfill \\ & =& \mathrm{-4}\hfill \end{array}$$

   We choose the negative value of $\tan \frac{\alpha }{2}$ because $\frac{\alpha }{2}$ lies in quadrant II, and tangent is negative in quadrant II.