9.3 Double-angle, half-angle, and reduction formulas

If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given $\tan \theta =-\frac{3}{4},$ such that $\theta$ is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because $\theta$ is in the second quadrant, the adjacent side is on the x -axis and is negative. Use the Pythagorean Theorem    to find the length of the hypotenuse:

Now we can draw a triangle similar to the one shown in [link] .

1. Let’s begin by writing the double-angle formula for sine.
   $$\sin (2\theta )=2\sin \theta \cos \theta$$

   We see that we to need to find $\sin \theta$ and $\cos \theta .$ Based on [link] , we see that the hypotenuse equals 5, so $\sin \theta =\frac{3}{5},$ and $\cos \theta =-\frac{4}{5}.$ Substitute these values into the equation, and simplify.

   Thus,

   $$\begin{array}{ccc}\hfill \sin (2\theta )& =& 2\left(\frac{3}{5}\right)\left(-\frac{4}{5}\right)\hfill \\ & =& -\frac{24}{25}\hfill \end{array}$$
2. Write the double-angle formula for cosine.
   $$\cos \left(2\theta \right)={\cos }^2\theta -{\sin }^2\theta$$

   Again, substitute the values of the sine and cosine into the equation, and simplify.

   $$\begin{array}{ccc}\hfill \cos (2\theta )& =& {\left(-\frac{4}{5}\right)}^2-{\left(\frac{3}{5}\right)}^2\hfill \\ & =& \frac{16}{25}-\frac{9}{25}\hfill \\ & =& \frac{7}{25}\hfill \end{array}$$
3. Write the double-angle formula for tangent.
   $$\tan (2\theta )=\frac{2\tan \theta }{1-{\tan }^2\theta }$$

   In this formula, we need the tangent, which we were given as $\tan \theta =-\frac{3}{4}.$ Substitute this value into the equation, and simplify.

   $$\begin{array}{ccc}\hfill \tan (2\theta )& =& \frac{2\left(-\frac{3}{4}\right)}{1-{\left(-\frac{3}{4}\right)}^2}\hfill \\ & =& \frac{-\frac{3}{2}}{1-\frac{9}{16}}\hfill \\ & =& -\frac{3}{2}\left(\frac{16}{7}\right)\hfill \\ & =& -\frac{24}{7}\hfill \end{array}$$