# Thread: [SOLVED] LHS = RHS - How on earth do I do this???

1. ## [SOLVED] LHS = RHS - How on earth do I do this???

Hey guys.

Could someone please help me with this question? How exactly do I prove $\frac{\pi}{4} -\frac{1}{4}\arctan(\frac{4}{3}) = \frac{\pi}{8} +\frac{1}{4}\arctan(\frac{3}{4})$

2. Hello, UltraGirl!

Prove: . $\frac{\pi}{4} -\frac{1}{4}\arctan\left(\frac{4}{3}\right) \;=\; \frac{\pi}{8} +\frac{1}{4}\arctan\left(\frac{3}{4}\right)$
Those two arctan expressions look similar . . . that's a clue.

. . $\begin{array}{cccccccc}\text{Let }\alpha &=& \arctan(\frac{4}{3}) & \Rightarrow & \tan\alpha &=& \frac{4}{3} \\ \\[-3mm] \text{Let }
\beta &=& \arctan(\frac{3}{4}) & \Rightarrow & \tan\beta &=& \frac{3}{4} \end{array}$

Look at the right triangles . . .

Code:
                  *                         *
* |                       * |
*   |                     * β |
*     | 4                 *     | 4
*       |                 *       |
* α       |               *         |
* - - - - - *             * - - - - - *
3                         3

See it?

The two angles are in the same right triangle.
. . They are complementary: . $\alpha + \beta \:=\:90^o$

That is: . $\arctan\left(\frac{4}{3}\right) + \arctan\left(\frac{3}{4}\right) \:=\:\frac{\pi}{2} \quad\Rightarrow\quad \arctan\left(\frac{4}{3}\right) \;=\;\frac{\pi}{2} - \arctan\left(\frac{3}{4}\right)$

Substitute into the left side:

. . $\frac{\pi}{4} - \frac{1}{4}\arctan\left(\frac{4}{3}\right)$

. . . . . $=\;\frac{\pi}{4} - \frac{1}{4}\left[\frac{\pi}{2}-\arctan\left(\frac{3}{4}\right)\right]$

. . . . . $=\;\frac{\pi}{4} - \frac{\pi}{8} + \frac{1}{4}\arctan\left(\frac{3}{4}\right)$

. . . . . $=\;\frac{\pi}{8} + \frac{1}{4}\arctan\left(\frac{3}{4}\right)$