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

• May 4th 2010, 10:25 AM
UltraGirl
[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 $\displaystyle \frac{\pi}{4} -\frac{1}{4}\arctan(\frac{4}{3}) = \frac{\pi}{8} +\frac{1}{4}\arctan(\frac{3}{4})$
• May 4th 2010, 12:54 PM
Soroban
Hello, UltraGirl!

Quote:

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

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

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: .$\displaystyle \alpha + \beta \:=\:90^o$

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

Substitute into the left side:

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

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

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

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