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

  1. #1
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    Exclamation [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})
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  2. #2
    Super Member

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    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 }<br />
\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)

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