# Trig Inverse and Differentiation

• Jun 15th 2011, 12:45 AM
BabyMilo
Trig Inverse and Differentiation
Prove $\displaystyle cos^{-1} 2x + sin^{-1} 2x = a$

$\displaystyle \frac{-1}{2}\leq x \leq \frac{1}{2}$

thank you.
• Jun 15th 2011, 01:36 AM
TheCoffeeMachine
Re: Trig Inverse and Differentiation
\displaystyle \begin{aligned}\bigg[\bigg(\sin^{-1}{2x}\bigg)+\bigg(\cos^{-1}{2x}\bigg)\bigg]' & = \bigg(\sin^{-1}{2x}\bigg)'+\bigg(\cos^{-1}{2x}\bigg)' \\& = \frac{(2x)'}{\sqrt{1-(2x)^2}}-\frac{(2x)'}{\sqrt{1-(2x)^2}} \\& = 0.\end{aligned}

Alternatively, you can find the exact value of $\displaystyle a$ by usual trigonometry.
• Jun 15th 2011, 01:51 AM
BabyMilo
Re: Trig Inverse and Differentiation
how do i find the value of a?
• Jun 15th 2011, 04:14 AM
Soroban
Re: Trig Inverse and Differentiation
Hello, BabyMilo!

Quote:

$\displaystyle \text{Prove: }\:\cos^{-1}(2x) + \sin^{-1}(2x) \:=\: a \qquad \text{-}\tfrac{1}{2}\:\leq\: x\: \leq\: \tfrac{1}{2}$

Let $\displaystyle \alpha \:=\:\cos^{-1}(2x) \quad\Rightarrow\quad \cos\alpha \,=\, 2x\;\;[1]$

Let $\displaystyle \beta \:=\:\sin^{-1}(2x) \quad\Rightarrow\quad \sin\beta \,=\,2x \;\;[2]$

From [1]: .$\displaystyle \cos\alpha \:=\:\frac{2x}{1} \:=\:\frac{adj}{hyp}$

Hence, $\displaystyle \alpha$ is in this right triangle.

Code:

                  *           1  *  *             *    *         * α      *       *  *  *  *  *             2x

From [2]: .$\displaystyle \sin\beta \:=\:\frac{2x}{1} \:=\:\frac{opp}{hyp}$

Hence, $\displaystyle \beta$ is in this right triangle:

Code:

            *           **       1  * *         *  * 2x         *  *       * β  *       * * * *

We see that the two triangles are congruent.

That is, $\displaystyle \alpha$ and $\displaystyle \beta$ are in the same right triangle!

Code:

                  *           1  * β*             *    *         * α      *       *  *  *  *  *             2x

Therefore: .$\displaystyle \alpha + \beta \:=\:\frac{\pi}{2} \quad\Rightarrow\quad \cos^{-1}(2x) + \sin^{-1}(2x) \:=\:\frac{\pi}{2}$