# Thread: Trig Inverse and Differentiation

1. ## Trig Inverse and Differentiation

Prove $cos^{-1} 2x + sin^{-1} 2x = a$

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

thank you.

2. ## Re: Trig Inverse and Differentiation

\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 $a$ by usual trigonometry.

3. ## Re: Trig Inverse and Differentiation

how do i find the value of a?

4. ## Re: Trig Inverse and Differentiation

Hello, BabyMilo!

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

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

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

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

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

Code:
                  *
1   *  *
*     *
* α      *
*  *  *  *  *
2x

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

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

Code:
            *
**
1  * *
*  * 2x
*   *
* β  *
* * * *

We see that the two triangles are congruent.

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

Code:
                  *
1   * β*
*     *
* α      *
*  *  *  *  *
2x

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