# Inverse Trigonometry Function

• Aug 1st 2008, 08:29 AM
Simplicity
Inverse Trigonometry Function
Question:

$\displaystyle f(x) = \sin ^{-1} \left(\frac{x}{a}\right) \implies f'(x) = \frac{1}{\sqrt{a^2-x^2}}$

$\displaystyle f(x) = \cos ^{-1} \left(\frac{x}{a}\right) \implies f'(x) = - \frac{1}{\sqrt{a^2-x^2}}$

These two seem to imply that $\displaystyle \sin ^{-1} x + \cos ^{-1} x= 0$. Why is this not, in fact, being implies? The answer lies in considering constants of integral.
__________________
Problem Faced:

I don't see why this isn't the case. For the identity:

$\displaystyle \sin ^2x + \cos ^2x = 0$ differentiated gives $\displaystyle 2\sin x\cos x - 2\sin x \cos x = 0$

So, when $\displaystyle \sin ^{-1} x + \cos ^{-1} x= 1$ is differentiated, it also gives the same result $\displaystyle \left(\frac{1}{\sqrt{a^2-x^2}}\right)$ but change of sign so why is that not a fact? (Worried) Thanks in advance.
• Aug 1st 2008, 09:01 AM
Soroban
Hello, Air!

Your thinking (and your notation) is a little fuzzy . . .

Quote:

$\displaystyle f(x) \:= \:\sin^{-1}\left(\frac{x}{a}\right) \quad\Rightarrow\quad f'(x) \:= \:\frac{1}{\sqrt{a^2-x^2}}$

$\displaystyle f(x) \:= \:\cos^{-1}\left(\frac{x}{a}\right) \quad\Rightarrow\quad f'(x) \:= \:\frac{{\color{red}-}1}{\sqrt{a^2-x^2}}$

These two seem to imply that $\displaystyle \sin ^{-1} x + \cos ^{-1} x= 0$. . . . . no

It says that the sum of their derivatives is zero.

$\displaystyle \text{Let: }\;\begin{array}{ccc}\alpha & = & \sin^{-1}\left(\frac{x}{a}\right) \\ \\[-3mm] \beta &=& \cos^{-1}\left(\frac{x}{a}\right) \end{array}$

We have this right triangle:
Code:

                        *                     * β|               a  *    |               *        | x             *          |         * α            |       * - - - - - - - - *

We see that $\displaystyle \alpha$ and $\displaystyle \beta$ are in the same right triangle.
. . They are complementary: .$\displaystyle \alpha + \beta \:=\:\frac{\pi}{2}$

So: .$\displaystyle g(x) \;=\;\sin^{-1}\left(\frac{x}{a}\right) + \cos^{-1}\left(\frac{x}{a}\right) \;=\;\frac{\pi}{2}$

Now differntiate . . .
. . and you'll see why the sum of the derivatives is zero.

• Aug 1st 2008, 09:06 AM
Jhevon
Quote:

Originally Posted by Air
Question:

$\displaystyle f(x) = \sin ^{-1} \left(\frac{x}{a}\right) \implies f'(x) = \frac{1}{\sqrt{a^2-x^2}}$

$\displaystyle f(x) = \cos ^{-1} \left(\frac{x}{a}\right) \implies f'(x) = {\color{red}-}~\frac{1}{\sqrt{a^2-x^2}}$ <------forgot the minus sign

These two seem to imply that $\displaystyle \sin ^{-1} x + \cos ^{-1} x= 0$. Why is this not, in fact, being implies? The answer lies in considering constants of integral.
__________________
Problem Faced:

I don't see why this isn't the case. For the identity:

$\displaystyle \sin ^2x + \cos ^2x = {\color{red}1}$ differentiated gives $\displaystyle 2\sin x\cos x - 2\sin x \cos x = 0$

So, when $\displaystyle \sin ^{-1} x + \cos ^{-1} x= 0$ is differentiated, it also gives the same result $\displaystyle \left(\frac{1}{\sqrt{a^2-x^2}}\right)$ but change of sign so why is that not a fact? (Worried) Thanks in advance.

not exactly sure what you're after, but what this implies is that $\displaystyle (\sin^{-1} x)' + (\cos^{-1} x)' = 0$ which is true. the example you gave with $\displaystyle \sin^2 x + \cos^2 x$ is a counter-example. since that quantity is 1, while the sum of the derivatives of each term is zero.

as far as going by the constants of integrals are concerned, this may be what they are after.

they seem to be hinting that since $\displaystyle (\sin^{-1} x)' + (\cos^{-1} x)' = 0$

then $\displaystyle \int \Bigg[ (\sin^{-1} x)' + (\cos^{-1} x)' \Bigg]~dx = \sin^{-1} x + \cos^{-1} x = \int 0~dx$

however, we cannot assume that $\displaystyle \int 0~dx = 0$ (this is what would imply $\displaystyle \sin^{-1} x + \cos^{-1} x = 0$). the fact is, $\displaystyle \int 0~dx = C$, an arbitrary constant, and thus, the implication in question does not follow