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We have been given a function to differentiate:

$$f(x) = \arcsin \left(\frac{5x + 12\sqrt{1-x^2}}{13}\right)$$

My teacher told me the method to substitute $ x= \sin\vartheta$ which would simplify the argument of $\arcsin$ to $$\frac{5}{13}\sin \vartheta + \frac{12}{13}\cos \vartheta $$ and further into $\sin(\vartheta + \alpha)$ where $ \alpha = \arctan\left(\frac{12}{5}\right)$ therefore the function gets reduced into $f(x) = \arctan\left(\frac{12}{5}\right) + \arcsin(x)$ giving $$f'(x) = \frac{1}{\sqrt{1-x^2}}$$. However when I substitute $x = \cos\vartheta $, the argument reduces to : $$\frac{5}{13}\cos\vartheta + \frac{12}{13}\sin\vartheta$$ and further into $\sin(\alpha + \vartheta)$ where $\alpha = \arctan\left(\frac{5}{12}\right)$ but this time the function gets reduced to $$f(x) = \arctan\left(\frac{5}{12}\right) + \arccos(x)$$ thus $$f'(x) = \frac{-1}{\sqrt{1-x^2}}$$ Hence we obtain two different derivatives for the same function and I cannot understand why. I tried plotting the argument and different simplification and got that they are not always equal but cannot figure out the reason.

Divyansh Tripathi
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  • $f(x)$ always takes values in $[-\pi/2,\pi/2].$

    Is that true for $\arcsin(x)+\arctan(12/5)?$

     – Thomas Andrews Apr 21 '20 at 09:15
  • No it is not the case. Does it mean the substitution is invalid? – Divyansh Tripathi Apr 21 '20 at 09:20
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    Use https://math.stackexchange.com/questions/672575/proof-for-the-formula-of-sum-of-arcsine-functions-arcsin-x-arcsin-y – lab bhattacharjee Apr 21 '20 at 09:26
  • What definition of $\arcsin x$ and $\sqrt x$ are used in this problem? If they are the "standard" real definitions, then the expression is not differentiable at $x=\frac5{13}$, where the argument of $\arcsin$ attain the value $1$. Observe that the derivative changes sign by transition over this point, so that either one or the other of your formulas are valid. – user Apr 21 '20 at 10:05
  • The definitions are standard. But instead of a specific point I wanted to know why is that we arrive at two different derivative of the same function and how do we know which is correct – Divyansh Tripathi Apr 21 '20 at 13:31
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    My comment answers exactly this question (probably you did not read it carefully). And the answer is: no of the formula is correct in the whole range $-1\le x\le 1$. Instead, one formula is valid on the left of $x=\frac5{13}$, and the other - on the right of it. – user Apr 21 '20 at 16:37
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    Thanks! Now I understand. I really did not read it carefully enough. – Divyansh Tripathi Apr 21 '20 at 16:52

3 Answers3

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You are right up to the point

$$f(x)=\arcsin[\sin(\theta +\alpha)],\>\>\>\>\>\theta \in [-\frac\pi2, \frac\pi2],\> \alpha =\arctan\frac{12}5$$

Note that the range of $f(x)=\arcsin()\in [-\frac\pi2, \frac\pi2]$ and $$\theta+\alpha \in [-\frac\pi2+\alpha, \frac\pi2+\alpha]$$

is outside the range for $ \theta>\frac\pi2-\alpha$, hence $f(x) \ne \theta+\alpha$. Instead, for the subdomain $ \theta\in[\frac\pi2-\alpha, \frac\pi2]$, the expression $\theta+\alpha -\pi \in [-\frac\pi2, 0]$ is in the range and

$$f(x) =\arcsin[\sin(\theta +\alpha)] =-\arcsin[\sin(\theta +\alpha-\pi)]=-(\theta +\alpha-\pi) $$

Therefore,

$$\begin{align} & f(x)= \theta +\alpha , \>\>\>\>\> \theta\in [-\frac\pi2, \frac\pi2-\alpha)\\ & f(x)= -(\theta +\alpha-\pi), \>\>\>\>\> \theta\in [\frac\pi2-\alpha,\frac\pi2]\\ \end{align}$$

and, with $\sin(\frac\pi2-\alpha)= \frac5{13}$,

$$\begin{align} & f’(x)=\theta’(x)= ( \arcsin x )’= \frac1{\sqrt{1-x^2}}, \>\>\>\>\> x\in [-1, \frac5{13})\\ & f’(x) =-\theta’(x)= -\frac1{\sqrt{1-x^2}}, \>\>\>\>\> x\in [\frac5{13},1]\\ \end{align}$$

i.e. it’s derivative has different function forms for the two sub domains. Similar analysis can be applied to the substitution $x=\cos \theta$, which leads to the same result above.

Quanto
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HINT.-$(5,12,13)$ is a Pythagorean triple so what is the relation between $\arctan\left(\frac{12}{5}\right)$ and $\arctan\left(\frac{5}{12}\right)$?

Piquito
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  • i see. therefore $\arctan\left(\frac{5}{12}\right) = \frac{\pi}{2} - \arctan\left(\frac{12}5\right)$ which means both simplifications are equivalent. Then how it is that their derivatives are different? – Divyansh Tripathi Apr 21 '20 at 15:27
  • Putting $\dfrac{5}{13}=\cos\alpha$, you get first $f(x)=\arcsin(\sin(\alpha+\vartheta))$ and secondly you get $f(x)=\arcsin(\cos(\alpha-\vartheta))$ – Piquito Apr 21 '20 at 16:07
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I assume that the reason for the "paradox" was clarified already in comments, but possibly there is still a need to go a little bit deeper into details:

Your error has its root in the assumption: $$ \arcsin(\sin x)=x. $$ However the equality is valid only in the range $-\frac\pi2\le x\le \frac\pi2$ whereas the correct expression for all real $x$ is: $$ \arcsin(\sin x)=(-1)^m\left(x-m\pi\right),\quad\text{with}\quad m=\left\lfloor\frac{x}\pi+\frac12\right\rfloor.\tag1 $$

In view of this the expression resulting from the substitution $x=\sin\vartheta$ reads: $$\arcsin\left(\sin\left(\vartheta+\arcsin\frac{12}{13}\right)\right)=\begin{cases} \frac\pi2-\arcsin\frac5{13}+\vartheta;& \vartheta\le \arcsin\frac{5}{13}\\ \frac\pi2+\arcsin\frac{5}{13}-\vartheta;& \vartheta\ge \arcsin\frac{5}{13} \end{cases} $$ or $$ f(x)=\begin{cases} \frac\pi2-\arcsin\frac5{13}+\arcsin x;& x\le\frac{5}{13}\\ \frac\pi2+\arcsin\frac{5}{13}-\arcsin x;& x\ge\frac{5}{13}. \end{cases} $$ Now it should be obvious that the derivative is discontinuous at $x=\frac5{13}$ taking the values $$ f'(x)=\pm\frac1{\sqrt{1-x^2}}, $$ to the left and to the right from the point $x=\frac5{13}$.

The same result will be obtained with the substitution $x=\cos\vartheta$ as well.

user
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