In our text book's(Higher Math 1st Paper-by S U Ahmed) differentiation chapter there is a section about replacing $x$(inside inverse trigonometric function) with trigonometric functions. A example problem was $\frac{d}{dx}\sin^{-1}\left(2x\sqrt{1-x^2}\right)$ and the solution given is;

Let \begin{align*} y &=\sin^{-1}\left(2x\sqrt{1-x^2}\right)\\ &= \sin^{-1}\left(2\sin \theta \sqrt{1-\sin^2 \theta}\right)\\ &=\sin^{-1}\left(2\sin \theta \cos \theta \right)\\ &=\sin^{-1}(\sin 2\theta )\\ &=2\theta\\ &=2 \sin^{-1}x \end{align*} Now, \begin{align*} \frac{d}{dx}\sin^{-1}\left(2x\sqrt{1-x^2}\right)&=\frac{d}{dx}2 \sin^{-1}x\\ &=\dfrac{2}{\sqrt{1-x^2}} \end{align*} But plotting two functions reveals the differentiation is not actually correct. If we differentiate by parts the answer would be $\frac{2\left(-2x^{2}+1\right)}{\sqrt{1-4x^{2}\left(1-x^{2}\right)}\sqrt{1-x^{2}}}$

[Plotted is Desmos1

Now my question is why this solution is wrong?

My guess: May be this is because replacing $x$ with $\sin \theta$ changes the range of $x$ from $(-\infty,\infty )$ to $[-1,1]$ and may be it causes some issue.


Your guess is wrong. The range of x in original equation is [-1,1] to make the thing inside square root positive.

The problem starts when author writes sin^-1(sin(2theta)) = 2theta. This is wrong. Read about inverse trigonometry functions. The wrong thing is this is true for only some values of theta. For example, put theta = 60 degrees

  • $\begingroup$ the range of $x$ is $[-0.7071,0.7071]$ as the domain of $\sin\alpha$ is $[-1,1]$ so $2x\sqrt{1-x^}=[-1,1]$ $\endgroup$ – Soyeb Jim Jun 13 '20 at 6:42
  • 1
    $\begingroup$ Let f(x) = 2*x*sqrt(1-x^2). Then x=+-1/sqrt(2) is the point of local maxima where the value of f(x) gets 1 before and after which its value decreases. $\endgroup$ – Aditya Kumar Jun 13 '20 at 11:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.