Why am I getting the correct value for $\sin\left(2\tan^{-1}\frac{4}{3}\right)$ even though the usage of the formula is incorrect?

The expression:

$$\sin\left(2\tan^{-1}\left(\frac{4}{3}\right)\right)$$

Way 1:

If I punch the above expression in my calculator, I get $\frac{24}{25}$.

Way 2:

$$\sin\left(2\tan^{-1}\left(\frac{4}{3}\right)\right)$$

$$\sin\left(\sin^{-1}\left(\frac{2\times\frac{4}{3}}{1+(\frac{4}{3})^{2}}\right)\right)$$

$$[\text{Using $2\tan^{-1}x=\sin^{-1}\frac{2x}{1+x^2}$}]$$

$$\frac{2\times\frac{4}{3}}{1+(\frac{4}{3})^{2}}$$

$$\frac{24}{25}$$

My comments:

One of the conditions of $2\tan^{-1}x=\sin^{-1}\frac{2x}{1+x^2}$ is that $|x|\leq1$. In this case $|\frac{4}{3}|\nleq1$, but still I'm getting the correct answer using the formula. Why is that?

Related:

• Why am I getting a different value for $\sin\left(2\tan^{-1}\frac{4}{3}\right)$ than my calculator?
• Why does the equation with $2 \arctan(x)$ and other Inverse Trigonometric functions have weird conditions?

Solution 1:

When $x > 1,$ as in this case, $\arctan x > \frac\pi4,$ and therefore $2 \arctan x > \frac\pi2.$ That is, $2 \arctan x$ is in the second quadrant. But $\arcsin$ can only produce angles in the first and fourth quadrants.

There is a similar difficulty when $x < -1.$ That's why the formula $2\tan^{-1}x=\sin^{-1}\frac{2x}{1+x^2}$ is good only for $\lvert x\rvert \leq 1.$

But when $x > 1,$ you need a different formula. One correct formula is

$$2\tan^{-1}x = \pi - \sin^{-1}\frac{2x}{1+x^2}.$$

And $\sin(\pi - \theta) = \sin \theta,$ so you get the same result in the end.

Postscript: The following is a kind of extended comment on the question. You could also view it as an answer to a question that was not literally asked.

Note that for any real number $x$ (without restriction) we have \begin{align} \sin\left(\tan^{-1} x\right) &= \frac{x}{\sqrt{1+x^2}}, \\ \cos\left(\tan^{-1} x\right) &= \frac{1}{\sqrt{1+x^2}}. \end{align}

So if $\tan^{-1} x = \alpha$ then \begin{align} \sin\left(2\tan^{-1} x\right) = \sin(2\alpha) &= 2\sin\alpha \cos\alpha \\ &= 2\left(\frac{x}{\sqrt{1+x^2}}\right)\left(\frac{1}{\sqrt{1+x^2}}\right) \\[1ex] &= \frac{2x}{1+x^2}. \end{align}

So this is in fact a general formula that does not require any restrictions or special cases in order to deal with the quadrants in which the trig functions fall.