<s>[INST] Why am I getting a different value for $\sin\left(2\tan^{-1}\frac{4}{3}\right)$ than my calculator? 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:
$$\sin\left(2\tan^{-1}\left(\frac{4}{3}\right)\right)$$
$$\sin\left(\tan^{-1}\left(\frac{2\times\frac{4}{3}}{1-(\frac{4}{3})^{2}}\right)\right)$$
$$[\text{Using the formula $2\arctan(x)=\arctan\left(\frac{2x}{1-x^2}\right)$}]$$
$$\sin\left(\tan^{-1}\left(\frac{-24}{7}\right)\right)$$
$$-\sin\left(\tan^{-1}\left(\frac{24}{7}\right)\right)$$
$$-\sin\left(\sin^{-1}\left(\frac{24}{25}\right)\right)$$
$$-\frac{24}{25}$$
Why am I getting a different answer than that of my calculator?
 [/INST] One has to keep in mind, which quadrant the angle lies in from the beginning, and not derive it from calculations. Your calculations will be prone to mistakes because trigonometric and inverse trigonometric functions are not one-to-one functions.
Notice that $\tan^{-1} 4/3 > \tan ^{-1} 1 = \pi/4$. So, $\theta = 2\tan^{-1} 4/3 > \pi/2$ and belongs to second quadrant. However $\tan^{-1} \text{(negative value)}$ lies in fourth quadrant. So the conversion
$$\sin\left( \underset{\text{II quadrant}}{2\tan^{-1}\left(\frac{4}{3}\right)}\right) \rightarrow \sin\left( \underset{\text{IV quadrant}} {\tan^{-1}\left(\frac{-24}{7}\right)} \right)$$
is incorrect. But it can be corrected as follows
$$\sin\left( \underset{\text{II quadrant}}{2\tan^{-1}\left(\frac{4}{3}\right)} \right) = \sin\left( \underset{\text{II quadrant}}{\pi + \tan^{-1}\left(\frac{-24}{7}\right)} \right)$$
Now calculation should give correct answer.
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