Question

Simplify $\sin^{-1} \left( \frac{x}{\sqrt{1 + x^2}} \right)$.

Hint:

Use trigonometric identities to simplify inverse trigonometric functions. In this case, $\sin^{-1} \left( \frac{x}{\sqrt{1 + x^2}} \right)$ simplifies to $\tan^{-1}(x)$.

Answer 1

The objective is to simplify the expression $\sin^{-1} \left( \frac{x}{\sqrt{1 + x^2}} \right)$. Let $\theta = \sin^{-1} \left( \frac{x}{\sqrt{1 + x^2}} \right)$. This implies $\sin(\theta) = \frac{x}{\sqrt{1 + x^2}}$. Using the Pythagorean identity $\cos^2(\theta) + \sin^2(\theta) = 1$, we substitute the value of $\sin(\theta)$: $\cos^2(\theta) + \left( \frac{x}{\sqrt{1 + x^2}} \right)^2 = 1$. This leads to $\cos^2(\theta) = 1 - \frac{x^2}{1 + x^2}$. Simplifying the right side yields $\cos^2(\theta) = \frac{1 + x^2 - x^2}{1 + x^2} = \frac{1}{1 + x^2}$. Therefore, $\cos(\theta) = \frac{1}{\sqrt{1 + x^2}}$. From $\sin(\theta) = \frac{x}{\sqrt{1 + x^2}}$ and $\cos(\theta) = \frac{1}{\sqrt{1 + x^2}}$, we can deduce that $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{x/\sqrt{1 + x^2}}{1/\sqrt{1 + x^2}} = x$. Thus, $\theta = \tan^{-1}(x)$. Consequently, $\sin^{-1} \left( \frac{x}{\sqrt{1 + x^2}} \right) = \tan^{-1}(x)$.