# Show that the expressions $\sin^{-1}(\frac{1}{\sqrt{x}})$ and $\frac{1}{\sqrt{x}}$ are the same for big values

How can you show that the expressions $$\sin^{-1}(\frac{1}{\sqrt{x}})$$ and $$\frac{1}{\sqrt{x}}$$ for big values are the same?

The opposite side of a triangle is given with $$1/\sqrt(x)$$, the angle between the hypotenuse and the the opposite side can be calculated by $$sin^{-1}(1/\sqrt(x))$$. For large x that seems to be correct, as a review with some inserted values in the calculator has given. Can one also show this connection differently? Maybe graphically?

• The wording is important - for large values they are approximately the same. Note that the Taylor expansion for arcsin involves a sum of odd powers, and for small values, those powers become negligible. – TheSimpliFire Mar 16 '19 at 8:00
• Hint: $\sin'(0)=1$. – Wolfgang Kais Mar 16 '19 at 8:04
• @TheSimpliFire if I understand you right, you mean: $sin^{-1}(x)=x-\frac{1}{6}x^3+\frac{3}{40}x^5+...$. Then this means for $sin^{-1}(1/\sqrt{x})$ something like that: $1/\sqrt(x)+-\frac{1}{6}(1/\sqrt(x))^3+...$ right? – P_Gate Mar 16 '19 at 8:16
• Yes, and when $x\to\infty$, $1/\sqrt x\to0$ so powers of three, five etc become extremely small. – TheSimpliFire Mar 16 '19 at 8:20
• If you agree with the approximation $\sin\theta\approx \theta$ when $\theta\to 0$, then the result you seek is just considering $\theta:=1/\sqrt x\to 0$ as $x\to\infty$ and applying $\arcsin$ on both sides. – learner Mar 16 '19 at 8:24

Consider the limit of their ratios: $$\lim_\limits{x\to +\infty} \frac{\arcsin \frac1{\sqrt{x}}}{\frac1{\sqrt{x}}}\stackrel{\frac1{\sqrt{x}}=t}{=}\lim_{t\to 0^+}\frac{\arcsin t}{t}\stackrel{L'H}=\lim_{t\to 0^+}\frac{\frac1{\sqrt{1-t^2}}}{1}=1.$$
If $$x$$ is large, then $$\frac1{\sqrt x}$$ is small and if $$y$$ is small, then $$\arcsin(y)$$ is approximately $$y$$ (because $$\arcsin(0)=0$$ and $$\arcsin'(0)=1$$). But they are never equal if $$y\neq0$$.
Not that: $$\lim_{x\rightarrow\infty}(\frac{1}{\sqrt{x}})=0$$
$$\Rightarrow \lim_{x\rightarrow\infty}(\sin^{-1}(\frac{1}{\sqrt{x}}))=\sin^{-1}(0)=0$$
Hence: $$\lim_{x\rightarrow\infty}(\frac{1}{\sqrt{x}})=\lim_{x\rightarrow\infty}(\sin^{-1}(\frac{1}{\sqrt{x}}))=0$$