Finding $\sin^{-1}(x)$ without using a calculator I don't understand how to compute $\sin^{-1} (0.6293)$, to figure out the angle without using a calculator. I understand how to find the answer if I use a calculator but I don't understand the necessary steps to solve the problem without a calculator. 
Am I wrong to assume you have to know at least the value of the opposite or hypotenuse length of the given triangle? Can you find the angle if you only have the sine and nothing else and you don't have a calculator?  
 A: No, you don't need the hypotenuse or any other side length to find $\sin^{-1}(x)$.
One way to do that will be to use the Maclaurin series for $\sin^{-1}(x)$,
$$
\sin^{-1}(x) = x + \frac{1}{6} x^3 + \frac{3}{40} x^5 + \frac{5}{112} x^7 + \frac{35}{1152}x^9+\cdots
$$
Using the first 4 terms gives $0.67998$, while the actual value is $0.68065$. That's an error of $0.0984\%$.
One way to simplify calculation is to find $x^2,x^4,x^6\cdots$, then find $1 + \frac{1}{6} x^2 + \frac{3}{40} x^4 + \frac{5}{112} x^6+\cdots$, and finally multiply with $x$.
A: There are several approaches to this problem, all of them are a pain in the neck without a calculator. The first approach you can try is to construct a triangle which could give you a decent triangle or if you know some calculus, you can use the Taylor series of $\arcsin x$ and use that. I'll illustrate the Taylor series approach. 
The Taylor series of $\arcsin x$ is 
$$x + \frac{x^3}{6} + \frac{x^5}{40} + O(x^7)$$. 
Using this information, if we plug in $.6293$ into the above then we'll get an "okay" approximation. The more terms you add, the better but it will get harder and harder to find the powers of $.6293$. So, we have
$$\arcsin(.6923) = .6293 + \frac{.6293^3}{6} + \frac{.6293^5}{40} + O(.6293^7) \approx .6782$$
The real answer is $\approx .6806$ so you be the judge. 
A: Probably your best bet would be a Taylor/Maclaurin series:
$$
\sin^{-1}(x)= x + 1/6\cdot x^3 + 3/40\cdot x^5 + 5/112\cdot x^7 + O(x^9)
$$
A: Some reverse engineering: Since $0.6293203$ is near ${\sqrt{2}\over2}\doteq0.7071$ we write the requested angle $\alpha$ in the form $\alpha={\pi\over4}-\beta$, where $\beta$ should be thought of as small. 
The condition $\sin\alpha=0.6293$ then becomes
$$0.6293=\sin\bigl({\pi\over4}-\beta\bigr)={\sqrt{2}\over2}(\cos\beta-\sin\beta)\doteq{\sqrt{2}\over2}(1-\beta)\ .$$
Solving for $\beta$ we obtain
$$\beta\doteq1-\sqrt{2}\cdot 0.6293=0.1100\ldots\ .$$
This finally gives
$$\alpha={\pi\over4}-\beta\doteq0.675\ ,$$
which is pretty good, considering that the exact value is $=0.68065\ldots$.
A: Since $f^{-1}(x)$ is just the reflection of $f(x)$ across $y=x$, we can get $\sin^{-1} x$ by finding an approximation to $\sin x$, and then letting $y=x$.
The Taylor series of $\sin x$ around $x=0$ is $x-\frac{x^3}{3!}+\frac{x^5}{5!} + o(x^6)$. Therefore, $\sin^{-1} x = \sin x $ $\approx y-\frac{y^3}{3!}+\frac{y^5}{5!}$. Substituting $x=0.6293$ in, we get $y - \frac{y^3}{3!} + \frac{y^5}{5!} - \sin 0.6293 = 0$, and we get three roots: $y=0.680635$, $y=2.66801$, and $y=3.39185$. 
Of course, since $\sin x$ ranges from $y=-1$ to $y=1$, $\sin^{-1} x$ ranges from $x=-1$ to $x=1$. Therefore, only the first solution is valid.
The actual value of $\sin^{-1} 0.6293$ is $0.680652$, compared to the approximation of $0.680635$, which has an error of $2 \cdot 10^{-4}$. If we add more terms to the Taylor series, our approximation will get more and more accurate.
