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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?

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  • $\begingroup$ Pls rewrite sin^-1(.6293) within dollar symbols and -1 between braces, ($sin^{-1}(0.6293)$). The change is too small for me to edit. It's the Latex format and is more readable. $\endgroup$
    – tpb261
    Jun 24 '14 at 3:52
  • $\begingroup$ As regards your question, are you looking for a geometrical construction or a theoretical answer? You can construct with opposite of .6293 units and hypotenuse of length 1 unit. Theoretically, maybe Taylor series is your best bet/approximation. $\endgroup$
    – tpb261
    Jun 24 '14 at 3:55
  • $\begingroup$ @tpb261 I tried to rewire it but i don't know if that's how it's supposed to look. A geometrical construction. What's a Taylor series? $\endgroup$
    – Jessica M.
    Jun 24 '14 at 3:59
  • $\begingroup$ In the old days when they didn't have calculators, they would use tables. They were accurate to about 4 decimal places. You could also use a slide rule to get one or two digits of accuracy. But they all had to be calculated using approximation methods, which were tedious. Now modern computers and calculators have these approximation methods built into them. $\endgroup$ Jun 24 '14 at 4:00
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    $\begingroup$ @Jessica: Have you taken any calculus? If not, sadly, the only answers you can get are (1) look at a table, (2) draw a triangle very very very accurately, or (3) learn calculus. Now, there are some special values which you can memorize, but there are no general by-hand before-calculus methods except the triangle one. A quick lookup on an inverse symbolic calculator suggests that $292/429$ would be a reasonable guess, but that's not really any better than using an actual calculator, and certainly much less accurate. $\endgroup$ Jun 24 '14 at 4:07
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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 $x$.

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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.

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Probably your best bet would be a Taylor/Maclaurin series:

$$ \sin^{-1}(x)= x + 1/6\cdot x^3 + 3/40\cdot x^5 + 5/112\cdot x^7 + O(x^9) $$

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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 $=0.68065\ldots$.

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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.

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