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1] We know that in this triangle, $h^2=a^2+a^2$. Therefore,

$$ h=\sqrt{a^2+a^2}=\sqrt{2a^2}=\sqrt{2}\:a\:. $$

As $h$ can be written as $\sqrt{2}\:a$, we now know how to write an expression for $\sin(45^\circ)$,

$$\sin(45^\circ)=\frac{a}{\sqrt{2}\:a}=\frac{1}{\sqrt{2}}\:.$$

Substitute $\sin(45^\circ)=\frac{1}{\sqrt{2}}$ into the expression $\sin^{-1} (\sin (45^\circ))$. Then

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

This means that

$$\alpha=\sin^{-1} \left ( \frac 1 {\sqrt 2} \right ) \tag 1$$

But how do I calculate $(1)$ without a calculator? It should be very simple, I think I'm just missing a concept here. I may have chosen the wrong method to calculate this expression, though.

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    **Hint:** How are the $\sin$ and the $\sin^{-1}$ function related to each other? – 2017-01-20
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    You should write $45^\circ$ rather than $45$. – 2017-01-20

4 Answers 4

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What is the function $\sin^{-1} $? $\sin^{-1}(x) $ is asking for the angle whose sine is $x $.

What is $\sin^{-1}(\sin(x)) $? That is asking for the angle whose sine is the sine of $x $. Can you see the wordplay? What is the angle whose sine is the sine of $x $? Well, breaking it into smaller parts, let us assume the sine of $x$ is $y $. We have $\sin^{-1}(y) $: what is the angle whose sine is $y $? Well, from definition, $y = \sin(x) $ so $x $ is the angle whose sine is $y $. Thus the answer to "what is the angle whose sine is $y $?" is $x $. But that is the same as asking "What is the angle whose sine is the sine of $x $?" which has $x $ as an answer.

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    This is not exactly true, for example $\sin^{-1}(\sin(360)) = 0 \neq 360$. – 2017-01-21
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If $\alpha$ is an angle in quadrant 1 or 4 -- in the sense that $-90^\circ \le \alpha \le 90^\circ$, the inverse sine function undoes the the sine function. Thus $$\sin^{-1}(\sin(45^\circ))= 45^\circ.$$

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Hint:

By definition we have, for $0\le y\le 90°$: $$ \sin^{-1}(x)=y \quad \iff \quad \sin y=x $$

In your case we have $x=\sin(45°)$, so: $$ \sin y=\sin (45°) \quad \iff \quad y=45° $$

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If I understood your question correctly you're asking how to calculate $arcsin \left(\frac{1}{\sqrt(2)}\right)$ without a calculator (since you are referring to formula (1)). I would suggest the unit circle. enter image description here

It's not that hard to memorize and a very important tool. But if you want to do it without memorizing you'll probably need to look at the origin of the sinus and cosines functions.