Question

How do you evaluate ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)$ without a calculator?

Answer 1

We use the basic trigonometric identities to find the value of ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)$ without a calculator. We use the fact that $\sin \dfrac{\pi }{6}=\dfrac{1}{2}.$ Also we use the fact that if $\sin x=y,\,x\in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right],$ then $x={{\sin }^{-1}}y,\,\left| y \right|\le 1.$ The converse is also true. Besides, we know that $\sin 30{}^\circ =\dfrac{1}{2}.$ We convert the value in degree measure to the value in radian measure.

Complete step by step solution:
Consider the given trigonometric function ${{\sin }^{-1}}\left( \dfrac{1}{2} \right).$
We can find the value of this trigonometric function easily by using $x=\sin y$ implies that ${{\sin }^{-1}}x=y.$
The converse of the above identity is also true, ${{\sin }^{-1}}x=y$ implies that $x=\sin y.$
Let us suppose that $x={{\sin }^{-1}}\left( \dfrac{1}{2} \right).$
Also, we know that $\dfrac{1}{2}\le 1.$
Now, from above, we can write $\sin x=\dfrac{1}{2}.$
In addition to this, we get $x\in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right].$
Now we have to find the value of $x$ in the interval $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$ for which the above supposition and also $\sin x=\dfrac{1}{2}$ are true.
And we have the sine value $\sin 30{}^\circ =\dfrac{1}{2}.$
Now, what we have to do is to find the radian value corresponding to $30{}^\circ .$
Since $1\,{}^\circ ={{\left( \dfrac{\pi }{180} \right)}^{c}},$ $30\,{}^\circ =30\times {{\left( \dfrac{\pi }{180} \right)}^{c}}.$
Since $30$ divides $180$ to give $60,$ we get $30{}^\circ ={{\left( \dfrac{\pi }{60} \right)}^{c}}.$
That means the value of $x$ in radian measure for which $\sin x=\dfrac{1}{2}$ is $\dfrac{\pi }{6}.$
In other words, $\sin \dfrac{\pi }{6}=\dfrac{1}{2}.$
And we know that the value $\dfrac{\pi }{6}$ belongs to the required interval $\left[ \dfrac{-\pi }{6},\dfrac{\pi }{6} \right].$
So, using the above given identity $\sin x=y,\,x\in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right],$ then $x={{\sin }^{-1}}y,\,\left| y \right|\le 1.$
We have now obtained the value of $x$ which satisfies the condition given above.
Therefore, ${{\sin }^{-1}}\dfrac{1}{2}=\dfrac{\pi }{6}.$

Note: Let us recall how we convert a value from degree measure to radian measure.
A circle subtends an angle at the centre. Radian measure of this angle is $2\pi$ and its degree measure is $360{}^\circ .$
So, we get $2{{\pi }^{c}}=360{}^\circ .$
This implies that ${{\pi }^{c}}=\dfrac{360}{2}=180{}^\circ .$
Therefore, $1{}^\circ ={{\left( \dfrac{\pi }{180} \right)}^{c}}.$
So, $30{}^\circ =30\times {{\left( \dfrac{\pi }{180} \right)}^{c}}.$
Hence, $30{}^\circ ={{\left( \dfrac{\pi }{6} \right)}^{c}},$ and so on.