Question

# The value of $\sin ({22^ \circ }30')$ isA. $\sqrt {\dfrac{{\sqrt 2 + 1}}{{2\sqrt 2 }}}$ B. $\sqrt {\sqrt 2 - 1}$ C. $\sqrt {\dfrac{{\sqrt 2 - 1}}{{2\sqrt 2 }}}$ D. None of these

Hint: In this type of question, since the angle is expressed both in degree and minutes, we will make use of the half angle formula of sin and solve it.

We have to find the value of $\sin ({22^ \circ }30')$
We know that 60 minutes = 1 degree
So, 30’=1/2 degree
So, we can express the given equation as sin $22{\dfrac{1}{2}^0}$ =sin${22.5^0}$
Making use of the half angle formula for sin, we can write $\sin \left( {\dfrac{\theta }{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos \theta }}{2}}$
Here, this is nothing but equal to $\dfrac{\theta }{2} = {22.5^0}$ $\Rightarrow \theta = {45^0}$
Here, we need only the positive value since ${22.5^0}$ lies in the first quadrant where sin is positive
$\Rightarrow \sin ({22.5^0}) = \sqrt {\dfrac{{1 - \cos {{45}^0}}}{2}}$
${\text{ }}\sqrt {\dfrac{{1 - \dfrac{1}{{\sqrt 2 }}}}{2}} \\ = \sqrt {\dfrac{{\sqrt 2 - 1}}{{2\sqrt 2 }}} \\ \Rightarrow \sin ({22^0}33') = \sqrt {\dfrac{{\sqrt 2 - 1}}{{2\sqrt 2 }}} \\$
So, option C is the correct answer for this question

Note: Whenever solving these types of problems first convert the value of angle given in minutes to degrees and solve it or we can also convert it into radians and solve it further.