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# 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

Last updated date: 16th May 2024
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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 $\sin ({22^ \circ }30')$
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
$\Rightarrow \sin ({22.5^0}) = \sqrt {\dfrac{{1 - \cos {{45}^0}}}{2}}$
${\text{ }}\sqrt {\dfrac{{1 - \dfrac{1}{{\sqrt 2 }}}}{2}} \\ = \sqrt {\dfrac{{\sqrt 2 - 1}}{{2\sqrt 2 }}} \\ \Rightarrow \sin ({22^0}33') = \sqrt {\dfrac{{\sqrt 2 - 1}}{{2\sqrt 2 }}} \\$