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Question: The value of \(\sin ({22^ \circ }30')\) is A. \(\sqrt {\dfrac{{\sqrt 2 + 1}}{{2\sqrt 2 }}} \) B...

The value of sin(2230)\sin ({22^ \circ }30') is
A. 2+122\sqrt {\dfrac{{\sqrt 2 + 1}}{{2\sqrt 2 }}}
B. 21\sqrt {\sqrt 2 - 1}
C. 2122\sqrt {\dfrac{{\sqrt 2 - 1}}{{2\sqrt 2 }}}
D. None of these

Explanation

Solution

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.

Complete step-by-step answer:
We have to find the value of sin(2230)\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 2212022{\dfrac{1}{2}^0} =sin22.50{22.5^0}
Making use of the half angle formula for sin, we can write sin(θ2)=±1cosθ2\sin \left( {\dfrac{\theta }{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos \theta }}{2}}
Here, this is nothing but equal to θ2=22.50\dfrac{\theta }{2} = {22.5^0} θ=450 \Rightarrow \theta = {45^0}
Here, we need only the positive value since 22.50{22.5^0} lies in the first quadrant where sin is positive
sin(22.50)=1cos4502\Rightarrow \sin ({22.5^0}) = \sqrt {\dfrac{{1 - \cos {{45}^0}}}{2}}
 1122 =2122 sin(22033)=2122  {\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.