Question

Find the value of $\sin \left( {{{15}^0}} \right)$
$
  {\text{A}}{\text{. }}\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }} \\
  {\text{B}}{\text{. }}\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }} \\
  {\text{C}}{\text{. }}\dfrac{{\sqrt 3 - 1}}{{\sqrt 2 }} \\
  {\text{D}}{\text{. None of these}} \\
$

Answer 1

Hint: To solve this question we have to write $\left( {{{15}^0} = {{45}^0} - {{30}^0}} \right)$ and we have to use the formula $\left( {\sin \left( {A - B} \right) = \sin A.\cos B - \cos A.\sin B} \right)$ and further solve it to get the required option.

Complete step-by-step solution:
We have to find out:
$\sin \left( {{{15}^0}} \right)$
We can write it as
$\sin \left( {{{15}^0}} \right) = \sin \left( {{{45}^0} - {{30}^0}} \right)$
Use the formulae $\left( {\because \sin \left( {A - B} \right) = \sin A.\cos B - \cos A.\sin B} \right)$
$
   \Rightarrow \sin {45^0}.\cos {30^0} - \cos {45^0}\sin {30^0} \\

\Rightarrow \sin {45^0} = \cos {45^0} = \dfrac{1}{{\sqrt 2 }} \\
 \Rightarrow \sin {30^0} = \dfrac{1}{2},\cos {30^0} = \dfrac{{\sqrt 3 }}{2} \\

   \Rightarrow \dfrac{1}{{\sqrt 2 }}.\dfrac{{\sqrt 3 }}{2} - \dfrac{1}{{\sqrt 2 }}.\dfrac{1}{2} \\
   \Rightarrow \dfrac{{\sqrt 3 }}{{2\sqrt 2 }} - \dfrac{1}{{2\sqrt 2 }} = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }} \\
 $
Hence option A is the correct option.

Note: Whenever we get this type of question the key concept of solving is we have use our brain to get required angle through known and easy angles like we don’t know value of ${15^0}$ but we know value of ${45^0}, {30^0}$ so we have to use them to get our required value.