Question

What is the value of $\sin {105^0} + \cos {105^0}$?
$
  {\text{A}}{\text{. }}\sin {50^0} \\
  {\text{B}}{\text{. cos}}{50^0} \\
  {\text{C}}{\text{. }}\dfrac{1}{{\sqrt 2 }} \\
  {\text{D}}{\text{. 0}} \\
 $

Answer 1

Hint- Here, we will be splitting the angle ${105^0}$ into the sum of ${60^0}$ and ${45^0}$ because from the trigonometric table we know the values of the trigonometric functions corresponding to ${60^0}$ and ${45^0}$.

“Complete step-by-step answer:”
As we know that $\sin \left( {A + B} \right) = \left( {\sin A} \right)\left( {\cos B} \right) + \left( {\cos A} \right)\left( {\sin B} \right)$
\[
  \sin {105^0} = \sin \left( {{{60}^0} + {{45}^0}} \right) \\
   \Rightarrow \sin {105^0} = \left( {\sin {{60}^0}} \right)\left( {\cos {{45}^0}} \right) + \left( {\cos {{60}^0}} \right)\left( {\sin {{45}^0}} \right){\text{ }} \to {\text{(1)}} \\
 \]
According to trigonometric table, $\sin {60^0} = \dfrac{{\sqrt 3 }}{2}$, $\cos {60^0} = \dfrac{1}{2}$ and $sin{45^0} = \cos {45^0} = \dfrac{1}{{\sqrt 2 }}$
\[
   \Rightarrow \sin {105^0} = \left( {\dfrac{{\sqrt 3 }}{2}} \right)\left( {\dfrac{1}{{\sqrt 2 }}} \right) + \left( {\dfrac{1}{2}} \right)\left( {\dfrac{1}{{\sqrt 2 }}} \right) = \dfrac{{\sqrt 3 }}{{2\sqrt 2 }} + \dfrac{1}{{2\sqrt 2 }} \\
   \Rightarrow \sin {105^0} = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}{\text{ }} \to {\text{(2)}} \\
 \]
Also we know that $\cos \left( {A + B} \right) = \left( {\cos A} \right)\left( {\cos B} \right) - \left( {\sin A} \right)\left( {\sin B} \right)$
$
  \cos {105^0} = \cos \left( {{{60}^0} + {{45}^0}} \right) \\
   \Rightarrow \cos {105^0} = \left( {\cos {{60}^0}} \right)\left( {\cos {{45}^0}} \right) - \left( {\sin {{60}^0}} \right)\left( {\sin {{45}^0}} \right){\text{ }} \to {\text{(3)}} \\
 $
According to trigonometric table, $\sin {60^0} = \dfrac{{\sqrt 3 }}{2}$, $\cos {60^0} = \dfrac{1}{2}$ and $sin{45^0} = \cos {45^0} = \dfrac{1}{{\sqrt 2 }}$
Putting the above values in equation (3), we get
$
   \Rightarrow \cos {105^0} = \left( {\dfrac{1}{2}} \right)\left( {\dfrac{1}{{\sqrt 2 }}} \right) - \left( {\dfrac{{\sqrt 3 }}{2}} \right)\left( {\dfrac{1}{{\sqrt 2 }}} \right) = \dfrac{1}{{2\sqrt 2 }} - \dfrac{{\sqrt 3 }}{{2\sqrt 2 }} \\
   \Rightarrow \cos {105^0} = \dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}{\text{ }} \to {\text{(4)}} \\
 $
The value of expression $\sin {105^0} + \cos {105^0}$ can be obtained by using equations (2) and (4), we get
$\sin {105^0} + \cos {105^0} = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }} + \dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }} = \dfrac{{\sqrt 3 + 1 + 1 - \sqrt 3 }}{{2\sqrt 2 }} = \dfrac{2}{{2\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }}$
Hence, option C is correct.

Note- In this particular problem, we doesn’t know the value of trigonometric functions corresponding to ${105^0}$ directly so in order to obtain that we split this angle and then use the formulas $\sin \left( {A + B} \right) = \left( {\sin A} \right)\left( {\cos B} \right) + \left( {\cos A} \right)\left( {\sin B} \right)$ and $\cos \left( {A + B} \right) = \left( {\cos A} \right)\left( {\cos B} \right) - \left( {\sin A} \right)\left( {\sin B} \right)$ to obtain the values of $\sin {105^0}$ and $\cos {105^0}$.