Question

The value of $\cos {15^0} - \sin {15^0}$ is
A). 0
B). $\dfrac{1}{{\sqrt 2 }}$
C). $ - \dfrac{1}{{\sqrt 2 }}$
D). $\dfrac{1}{{2\sqrt 2 }}$

Answer 1

Hint- Here we have to find the value of \[\cos {15^0}\] and $\sin {15^0}$ . Since we don’t know these values directly but we can convert it into the known values of 30° and 45° with the help of trigonometric identities then we can get the required value very easily.

Complete step-by-step solution -
$
  \cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B \\
  \sin (A - B) = \sin A\cos B - \cos A\sin B \\
 $
First we will find the value of $\sin {15^0}$
The angle ${15^0}$ can be written as the difference of two know angles as $\left( {{{45}^0} - {{30}^0}} \right)$
Therefore
\[\sin {15^0} = \sin \left( {{{45}^0} - {{30}^0}} \right)\]
Now we will solve the above equation using trigonometric identities
$ \Rightarrow \sin \left( {{{45}^0} - {{30}^0}} \right)$
Using the trigonometric identity $\sin (A - B) = \sin A\cos B - \cos A\sin B$ and comparing with the above equation, we get $A = {45^0},B = {30^0}$
Substituting these values in the above formula, we get
$
   \Rightarrow \sin (A - B) = \sin A\cos B - \cos A\sin B \\
   \Rightarrow \sin ({45^0} - {30^0}) = \sin {45^0} \times \cos {30^0} - \cos {45^0}\sin {30^0} \\
$
As we know that $\left[ \
  \sin {30^0} = \dfrac{1}{2},\sin {45^0} = \dfrac{1}{{\sqrt 2 }},
  \cos {30^0} = \dfrac{{\sqrt 3 }}{2},\cos {45^0} = \dfrac{1}{{\sqrt 2 }} \right]$
Substituting these values in the above equation, we get
$
   \Rightarrow \sin ({45^0} - {30^0}) = \sin {45^0} \times \cos {30^0} - \cos {45^0}\sin {30^0} \\
   \Rightarrow \sin ({45^0} - {30^0}) = \dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2} - \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2} \\
   \Rightarrow \sin ({15^0}) = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }} \\
$
Now, we will find the value of $\cos {15^0}$
The angle ${15^0}$ can be written as the difference of two know angles as $\left( {{{45}^0} - {{30}^0}} \right)$
Therefore
\[\cos {15^0} = \cos \left( {{{45}^0} - {{30}^0}} \right)\]
Now we will solve the above equation using trigonometric identities
$ \Rightarrow \cos \left( {{{45}^0} - {{30}^0}} \right)$
Using the trigonometric identity $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$ and comparing with the above equation, we get $A = {45^0},B = {30^0}$
Substituting these values in the above formula, we get
$
   \Rightarrow \cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B \\
   \Rightarrow \cos ({45^0} - {30^0}) = \cos {45^0} \times \cos {30^0} + \sin {45^0}\sin {30^0} \\
$
As we know that $\left[
  \sin {30^0} = \dfrac{1}{2},\sin {45^0} = \dfrac{1}{{\sqrt 2 }},
  \cos {30^0} = \dfrac{{\sqrt 3 }}{2},\cos {45^0} = \dfrac{1}{{\sqrt 2 }} \right]$
Substituting these values in the above equation, we get
$
   \Rightarrow \cos ({45^0} - {30^0}) = \cos {45^0} \times \cos {30^0} + \sin {45^0}\sin {30^0} \\
   \Rightarrow \cos ({45^0} - {30^0}) = \dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2} + \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2} \\
   \Rightarrow \cos ({15^0}) = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }} \\
$
Since, we get the value of $\sin {15^0}$ and $\cos {15^0}$
Therefore
$ \Rightarrow \cos {15^0} - \sin {15^0}$
Substituting the value of $\sin {15^0}$ and $\cos {15^0}$ in the above equation, we get
$
   \Rightarrow \cos {15^0} - \sin {15^0} \\
   \Rightarrow \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }} - \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }} \\
  \Rightarrow \dfrac {2}{2\sqrt{2}} \\
   \Rightarrow \dfrac{1}{{\sqrt 2 }} \\
$
Hence, the correct option is B.

Note- In order to solve these types of questions, you need to remember all trigonometric identities and then need to apply identity according to the question. In the above question we have found the value of $\sin {15^0}$ and $\cos {15^0}$ and then we have to take their difference. So, accordingly we use the identity we needed. In other questions, we may need other identities so learn all identities and values of the important angles from the table.