Question

The value of $\sqrt 2 (\cos {15^0} - \sin {15^0})$ is equal to:
A. $\sqrt 3 $
B. $\sqrt 2 $
C. 1
D. 2
E. $2\sqrt 3 $

Answer 1

Hint: Here we will simplify the given equation into any standard formula and then by applying the formula the value can be calculated.

Complete step-by-step answer:
The given equation is $\sqrt 2 (\cos {15^0} - \sin {15^0})$. In this equation divide and multiply by $\sqrt 2 $ we get
$\sqrt 2 \times \sqrt 2 \left( {\dfrac{{\cos {{15}^0}}}{{\sqrt 2 }} - \dfrac{{\sin {{15}^0}}}{{\sqrt 2 }}} \right)$.
Now as we know $\cos {45^0} = \sin {45^0} = \dfrac{1}{{\sqrt 2 }}$.So substituting this value we get
$2\left( {\cos {{45}^0}\cos {{15}^0} - \sin {{45}^0}.\sin {{15}^0}} \right)$
Now as we know $\cos (A + B) = \cos A\cos B - \sin A\sin B$, so using this property
$ \Rightarrow 2\left( {\cos {{45}^0}\cos {{15}^0} - \sin {{45}^0}.\sin {{15}^0}} \right) = 2\cos ({45^0} + {15^0}) = 2\cos {60^0}$
The value of $\cos {60^0} = \dfrac{1}{2}$.
Therefore, $\sqrt 2 (\cos {15^0} - \sin {15^0}) = 2\cos {60^0} = 2 \times \dfrac{1}{2} = 1$
Hence option ‘C’ is correct.

Note: In such types of questions the key concept we have to remember is to always convert the equation in the standard formula of cos (A+B), or sin (A-B), then we will easily calculate the required value.