Question

Prove that $\sqrt 2 \cos \left( {\dfrac{\pi }{4} - A} \right) = \cos A + \sin A$ .

Answer 1

Hint:Observe that the given trigonometric ratio is expressed as difference of the angles in which it is measured. Directly use the formula that gives the value of the difference of the angles. Also, we will use the basic values of the different angles for the basic trigonometric ratios. Thus, we can prove the given relation.

Complete step-by-step answer:
First thing to observe is that the given angle is the difference between two angles $\dfrac{\pi }{4}$ and $A$.
We have a direct formula for the difference of two angles in case of all the trigonometric ratios.
Here the trigonometric ratio is $\cos $ so we will use the direct formula to simplify the given equation.
We will start with the left-hand side.
Note that the term $\sqrt 2 $ is a constant.
We know that,
$\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$ … (1)
In this case $A = \dfrac{\pi }{4}$ and $B = A$.
Substituting these values in the equation (1) we get:
$\sqrt 2 \cos \left( {\dfrac{\pi }{4} - A} \right) = \sqrt 2 \left( {\cos \dfrac{\pi }{4}\cos A + \sin \dfrac{\pi }{4}\sin A} \right)$ … (2)
Now on the right-hand side we have angles whose values we already know.
We know that $\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ and similarly $\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$.
Substitute these values in equation (2) we get:

$\sqrt 2 \cos \left( {\dfrac{\pi }{4} - A} \right) = \sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }}\cos A + \dfrac{1}{{\sqrt 2 }}\sin A} \right)$
We can take $\dfrac{1}{{\sqrt 2 }}$ common from the bracket and rearrange the terms after cancelling the term $\sqrt 2 $ and $\dfrac{1}{{\sqrt 2 }}$.
We finally get,

$\sqrt 2 \cos \left( {\dfrac{\pi }{4} - A} \right) = \left( {\cos A + \sin A} \right)$
Which is the same as the right-hand side.
Thus, the left-hand side and right-hand side are the same.
Therefore, we proved that $\sqrt 2 \cos \left( {\dfrac{\pi }{4} - A} \right) = \left( {\cos A + \sin A} \right)$.

Note:Here we had to use the correct formula and split the terms properly. We have to use the basic values of the angles so that the constant will get cancelled. The important thing is sign while using the formula as the sign is exactly positive when there is a difference of the angles in the cosecant trigonometric function..