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

Prove that 2cos(π4A)=cosA+sinA\sqrt 2 \cos \left( {\dfrac{\pi }{4} - A} \right) = \cos A + \sin A .

Explanation

Solution

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 π4\dfrac{\pi }{4} and AA.
We have a direct formula for the difference of two angles in case of all the trigonometric ratios.
Here the trigonometric ratio is cos\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 2\sqrt 2 is a constant.
We know that,
cos(AB)=cosAcosB+sinAsinB\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B … (1)
In this case A=π4A = \dfrac{\pi }{4} and B=AB = A.
Substituting these values in the equation (1) we get:
2cos(π4A)=2(cosπ4cosA+sinπ4sinA)\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 cosπ4=12\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} and similarly sinπ4=12\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}.
Substitute these values in equation (2) we get:

2cos(π4A)=2(12cosA+12sinA)\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 12\dfrac{1}{{\sqrt 2 }} common from the bracket and rearrange the terms after cancelling the term 2\sqrt 2 and 12\dfrac{1}{{\sqrt 2 }}.
We finally get,

2cos(π4A)=(cosA+sinA)\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 2cos(π4A)=(cosA+sinA)\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..