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Question: How do you prove \(\sin \left( {\dfrac{\pi }{4} + x} \right) = \cos \left( {\dfrac{\pi }{4} - x} \ri...

How do you prove sin(π4+x)=cos(π4x)\sin \left( {\dfrac{\pi }{4} + x} \right) = \cos \left( {\dfrac{\pi }{4} - x} \right) ?

Explanation

Solution

Here, we are given a trigonometric equation and we need to prove it. Since we are asked to prove, we can start with the left-hand side of the equation. We will be taking the help of sin(A+B)sin(A+B) and cos(AB)cos(A-B) formula. Also, we need to apply the required trigonometric identities to prove the given equation.

Formula to be used:
The required trigonometric identities that are applied to the given problem are as follows.
sin(a+b)=sinacosb+cosasinb\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b
cos(a+b)=cosacosbsinasinb\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b

Complete step-by-step answer:
The given trigonometric equation is sin(π4+x)=cos(π4x)\sin \left( {\dfrac{\pi }{4} + x} \right) = \cos \left( {\dfrac{\pi }{4} - x} \right) and we are asked to prove it.
To prove the given equation, we shall start with the left-hand side of the equation and the result of the left-hand side and the right-hand side of the equation must be the same.
Hence, we shall start with the left-hand side of the equation sin(π4+x)=cos(π4x)\sin \left( {\dfrac{\pi }{4} + x} \right) = \cos \left( {\dfrac{\pi }{4} - x} \right)
sin(π4+x)\sin \left( {\dfrac{\pi }{4} + x} \right) =sinπ4cosx+cosπ4sinx = \sin \dfrac{\pi }{4}\cos x + \cos \dfrac{\pi }{4}\sin x (Here we have applied sin(a+b)=sinacosb+cosasinb\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b)
=12cosx+12sinx= \dfrac{1}{{\sqrt 2 }}\cos x + \dfrac{1}{{\sqrt 2 }}\sin x (We know that the values of sinπ4=cosπ4=12\sin \dfrac{\pi }{4} = \cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} )
Now, we shall replace 12\dfrac{1}{{\sqrt 2 }} by cosπ4\cos \dfrac{\pi }{4} in the first term and 12\dfrac{1}{{\sqrt 2 }}by ssinπ4\sin \dfrac{\pi }{4} in the second term.
Thus, we have sin(π4+x)\sin \left( {\dfrac{\pi }{4} + x} \right) =cosπ4cosx+sinπ4sinx = \cos \dfrac{\pi }{4}\cos x + \sin \dfrac{\pi }{4}\sin x
sin(π4+x)=cos(π4x)\Rightarrow \sin \left( {\dfrac{\pi }{4} + x} \right) = \cos \left( {\dfrac{\pi }{4} - x} \right) (Here we applied cos(a+b)=cosacosbsinasinb\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b)
Thus, we have got the result that equals the right-hand of the given equation.
Hence, the given trigonometric equation sin(π4+x)=cos(π4x)\sin \left( {\dfrac{\pi }{4} + x} \right) = \cos \left( {\dfrac{\pi }{4} - x} \right) is proved.

Note: First, we shall analyze the given equation is a trigonometric equation. When we are given a trigonometric equation to prove, we should always start with either the left-hand side or right-hand side of the equation. Also, we need to analyze what trigonometric identities are to be used. Therefore, we have proved the given trigonometric equation.