How do you prove $\sin \left( {\dfrac{\pi }{4} + x} \right) = \cos \left( {\dfrac{\pi }{4} - x} \right)$ ?
Answer
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Hint: 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 $sin(A+B)$ and $cos(A-B)$ 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.
$\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b$
$\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b$
Complete step-by-step answer:
The given trigonometric equation is $\sin \left( {\dfrac{\pi }{4} + x} \right) = \cos \left( {\dfrac{\pi }{4} - x} \right)$ 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 $\sin \left( {\dfrac{\pi }{4} + x} \right) = \cos \left( {\dfrac{\pi }{4} - x} \right)$
$\sin \left( {\dfrac{\pi }{4} + x} \right)$ $ = \sin \dfrac{\pi }{4}\cos x + \cos \dfrac{\pi }{4}\sin x$ (Here we have applied $\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b$)
$ = \dfrac{1}{{\sqrt 2 }}\cos x + \dfrac{1}{{\sqrt 2 }}\sin x$ (We know that the values of $\sin \dfrac{\pi }{4} = \cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ )
Now, we shall replace $\dfrac{1}{{\sqrt 2 }}$ by $\cos \dfrac{\pi }{4}$ in the first term and $\dfrac{1}{{\sqrt 2 }}$by s$\sin \dfrac{\pi }{4}$ in the second term.
Thus, we have $\sin \left( {\dfrac{\pi }{4} + x} \right)$$ = \cos \dfrac{\pi }{4}\cos x + \sin \dfrac{\pi }{4}\sin x$
$ \Rightarrow \sin \left( {\dfrac{\pi }{4} + x} \right) = \cos \left( {\dfrac{\pi }{4} - x} \right)$ (Here we applied $\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 \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.
Formula to be used:
The required trigonometric identities that are applied to the given problem are as follows.
$\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b$
$\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b$
Complete step-by-step answer:
The given trigonometric equation is $\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 \left( {\dfrac{\pi }{4} + x} \right) = \cos \left( {\dfrac{\pi }{4} - x} \right)$
$\sin \left( {\dfrac{\pi }{4} + x} \right)$ $ = \sin \dfrac{\pi }{4}\cos x + \cos \dfrac{\pi }{4}\sin x$ (Here we have applied $\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b$)
$ = \dfrac{1}{{\sqrt 2 }}\cos x + \dfrac{1}{{\sqrt 2 }}\sin x$ (We know that the values of $\sin \dfrac{\pi }{4} = \cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ )
Now, we shall replace $\dfrac{1}{{\sqrt 2 }}$ by $\cos \dfrac{\pi }{4}$ in the first term and $\dfrac{1}{{\sqrt 2 }}$by s$\sin \dfrac{\pi }{4}$ in the second term.
Thus, we have $\sin \left( {\dfrac{\pi }{4} + x} \right)$$ = \cos \dfrac{\pi }{4}\cos x + \sin \dfrac{\pi }{4}\sin x$
$ \Rightarrow \sin \left( {\dfrac{\pi }{4} + x} \right) = \cos \left( {\dfrac{\pi }{4} - x} \right)$ (Here we applied $\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b$)
Thus, we have got the result that equals the right-hand of the given equation.
Hence, the given trigonometric equation $\sin \left( {\dfrac{\pi }{4} + x} \right) = \cos \left( {\dfrac{\pi }{4} - x} \right)$ 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.
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