Question
Question: How do you prove \(\sin \left( {\dfrac{\pi }{4} + x} \right) = \cos \left( {\dfrac{\pi }{4} - x} \ri...
How do you prove sin(4π+x)=cos(4π−x) ?
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) 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(a+b)=sinacosb+cosasinb
cos(a+b)=cosacosb−sinasinb
Complete step-by-step answer:
The given trigonometric equation is sin(4π+x)=cos(4π−x) 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(4π+x)=cos(4π−x)
sin(4π+x) =sin4πcosx+cos4πsinx (Here we have applied sin(a+b)=sinacosb+cosasinb)
=21cosx+21sinx (We know that the values of sin4π=cos4π=21 )
Now, we shall replace 21 by cos4π in the first term and 21by ssin4π in the second term.
Thus, we have sin(4π+x) =cos4πcosx+sin4πsinx
⇒sin(4π+x)=cos(4π−x) (Here we applied cos(a+b)=cosacosb−sinasinb)
Thus, we have got the result that equals the right-hand of the given equation.
Hence, the given trigonometric equation sin(4π+x)=cos(4π−x) 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.