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Question

Question: How do you simplify \(\cos \left( {x - \dfrac{\pi }{2}} \right)\)?...

How do you simplify cos(xπ2)\cos \left( {x - \dfrac{\pi }{2}} \right)?

Explanation

Solution

Here we can proceed by applying the formula cos(AB)=cosAcosB+sinAsinB\cos (A - B) = \cos A\cos B + \sin A\sin B
And then we can easily substitute the value of cosπ2=0\cos \dfrac{\pi }{2} = 0 and sinπ2=1\sin \dfrac{\pi }{2} = 1 and get our desired result.

Complete step-by-step answer:
Here we are given to simplify the value of the term cos(xπ2)\cos \left( {x - \dfrac{\pi }{2}} \right)
We know that there are different formulas of all the trigonometric functions and we need to just know which formula must be applied and when to apply which formula. So as we are given the term cos(xπ2)\cos \left( {x - \dfrac{\pi }{2}} \right) which is of the form cos(AB)\cos (A - B) so we can apply the formula of this trigonometric function which is:
cos(AB)=cosAcosB+sinAsinB\cos (A - B) = \cos A\cos B + \sin A\sin B (1) - - - - (1)
So by comparing this cos(AB)\cos (A - B) with the term cos(xπ2)\cos \left( {x - \dfrac{\pi }{2}} \right) we can get the values of A and BA{\text{ and }}B as:
A=x B=π2  A = x \\\ B = \dfrac{\pi }{2} \\\
Now substituting the values of them in equation (1) we will get:
cos(xπ2)=cosxcosπ2+sinxsinπ2\cos (x - \dfrac{\pi }{2}) = \cos x\cos \dfrac{\pi }{2} + \sin x\sin \dfrac{\pi }{2} (2) - - - - (2)
We also know that cosπ2=0\cos \dfrac{\pi }{2} = 0 and sinπ2=1\sin \dfrac{\pi }{2} = 1
Now substituting these values in equation (2) we will get:
cos(xπ2)=cosx(0)+sinx(1)\cos (x - \dfrac{\pi }{2}) = \cos x(0) + \sin x(1)
Simplifying it we get:
cos(xπ2)=sinx\cos (x - \dfrac{\pi }{2}) = \sin x
Hence on simplifying we have got the value of cos(xπ2)=sinx\cos (x - \dfrac{\pi }{2}) = \sin x and therefore we must remember that in order to solve such problems that deals with the trigonometric functions we need to always remember the trigonometric formula as without them, the problem is quite difficult to solve. We also have another formula in sin,cos,tan,cot,sec,cosec\sin ,\cos ,\tan ,\cot ,\sec ,{\text{cosec}}
Moreover when we are asked to find the value of the tangent or cotangent of any angle, we must know the basic values of the sine and cosine of the angles like 0,30,45,60,900^\circ ,30^\circ ,45^\circ ,60^\circ ,90^\circ and then we can easily calculate the same angles of the tangent, cotangent, secant, and cosecant of that same angle.

Hence we get the value of cos(xπ2)=sinx\cos (x - \dfrac{\pi }{2}) = \sin x

Note: We can also solve this by another method where we can use two formula:

  1. cos(π2x)=sinx\cos (\dfrac{\pi }{2} - x) = \sin x
  2. cos(x)=cosx\cos ( - x) = \cos x
    So we can write this given problem as:
    cos(xπ2)=cos((π2x))=cos(π2x)=sinx\cos (x - \dfrac{\pi }{2}) = \cos ( - (\dfrac{\pi }{2} - x)) = \cos (\dfrac{\pi }{2} - x) = \sin x