Solveeit Logo
Login

Question

Question: How do you prove \[\cos \left( {x - \dfrac{\pi }{2}} \right) = \sin x\] ?...

How do you prove cos⁡(x−π2)=sin⁡x\cos \left( {x - \dfrac{\pi }{2}} \right) = \sin x ?

Explanation

Solution

Hint : In order to prove the statement above we will use the expansion of the formula cos⁡(A−B)\cos \left( {A - B} \right) . Such that cos⁡(A−B)=cos⁡Acos⁡B+sin⁡A.sin⁡B\cos \left( {A - B} \right) = \cos A\cos B + \sin A.\sin B
In this formula we will put the value of A and B and then we will write the values of sin and cos function to get the proof done.

Complete step-by-step answer :
We are given that,
LHS= cos⁡(x−π2)\cos \left( {x - \dfrac{\pi }{2}} \right)
Now we know that cos⁡(A−B)=cos⁡Acos⁡B+sin⁡A.sin⁡B\cos \left( {A - B} \right) = \cos A\cos B + \sin A.\sin B
Then comparing the formula with the given angles we can write,
cos⁡(x−π2)=cos⁡xcos⁡π2+sin⁡x.sin⁡π2\cos \left( {x - \dfrac{\pi }{2}} \right) = \cos x\cos \dfrac{\pi }{2} + \sin x.\sin \dfrac{\pi }{2}
Now we know that cos⁡π2=0\cos \dfrac{\pi }{2} = 0 and sin⁡π2=1\sin \dfrac{\pi }{2} = 1
Putting these values in the formula above,
cos⁡(x−π2)=cos⁡x.0+sin⁡x.1\cos \left( {x - \dfrac{\pi }{2}} \right) = \cos x.0 + \sin x.1
On solving this further,
cos⁡(x−π2)=sin⁡x\cos \left( {x - \dfrac{\pi }{2}} \right) = \sin x
Hence proved the statement.
So, the correct answer is “ cos⁡(x−π2)=sin⁡x\cos \left( {x - \dfrac{\pi }{2}} \right) = \sin x ”.

Note : Note that these functions sine, cosine, tangent are the basic trigonometric functions. Rather cosec, sec and cot are the functions that are reciprocals of basic functions. The formula used above is a trigonometric formula when two angles with some relation are given. If sometimes angle is given in the form that can be split in standard angles. At that time we can use this formula to obtain the value.
For example if the given angle is 125∘{125^ \circ } then we can write this angle as 125∘=90∘+30∘{125^ \circ } = {90^ \circ } + {30^ \circ } .