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Question: Prove the following: \[\dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = \tan \dfrac{{x - y}}{2}\]...

Prove the following:
sin⁡x−sin⁡ycos⁡x+cos⁡y=tan⁡x−y2\dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = \tan \dfrac{{x - y}}{2}

Explanation

Solution

We can take the LHS of the given equation. Then we can simplify its numerator using the trigonometric identity sin⁡(A)−sin⁡(B)=2cos⁡(A+B2)sin⁡(A−B2)\sin \left( A \right) - \sin \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right). We can simplify the denominator using the identity cos⁡(A)+cos⁡(B)=2cos⁡(A+B2)cos⁡(A−B2)\cos \left( A \right) + \cos \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right). On doing further calculations, we will obtain the RHS of the equation. We can say the given equation is true when L.H.S=R.H.S

Complete step-by-step answer:
We need to prove that sin⁡x−sin⁡ycos⁡x+cos⁡y=tan⁡x−y2\dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = \tan \dfrac{{x - y}}{2}
Let us look at the LHS,
LHS=sin⁡x−sin⁡ycos⁡x+cos⁡yLHS = \dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} … (1)
We can consider the numerator of the LHS
We know that sin⁡(A)−sin⁡(B)=2cos⁡(A+B2)sin⁡(A−B2)\sin \left( A \right) - \sin \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)
We can substitute the values,
⇒sin⁡(x)−sin⁡(y)=2cos⁡(x+y2)sin⁡(x−y2)\Rightarrow \sin \left( x \right) - \sin \left( y \right) = 2\cos \left( {\dfrac{{x + y}}{2}} \right)\sin \left( {\dfrac{{x - y}}{2}} \right)… (2)
We can consider the denominator of the LHS
We know that cos⁡(A)+cos⁡(B)=2cos⁡(A+B2)cos⁡(A−B2)\cos \left( A \right) + \cos \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)
We can substitute the values,
⇒cos⁡(x)+cos⁡(y)=2cos⁡(x+y2)cos⁡(x−y2)\Rightarrow \cos \left( x \right) + \cos \left( y \right) = 2\cos \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)… (3)
We can substitute (3) and (2) in (1)
⇒LHS=2cos⁡(x+y2)sin⁡(x−y2)2cos⁡(x+y2)cos⁡(x−y2)\Rightarrow LHS = \dfrac{{2\cos \left( {\dfrac{{x + y}}{2}} \right)\sin \left( {\dfrac{{x - y}}{2}} \right)}}{{2\cos \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)}}
On further simplification, we get,
⇒LHS=sin⁡(x−y2)cos⁡(x−y2)\Rightarrow LHS = \dfrac{{\sin \left( {\dfrac{{x - y}}{2}} \right)}}{{\cos \left( {\dfrac{{x - y}}{2}} \right)}}
We know that tan⁡A=sin⁡Acos⁡A\tan A = \dfrac{{\sin A}}{{\cos A}}. So, we get,
⇒LHS=tan⁡(x−y2)\Rightarrow LHS = \tan \left( {\dfrac{{x - y}}{2}} \right)
RHS is also equal to tan⁡(x−y2)\tan \left( {\dfrac{{x - y}}{2}} \right). So, we can write,
L.H.S.=R.H.S
Hence the equation is proved.

Note: We must be familiar with the following trigonometric identities used in this problem.
1.cos⁡(A)+cos⁡(B)=2cos⁡(A+B2)cos⁡(A−B2)\cos \left( A \right) + \cos \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)

2.cos⁡(A)−cos⁡(B)=−2sin⁡(A+B2)sin⁡(A−B2)\cos \left( A \right) - \cos \left( B \right) = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)
3.sin⁡(A)+sin⁡(B)=2sin⁡(A+B2)cos⁡(A−B2)\sin \left( A \right) + \sin \left( B \right) = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)
4.sin⁡(A)−sin⁡(B)=2cos⁡(A+B2)sin⁡(A−B2)\sin \left( A \right) - \sin \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)
5.sin⁡(−x)=−sin⁡(x)\sin \left( { - x} \right) = - \sin \left( x \right)
6.cos⁡(−x)=cos⁡(x)\cos \left( { - x} \right) = \cos \left( x \right)
We must know the values of trigonometric functions at common angles. Adding π\pi or multiples of π\pi with the angle retains the ratio and adding π2\dfrac{\pi }{2} or odd multiples of π2\dfrac{\pi }{2} will change the ratio.