Question

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

Answer 1

Hint: We can take the LHS of the given equation. Then we can simplify its numerator using the trigonometric identity \[\sin \left( A \right) - \sin \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)\]. We can simplify the denominator using the identity \[\cos \left( A \right) + \cos \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\]. 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 \[\dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = \tan \dfrac{{x - y}}{2}\]
Let us look at the LHS,
$LHS = \dfrac{{\sin x - \sin y}}{{\cos x + \cos y}}$ … (1)
We can consider the numerator of the LHS
We know that \[\sin \left( A \right) - \sin \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)\]
We can substitute the values,
 \[ \Rightarrow \sin \left( x \right) - \sin \left( y \right) = 2\cos \left( {\dfrac{{x + y}}{2}} \right)\sin \left( {\dfrac{{x - y}}{2}} \right)\]… (2)
We can consider the denominator of the LHS
We know that \[\cos \left( A \right) + \cos \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\]
We can substitute the values,
 \[ \Rightarrow \cos \left( x \right) + \cos \left( y \right) = 2\cos \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)\]… (3)
We can substitute (3) and (2) in (1)
$ \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,
$ \Rightarrow LHS = \dfrac{{\sin \left( {\dfrac{{x - y}}{2}} \right)}}{{\cos \left( {\dfrac{{x - y}}{2}} \right)}}$
We know that \[\tan A = \dfrac{{\sin A}}{{\cos A}}\]. So, we get,
\[ \Rightarrow LHS = \tan \left( {\dfrac{{x - y}}{2}} \right)\]
RHS is also equal to \[\tan \left( {\dfrac{{x - y}}{2}} \right)\]. 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.\[\cos \left( A \right) + \cos \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\]

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