Question

Verify the identity\[\dfrac{{\sin x \pm \sin y}}{{\cos x + \cos y}} = \tan \left( {\dfrac{{x \pm y}}{2}} \right)\].

Answer 1

Hint: To prove an identity we have to use logical steps to show that one side of the equation can be transformed into the other side of the equation. Any values should not be plugged into identities to prove anything. There are infinitely many values to plug in. to verify the given identity trigonometric sum identities should be used.

Complete step-by-step solution:
The given identities can be proved by trigonometric sum identities which are as follows:
\[
  \sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) \\
  \sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right) \\
  \cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) \\
  \cos A - \cos B = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right) \\
 \]
In the given identity \[\dfrac{{\sin x \pm \sin y}}{{\cos x + \cos y}} = \tan \left( {\dfrac{{x \pm y}}{2}} \right)\] considering the left hand side we have two cases one is positive and another is negative. We need to check for both positive and negative and prove for right hand side.
Now considering the left hand side positive case,
\[\dfrac{{\sin x + \sin y}}{{\cos x + \cos y}} = \frac{{2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)}}{{2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)}}\]
This can be further written as,
\[ \Rightarrow \dfrac{{\sin \left( {\dfrac{{A + B}}{2}} \right)}}{{\cos \left( {\dfrac{{A + B}}{2}} \right)}}\]
We know that \[\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta \]hence we this can be further solved as,
\[ \Rightarrow \tan \left( {\dfrac{{A + B}}{2}} \right)\]
Now considering the left hand side negative case,
\[\dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = \dfrac{{2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)}}{{2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)}}\]
This can be further written as,
\[ \Rightarrow \dfrac{{\sin \left( {\dfrac{{A - B}}{2}} \right)}}{{\cos \left( {\dfrac{{A - B}}{2}} \right)}}\]
We know that \[\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta \]hence we this can be further solved as,
\[ \Rightarrow \tan \left( {\dfrac{{A - B}}{2}} \right)\]
Hence, combining both left hand side and right hand side we have,
\[\dfrac{{\sin x \pm \sin y}}{{\cos x + \cos y}} = \tan \left( {\dfrac{{x \pm y}}{2}} \right)\]
Hence it is proved.

Note: There is no set method that can be applied to verifying identities; there are, however, a few different ways to start based on the identity which is to be verified. Trigonometric identities are used in both course texts and in real life applications to abbreviate trigonometric expressions. It is important to remember that merely verifying an identity or altering an expression is not an end in itself but rather that identities are used to simplify expressions according to the task at hand.