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Question: Prove that \(\dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = \tan \dfrac{{x - y}}{2}\)...

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}

Explanation

Solution

Here, we have to prove sin⁡x−sin⁡ycos⁡x+cos⁡y=tan⁡x−y2\dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = \tan \dfrac{{x - y}}{2}. Take the LHS of the equation and try to prove LHS equal to RHS. For sin⁡A−sin⁡B\sin A - \sin B, we have formula sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right) and for cos⁡A+cos⁡B\cos A + \cos B we have formula cos⁡A+cos⁡B=2cos⁡(A+B2)cos⁡(A−B2)\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right). Using these formulas, we will get LHS equal to RHS.

Complete step-by-step answer:
In this question, we are given a trigonometric equation and we need to prove that it is correct.
Given equation: sin⁡x−sin⁡ycos⁡x+cos⁡y=tan⁡x−y2\dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = \tan \dfrac{{x - y}}{2}- - - - - - - - - - - - (1)
Here, LHS=sin⁡x−sin⁡ycos⁡x+cos⁡y = \dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} and RHS=tan⁡x−y2 = \tan \dfrac{{x - y}}{2}. So, we need to prove that LHS is equal to RHS.
For proving this, let us take the LHS part of the equation (1). Therefore, we get
⇒\RightarrowLHS=sin⁡x−sin⁡ycos⁡x+cos⁡y = \dfrac{{\sin x - \sin y}}{{\cos x + \cos y}}- - - - - - - - - - - - (2)
Here, in numerator we have sin⁡x−sin⁡y\sin x - \sin y. Now we know the formula that
sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)
And in denominator, we have cos⁡x+cos⁡y\cos x + \cos y. Now, we know the formula that
cos⁡A+cos⁡B=2cos⁡(A+B2)cos⁡(A−B2)\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)
Therefore, putting these values in equation (2), we get
⇒\RightarrowLHS=2cos⁡(x+y2)sin⁡(x−y2)2cos⁡(x+y2)cos⁡(x−y2) = \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)}}- - - - - - - - - - - - - (3)
Here, 2 and cos⁡(x+y2)\cos \left( {\dfrac{{x + y}}{2}} \right) gets cancelled. Therefore, equation (3) will become
⇒\RightarrowLHS=sin⁡(x−y2)cos⁡(x−y2) = \dfrac{{\sin \left( {\dfrac{{x - y}}{2}} \right)}}{{\cos \left( {\dfrac{{x - y}}{2}} \right)}}- - - - - - - - - (4)
Now, we know that
sin⁡θcos⁡θ=tan⁡θ\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta
Therefore, putting this value in equation (4), we get
⇒\RightarrowLHS=tan⁡(x−y2) = \tan \left( {\dfrac{{x - y}}{2}} \right)
⇒LHS=RHS\Rightarrow LHS = RHS
Hence, we have got LHS equal to RHS.
Therefore, we have proved 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}.

Note: This question is a simple formula based question. For solving trigonometric questions, always keep the important formulas and results in mind. Other important formulas are
cos⁡A+cos⁡B=2cos⁡(A+B2)cos⁡(A−B2)\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)
cos⁡A−cos⁡B=−2sin⁡(A+B2)sin⁡(A−B2)\cos A - \cos B = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)
sin⁡A+sin⁡B=2sin⁡(A+B2)cos⁡(A−B2)\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)
sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)