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Question: In any \(\Delta ABC\), prove that \(\dfrac{\left( a-b \right)}{c}\cos \dfrac{C}{2}=\sin \left( \df...

In any ΔABC\Delta ABC, prove that
(a−b)ccos⁡C2=sin⁡(A−B2)\dfrac{\left( a-b \right)}{c}\cos \dfrac{C}{2}=\sin \left( \dfrac{A-B}{2} \right)

Explanation

Solution

Hint: Try to simplify the left-hand side of the equation given in the question by the application of the sine rule of a triangle followed by the use of the formula of sin2A and the formula of (sinA-sinB).

Complete step-by-step answer:
Before starting with the solution, let us draw a diagram for better visualisation.

Now starting with the left-hand side of the equation that is given in the question.
We know, according to the sine rule of the triangle: asin⁡A=bsin⁡B=csin⁡C=k\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k and in other terms, it can be written as:
a=ksin⁡A b=ksin⁡B c=ksin⁡C \begin{aligned} & a=k\sin A \\\ & b=k\sin B \\\ & c=k\sin C \\\ \end{aligned}
So, applying this to our expression, we get
(a−b)ccos⁡C2\dfrac{\left( a-b \right)}{c}\cos \dfrac{C}{2}
=ksin⁡A−ksin⁡Bksin⁡C×cos⁡C2=\dfrac{k\sin A-k\sin B}{k\sin C}\times \cos \dfrac{C}{2}
Now we will take kk common from all the terms. On doing so, we get
=(sin⁡A−sin⁡B)sin⁡C×cos⁡C2=\dfrac{\left( \sin A-\sin B \right)}{\sin C}\times \cos \dfrac{C}{2}
To further simplify the expression, we use the formula sin⁡2X=2sin⁡Xcos⁡X\sin 2X=2\sin X\cos X .
=(sin⁡A−sin⁡B)2cos⁡C2sin⁡C2×cos⁡C2=\dfrac{\left( \sin A-\sin B \right)}{2\cos \dfrac{C}{2}\sin \dfrac{C}{2}}\times \cos \dfrac{C}{2}
=(sin⁡A−sin⁡B)2sin⁡C2=\dfrac{\left( \sin A-\sin B \right)}{2\sin \dfrac{C}{2}}
Now we know 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) . On using this in our expression, we get
=cos⁡(A+B2)sin⁡(A−B2)sin⁡C2=\dfrac{\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)}{\sin \dfrac{C}{2}}
Now as ABC is a triangle, we can say:
∠A+∠B+∠C=180∘\angle A+\angle B+\angle C=180{}^\circ
⇒∠A+∠B=180∘−∠C\Rightarrow \angle A+\angle B=180{}^\circ -\angle C
So, substituting the value of A+B in our expression. On doing so, we get
=cos⁡(180∘−C2)sin⁡(A−B2)sin⁡C2=\dfrac{\cos \left( \dfrac{180{}^\circ -C}{2} \right)\sin \left( \dfrac{A-B}{2} \right)}{\sin \dfrac{C}{2}}
=cos⁡(90∘−C2)sin⁡(A−B2)sin⁡C2=\dfrac{\cos \left( 90{}^\circ -\dfrac{C}{2} \right)\sin \left( \dfrac{A-B}{2} \right)}{\sin \dfrac{C}{2}}
We know cos⁡(90∘−X)=sin⁡X\cos \left( 90{}^\circ -X \right)=\sin X . Using this in our expression, we get
=sinC2sin⁡(A−B2)sin⁡C2=\dfrac{sin\dfrac{C}{2}\sin \left( \dfrac{A-B}{2} \right)}{\sin \dfrac{C}{2}}
=sin⁡(A−B2)=\sin \left( \dfrac{A-B}{2} \right)
The left-hand side of the equation given in the question is equal to the right-hand side of the equation. Hence, we can say that we have proved the equation given in the question.

Note: Be careful about the calculation and the signs while opening the brackets. Also, you need to learn the sine rule and the cosine rule as they are used very often. The k in the sine rule is twice the radius of the circumcircle of the triangle, i.e., sine rule can also be written as asin⁡A=bsin⁡B=csin⁡C=k=2R=abc2Δ\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k=2R=\dfrac{abc}{2\Delta } , where Δ\Delta represents the area of the triangle.