Solveeit Logo
Login

Question

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

In any ΔABC\Delta ABC , prove that
a(cos⁡C−cos⁡B)=2(b−c)cos⁡2A2a\left( \cos C-\cos B \right)=2\left( b-c \right){{\cos }^{2}}\dfrac{A}{2}

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 (cosC-cosB) and the formula of 2cosAsinB.

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(cos⁡C−cos⁡B)a\left( \cos C-\cos B \right)
=ksin⁡A(cos⁡C−cos⁡B)=k\sin A\left( \cos C-\cos B \right)
Now we know that cosC−cosB⁡=−2sin⁡(C+B2)sin⁡(B−C2)cosC-\operatorname{cosB}=-2\sin \left( \dfrac{C+B}{2} \right)\sin \left( \dfrac{B-C}{2} \right) . On using this in our expression, we get
=ksin⁡A(−2sin⁡(C+B2)sin⁡(C−B2))=k\sin A\left( -2\sin \left( \dfrac{C+B}{2} \right)\sin \left( \dfrac{C-B}{2} \right) \right)
We also know that −sin⁡X=sin⁡(−X)-\sin X=\sin \left( -X \right) . On using this in our expression, we get
=2sin⁡(C+B2)sin⁡(B−C2)ksin⁡A=2\sin \left( \dfrac{C+B}{2} \right)\sin \left( \dfrac{B-C}{2} \right)k\sin A
Now, when we use the formula sin⁡A=2sin⁡A2cos⁡A2\sin A=2\sin \dfrac{A}{2}\cos \dfrac{A}{2} , we get
=4sin⁡(C+B2)sin⁡(B−C2)ksin⁡A2cos⁡A2=4\sin \left( \dfrac{C+B}{2} \right)\sin \left( \dfrac{B-C}{2} \right)k\sin \dfrac{A}{2}\cos \dfrac{A}{2}
Now as ABC is a triangle, we can say:
∠A+∠B+∠C=180∘\angle A+\angle B+\angle C=180{}^\circ
⇒∠C+∠B=180∘−∠A\Rightarrow \angle C+\angle B=180{}^\circ -\angle A
So, substituting the value of B+C in our expression. On doing so, we get
=4sin⁡(180∘−A2)sin⁡(B−C2)ksin⁡(180∘−B−C2)cos⁡A2=4\sin \left( \dfrac{180{}^\circ -A}{2} \right)\sin \left( \dfrac{B-C}{2} \right)k\sin \left( \dfrac{180{}^\circ -B-C}{2} \right)\cos \dfrac{A}{2}
=4sin⁡(90∘−A2)sin⁡(B−C2)ksin⁡(90∘−B+C2)cos⁡A2=4\sin \left( 90{}^\circ -\dfrac{A}{2} \right)\sin \left( \dfrac{B-C}{2} \right)k\sin \left( 90{}^\circ -\dfrac{B+C}{2} \right)\cos \dfrac{A}{2}
We know sin⁡(90∘−X)=cos⁡X\sin \left( 90{}^\circ -X \right)=\cos X . Using this in our expression, we get
=4cos⁡A2sin⁡(B−C2)kcos⁡(B+C2)cos⁡A2=4\cos \dfrac{A}{2}\sin \left( \dfrac{B-C}{2} \right)k\cos \left( \dfrac{B+C}{2} \right)\cos \dfrac{A}{2}
=4sin⁡(B−C2)cos⁡(B+C2)kcos⁡2A2=4\sin \left( \dfrac{B-C}{2} \right)\cos \left( \dfrac{B+C}{2} \right)k{{\cos }^{2}}\dfrac{A}{2}
According to the formula: 2sin⁡Xcos⁡Y=sin⁡(X+Y)+sin(X−Y)2\sin X\cos Y=\sin \left( X+Y \right)+sin\left( X-Y \right) , we get
=2(sin⁡(B−C+B+C2)+sin(B−C−B−C2))kcos⁡2A2=2\left( \sin \left( \dfrac{B-C+B+C}{2} \right)+sin\left( \dfrac{B-C-B-C}{2} \right) \right)k{{\cos }^{2}}\dfrac{A}{2}
=2(ksin⁡B+ksin⁡(−C))cos⁡2A2=2\left( k\sin B+k\sin \left( -C \right) \right){{\cos }^{2}}\dfrac{A}{2}
We know that −sin⁡X=sin⁡(−X)-\sin X=\sin \left( -X \right) . On using this in our expression, we get
=2(ksin⁡B−ksinC⁡)cos⁡2A2=2\left( k\sin B-k\operatorname{sinC} \right){{\cos }^{2}}\dfrac{A}{2}
Now using the sine rule as mentioned above, we get
=2(b−c)cos⁡2A2=2\left( b-c \right){{\cos }^{2}}\dfrac{A}{2}
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.