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Question: In any \(\Delta ABC\) , prove that \({{a}^{3}}\sin \left( B-C \right)+{{b}^{3}}\sin \left( C-A \ri...

In any ΔABC\Delta ABC , prove that
a3sin(BC)+b3sin(CA)+c3sin(AB)=0{{a}^{3}}\sin \left( B-C \right)+{{b}^{3}}\sin \left( C-A \right)+{{c}^{3}}\sin \left( A-B \right)=0

Explanation

Solution

Hint: Try to simplify the left-hand side of the equation by applying the sine rule followed by the use of the identity: sin(A+B)sin(AB)=sin2Asin2B\sin \left( A+B \right)\sin \left( A-B \right)={{\sin }^{2}}A-{{\sin }^{2}}B .

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

Now let’s start with the simplification of the left-hand side of the equation given in the question.
Now we know, according to the sine rule of the triangle: asinA=bsinB=csinC=k\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k and in other terms, it can be written as:
a=ksinA b=ksinB c=ksinC \begin{aligned} & a=k\sin A \\\ & b=k\sin B \\\ & c=k\sin C \\\ \end{aligned}
So, applying this to our expression, we get
a3sin(BC)+b3sin(CA)+c3sin(AB){{a}^{3}}\sin \left( B-C \right)+{{b}^{3}}\sin \left( C-A \right)+{{c}^{3}}\sin \left( A-B \right)
a2ksinAsin(BC)+b2ksinBsin(CA)+c2ksinCsin(AB){{a}^{2}}ksinA\sin \left( B-C \right)+{{b}^{2}}ksinB\sin \left( C-A \right)+{{c}^{2}}ksinC\sin \left( A-B \right)

Now as ABC is a triangle, we can say:
A+B+C=180\angle A+\angle B+\angle C=180{}^\circ
A=180CB\Rightarrow \angle A=180{}^\circ -\angle C-\angle B
So, substituting the values in our expression. On doing so, we get
=a2ksin(180BC)sin(BC)+b2ksin(180CA)sin(CA)+c2ksin(180AB)sin(AB)={{a}^{2}}ksin\left( 180{}^\circ -B-C \right)\sin \left( B-C \right)+{{b}^{2}}ksin\left( 180{}^\circ -C-A \right)\sin \left( C-A \right)+{{c}^{2}}ksin\left( 180{}^\circ -A-B \right)\sin \left( A-B \right)

We know sin(180X)=sinX\sin \left( 180{}^\circ -X \right)=\sin X . Using this in our expression, we get
=a2ksin(B+C)sin(BC)+b2ksin(C+A)sin(CA)+c2ksin(A+B)sin(AB)={{a}^{2}}ksin\left( B+C \right)\sin \left( B-C \right)+{{b}^{2}}ksin\left( C+A \right)\sin \left( C-A \right)+{{c}^{2}}ksin\left( A+B \right)\sin \left( A-B \right)

Now using the identity: sin(A+B)sin(AB)=sin2Asin2B\sin \left( A+B \right)\sin \left( A-B \right)={{\sin }^{2}}A-{{\sin }^{2}}B , we get
=ka2(sin2Bsin2C)+kb2(sin2Csin2A)+kc2(sin2Asin2B)=k{{a}^{2}}\left( {{\sin }^{2}}B-{{\sin }^{2}}C \right)+k{{b}^{2}}\left( {{\sin }^{2}}C-{{\sin }^{2}}A \right)+k{{c}^{2}}\left( {{\sin }^{2}}A-{{\sin }^{2}}B \right)

Now we can also interpret the sine rule as:
asinB=bsinA bsinC=csinB csinA=asinC \begin{aligned} & a\sin B=b\sin A \\\ & b\sin C=c\sin B \\\ & c\sin A=a\sin C \\\ \end{aligned}
So, using this in our expression, we get
=k(a2sin2Ba2sin2C+b2sin2Cb2sin2A+c2sin2Ac2sin2B)=k\left( {{a}^{2}}{{\sin }^{2}}B-{{a}^{2}}{{\sin }^{2}}C+{{b}^{2}}{{\sin }^{2}}C-{{b}^{2}}{{\sin }^{2}}A+{{c}^{2}}{{\sin }^{2}}A-{{c}^{2}}{{\sin }^{2}}B \right)
=k(b2sin2Aa2sin2C+c2sin2Bb2sin2A+a2sin2Cc2sin2B)=k\left( {{b}^{2}}{{\sin }^{2}}A-{{a}^{2}}{{\sin }^{2}}C+{{c}^{2}}{{\sin }^{2}}B-{{b}^{2}}{{\sin }^{2}}A+{{a}^{2}}{{\sin }^{2}}C-{{c}^{2}}{{\sin }^{2}}B \right)

It is clearly seen that all the terms are cancelled. So, we get
=k×0=0=k\times 0=0

The left-hand side of the equation given in the question is equal to the right-hand side of the equation which is equal to zero. 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 asinA=bsinB=csinC=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.