Question
Question: In any \(\Delta ABC\) , prove that \({{a}^{3}}\sin \left( B-C \right)+{{b}^{3}}\sin \left( C-A \ri...
In any ΔABC , prove that
a3sin(B−C)+b3sin(C−A)+c3sin(A−B)=0
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(A−B)=sin2A−sin2B .
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: sinAa=sinBb=sinCc=k and in other terms, it can be written as:
a=ksinAb=ksinBc=ksinC
So, applying this to our expression, we get
a3sin(B−C)+b3sin(C−A)+c3sin(A−B)
a2ksinAsin(B−C)+b2ksinBsin(C−A)+c2ksinCsin(A−B)
Now as ABC is a triangle, we can say:
∠A+∠B+∠C=180∘
⇒∠A=180∘−∠C−∠B
So, substituting the values in our expression. On doing so, we get
=a2ksin(180∘−B−C)sin(B−C)+b2ksin(180∘−C−A)sin(C−A)+c2ksin(180∘−A−B)sin(A−B)
We know sin(180∘−X)=sinX . Using this in our expression, we get
=a2ksin(B+C)sin(B−C)+b2ksin(C+A)sin(C−A)+c2ksin(A+B)sin(A−B)
Now using the identity: sin(A+B)sin(A−B)=sin2A−sin2B , we get
=ka2(sin2B−sin2C)+kb2(sin2C−sin2A)+kc2(sin2A−sin2B)
Now we can also interpret the sine rule as:
asinB=bsinAbsinC=csinBcsinA=asinC
So, using this in our expression, we get
=k(a2sin2B−a2sin2C+b2sin2C−b2sin2A+c2sin2A−c2sin2B)
=k(b2sin2A−a2sin2C+c2sin2B−b2sin2A+a2sin2C−c2sin2B)
It is clearly seen that all the terms are cancelled. So, we get
=k×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 sinAa=sinBb=sinCc=k=2R=2Δabc , where Δ represents the area of the triangle.