Question

a^3sin(B-C) +b^3sin(C-A) +c^3sin(A-B) =0

Answer 1

Answer to Question #90824 – Math – Trigonometry

Question

$$a^3 \sin(B-C) + b^3 \sin(C-A) + c^3 \sin(A-B) = 0$$

Solution

Take L.H.S,

$$= a^3 \sin(B - C) + b^3 \sin(C - A) + c^3 \sin(A - B)$$

Using Sine rule,

$$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k \text{ [where } k \text{ is a constant]}...(\text{i})$$

Using Cosine rule,

$$\cos A = \frac{b^2 + c^2 - a^2}{2bc}, \cos B = \frac{a^2 + c^2 - b^2}{2ac}, \cos C = \frac{b^2 + a^2 - c^2}{2ba}...(\text{ii})$$

Consider,

$$a^3 \sin(B - C) = a^3 \left[ \sin B \cos C - \cos B \sin C \right]$$

Plug values from (i) and (ii):

$$\begin{aligned} &= a^3 \left[ kb \times \frac{b^2 + a^2 - c^2}{2ba} - \frac{a^2 + c^2 - b^2}{2ac} \times kc \right] \\ &= a^3 \left[ k \times \frac{b^2 + a^2 - c^2}{2a} - \frac{a^2 + c^2 - b^2}{2a} \times k \right] \\ &= ka^3 \left[ \frac{b^2 + a^2 - c^2}{2a} - \frac{a^2 + c^2 - b^2}{2a} \right] \\ &= ka^3 \left[ \frac{2b^2 - 2c^2}{2a} \right] \\ &= ka^2 \left[ b^2 - c^2 \right] \end{aligned}$$

Similarly, using values from (i) and (ii) to get,

$$b ^ {3} \sin (C - A) = k b ^ {2} \left[ c ^ {2} - a ^ {2} \right]$$ $$c ^ {3} \sin (A - B) = k c ^ {2} \left[ a ^ {2} - b ^ {2} \right]$$

Now putting all these values in L.H.S., we get,

$$\begin{array}{l} = k a ^ {2} \left[ b ^ {2} - c ^ {2} \right] + k b ^ {2} \left[ c ^ {2} - a ^ {2} \right] + k c ^ {2} \left[ a ^ {2} - b ^ {2} \right] \\ = k \left[ a ^ {2} b ^ {2} - a ^ {2} c ^ {2} + b ^ {2} c ^ {2} - b ^ {2} a ^ {2} + c ^ {2} a ^ {2} - c ^ {2} b ^ {2} \right] \\ = k \left[ a ^ {2} b ^ {2} - a ^ {2} c ^ {2} + b ^ {2} c ^ {2} - a ^ {2} b ^ {2} + a ^ {2} c ^ {2} - b ^ {2} c ^ {2} \right] \\ = k [ 0 ] \\ = 0 \\ = R.H.S \\ \end{array}$$

Hence, proved.

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