Trigonometry
Complex number representation of triangle identity
MJAT_TS1_P2
Grade 12
Question:
In $\triangle ABC$, let $\ell = a^3\cos 3B + 3a^2b\cos(A-2B) + 3ab^2\cos(2A-B) + b^3\cos 3A$. Then the value of $\dfrac{\ell}{c^3}$ equals
Step-by-Step Solution
Key Concept: Recognise the expression as the real part of $(ae^{-iB} + be^{iA})^3$. Use the sine rule: $a = 2R\sin A$, $b = 2R\sin B$, $c = 2R\sin C$. Then $ae^{-iB}+be^{iA} = $ something real related to $c$.
$(ae^{-iB}+be^{iA})^3$ expanded $= a^3e^{-3iB}+3a^2be^{-iB}e^{iA}\cdot e^{i(A)}+\ldots = c^3$ (real). Taking real part: $a^3\cos 3B + 3a^2b\cos(A-2B)+3ab^2\cos(2A-B)+b^3\cos 3A = c^3$. So $\ell = c^3$ and $\dfrac{\ell}{c^3} = \mathbf{1}$.
Correct Answer: 1