Answer to Question #137558 in Trigonometry for Axwell Alesso Lee

Question

Using complex numbers, prove that the, angles A, B and C of a planar triangle satisfy the relations

(i) cos^2 A + cos^2 B + cos^2 C = 1 − 2 cos A cos B cos C

(ii) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C.

Answer 1

$\displaystyle(i)\\\textsf{Prove that}\\ \sin(2A) + \sin(2B) + \sin(2C) = 4\sin A \sin B \sin C \\ \textsf{Since the figure is a planar triangle}\\ A + B + C = \pi\\ \textsf{It is known that,}\\ \sin P + \sin Q = 2\sin\left(\frac{P + Q}{2}\right)\cos\left(\frac{P - Q}{2}\right) \\ \begin{aligned} \sin(2A) + \sin(2B) + \sin(2C) &= 2\sin(A + B)\cos(A - B) + \sin(2C)\\ &= 2\sin(\pi - C)\cos(A - B) + 2\sin(C)\cos(C)\\ &= 2\sin(C)\cos(A - B) + 2\sin(C)\cos(C)\\ &= 2\sin(C)(\cos(A - B) + \cos(C)) \end{aligned} \\ \textsf{It is also known that,}\\ \cos P + \cos Q = 2\cos\left(\frac{P + Q}{2}\right)\cos\left(\frac{P - Q}{2}\right) \\ \begin{aligned} \implies \sin(2A) + \sin(2B) + \sin(2C) &= 2\sin(C)\left(2\cos\left(\frac{A + C - B}{2}\right)\cos\left(\frac{A -B - C}{2}\right) \right)\\ &= 4\sin C \cos\left(\frac{\pi}{2} - B\right) \cos\left(\frac{\pi}{2} - A\right) \\ &= 4\sin C \sin B \sin A\\ &= 4\sin A \sin B \sin C \end{aligned}\\ (ii)\\\begin{aligned} \cos(2A) + \cos(2B) + \cos(2C) \\&= 2\cos\left(\frac{2(A + B)}{2}\right)\cos\left(\frac{2(A - B)}{2}\right) + \cos(2C) \\&= 2\cos(\pi - C)\cos(A - B) + 2\cos^2 C - 1 \\&= 2\cos C (\cos C - \cos(A - B)) - 1 \end{aligned}\\ \textsf{It is also known that,}\\ \cos P - \cos Q = 2\sin\left(\frac{P + Q}{2}\right)\sin\left(\frac{P - Q}{2}\right) \\ \begin{aligned} \cos(2A) + \cos(2B) + \cos(2C) \\&= 2\cos C \left(2\sin\left(\frac{A + C - B}{2}\right)\sin\left(\frac{C - A+ B}{2}\right)\right) - 1 \\&= 2\cos C \left(2\sin\left(\frac{\pi}{2} - B\right)\sin\left(\frac{\pi}{2} - A\right)\right) - 1 \\&= 2\cos C(2\cos B \times 2\cos A) - 1 = 4\cos A \cos B \cos C - 1 \end{aligned}$