Easier way to solve this problem of trigonometry.

Question

Prove that $\sin x \sin y \sin(x-y) + \sin y \sin z \sin(y-z) + \sin z \sin x \sin(z-x) + \sin(x-y)\sin(y-z)\sin(z-x) = 0$ . When I expanded them ,it became horrendous. Is there any easy way or trick to prove this? Plz help.

Answer 1

HINT:

$$2\sin x\sin y=\cos(x-y)-\cos(x+y)$$

$$\implies2(2\sin x\sin y)\sin(x-y)=2\sin(x-y)[\cos(x-y)-\cos(x+y)]=\sin(2x-2y)-(\sin2x-\sin2y)$$

$$\sum 2(2\sin x\sin y)\sin(x-y)=\sum\sin(2x-2y)$$

Now setting $x-y=A$ etc., $$\sin2A+\sin2B+\sin2C=2\sin(A+B)\cos(A-B)+2\sin C\cos C$$

As $A+B+C=0\implies \sin(A+B)=\sin(-C)=-\sin C,\cos(A+B)=\cdots=\cos C$