Question

Value of $cos(A-B)+cos(B-C)+cos(C-A)$ is ?

• A. $-\frac{3}{2}$
• B. $\frac{\sqrt{3}}{2}+1$
• C. $1+\sqrt{3}$
• D. $\sqrt{3}+\frac{1}{2}$

Answer 1

The correct option is A $-\frac{3}{2}$

Value of $\cos (A-B)+\cos (B-C)+\cos (C-A)$

$=\cos A\cos B+\sin A\sin B+\cos B\cos C+\sin B\sin C+\cos C\cos A+\sin C\sin A------\left(1\right)$

Now, if $\cos A+\cos B+\cos C=0$ and $\sin A+\sin B+\sin C=0$

So:- $(\cos A+\cos B+\cos C{)}^2=0$

$={\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C+2\cos A\cos B+2\cos B\cos C+2\cos A\cos C=0----\left(2\right)$

Similarly ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C+2\sin A\sin B+2\sin B\sin C+2\sin A\sin C=0-----\left(3\right)$

And $\left(2\right)$ & $\left(3\right)\Rightarrow$

$\Rightarrow {\sin }^{2}A+{\cos }^{2}A+{\sin }^{2}B+{\cos }^{2}B+{\sin }^{2}C+{\cos }^{2}C+2(\sin A\sin B+\sin B\sin C+\sin A\sin C+\cos A\cos B+\cos B\cos C+\cos C\cos A)=0$

By $\left(1\right)$

$\left(\cos \right(A-B)+\cos (B-C)+\cos (C-A\left)\right)=-3/2$