Question

I: lf $A+B+C=\frac{3\pi }{2}$, then $\cos 2A+\cos 2B+\cos 2C+4\sin A\sin B\sin C=0$.
II. lf $A+B+C=0$, then $\sin 2A+\sin 2B+\sin 2C+4\sin A\sin B\sin C=0$

• A. only I is true
• B. only II is true
• C. both I and II are true
• D. neither I nor II are true

Answer 1

Answer: (B)

only II is true

I: $A+B+C=\frac{3\pi }{2}\Rightarrow 2A+2B+2C=3\pi$

$\cos 2A+\cos 2B+\cos 2C+4\sin A\sin B\sin C$

$=2\cos (A+B)\cos (A-B)+\cos (3\pi -2(A+B)+4\sin A\sin B\sin C$

$=2\cos (A+B)\cos (A-B)-\cos 2(A+B)+4\sin A\sin B\sin C$

$=2\cos (A+B)\cos (A-B)-2{\cos }^2(A+B)+1+4\sin A\sin B\sin C$

$=2\cos (\frac{3\pi }{2}-C)\left(\cos \right(A-B)-cos(A+B)+1+4\sin A\sin B\sin C$

$=-2\sin \left(C\right)\left(2\sin A\sin B\right)+1+4\sin A\sin B\sin C$

$=-4\sin A\sin B\sin C+1+4\sin A\sin B\sin C=1$

II: $A+B+C=0\Rightarrow 2A+2B+2C=0$

$\sin 2A+\sin 2B+\sin 2C+4\sin A\sin B\sin C$

$=2\sin (A+B)\cos (A-B)+\sin 2(-(A+B\left)\right)+4\sin A\sin B\sin C$

$=2\sin (A+B)\cos (A-B)-2\sin (A+B)\cos (A+B)+4\sin A\sin B\sin C$

$=2\sin (A+B)\left(\cos \right(A-B)-\cos (A+B\left)\right)+4\sin A\sin B\sin C$

$=2\sin (-C)\left(2\sin A\sin B\right)+4\sin A\sin B\sin C$

$=-4\sin A\sin B\sin C+4\sin A\sin B\sin C=0$

Hence statement I is incorrect and II is correct.