Question

I : lf $A+B+C={180}^{\circ }$, then $\cos 2A+\cos 2B+\cos 2C=1-4\cos A\cos B\cos C$
II: If $A+B+C=0^{\circ }$, then $\cos A+\cos B+\cos C=-1+4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$

• 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: As $A+B+C=\pi \Rightarrow C=\pi -(A+B)$

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

$=2\cos (A+B)\cos (A-B)+\cos (2\pi -2(A+B\left)\right)$

$=2\cos (A+B)\cos (A-B)+\cos 2(A+B)$

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

Again using $A+B+C=\pi \Rightarrow \frac{A+B}{2}=\frac{\pi }{2}-\frac{C}{2}$

We get

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

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

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

II: As $A+B+C=\pi \Rightarrow C=\pi -(A+B)$

$\cos A+\cos B+\cos C$

$=2\cos \frac{A+B}{2}\cos \frac{A-B}{2}+\cos (-(A+B\left)\right)$

$=2\cos \frac{A+B}{2}\cos \frac{A-B}{2}+2{\cos }^2\frac{A+B}{2}-1$

$=2\cos \frac{A+B}{2}(\cos \frac{A-B}{2}+\cos \frac{A+B}{2})-1$

Again using $A+B+C=0\Rightarrow \frac{(A+B)}{2}=-\frac{C}{2}$

We get

$\cos A+\cos B+\cos C$

$=2\cos (-\frac{C}{2})\left(2\cos \frac{A}{2}\cos \frac{B}{2}\right)-1$

$=-1+4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$

Hence I is incorrect and II is correct.