Question

If $A+B+C=\frac{3\pi }{2}$ then $\cos 2A+\cos 2B+\cos 2C=$

• A. $1-4\cos A\cos B\cos C$
• B. $4\sin A\sin B\sin C$
• C. $1+2\cos A\cos B\cos C$
• D. $1-4\sin A\sin B\sin C$

Answer 1

The correct option is D $1-4\sin A\sin B\sin C$

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

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

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

$[\because \cos C+\cos D=2\cos \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)]$

$[\because A+B=\frac{3\pi }{2}-C]$

$[\because \cos 2C=1-2{\sin }^{2}C]$

$2\cos (\frac{3\pi }{2}-C)\cos (A-B)+[1-2{\sin }^{2}C]$

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

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

$=1-2\sin \left(C\right)\left[\cos \right(A-B)-\cos (A+B\left)\right]$

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

$[\because \cos (A-B)-\cos (A+B)=2\sin A\sin B]$