Question

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

• 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 $1-4\sin A\sin B\sin C$

Solution -

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

$cos2A+cos2B+cos2C$

$cos2A+cos2B=2cos(A+B)cos(A-B)$

$=2cos(\frac{3\pi }{2}-C)cos(A-B)$

$=-2sinCcos(A-B)$

$cos2C=1-2si{n}^{2}C$

$cos2A+cos2B+cos2C=-2sinCcos(A-B)+1-2si{n}^{2}C$

$=1-2sinC\left(cos\right(A-B)+sinC)$

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

$=1-2sinC\left(cos\right(A-B)-cos(A+B\left)\right)$

$=1-2sinC\times 2sinA\times 2sinB$

$=1-1sinAsinBsinC$

D is correct