Question

If $A+B+C={180}^0$ then the value of $cos2A+cos2B+cos2C$ is

• A. $1+4cosAcosBcosC$
• B. $-1-4cosAcosBcosC$
• C. $-1+4cosAcosBcosC$
• D. $1-4cosAcosBcosC$

Answer 1

The correct option is B $-1-4cosAcosBcosC$

Given:$A+B+C={180}^{\circ }$

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

Using transformation angle formula $\cos C+\cos D=2\cos \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$ we have

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

Using $\cos 2\theta =2{\cos }^2\theta -1$

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

$=2\cos ({180}^{\circ }-C)\cos (A-B)+2{\cos }^{2}C-1$ since $A+B={180}^{\circ }-C$

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

$=-2\cos C[\cos (A-B)-\cos C]-1$

$=-2\cos C[\cos (A-B)-\cos ({180}^{\circ }-(A+B))]-1$ since $C={180}^{\circ }-(A+B)$

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

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

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

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