Question

If $\sin A=x,\cos B=y$ and $A+B+C=0$, then $x^2+2xy\sin C+y^2$ is equal to

• A. ${\sin }^{2}C$
• B. ${\cos }^{2}C$
• C. $1+{\sin }^{2}C$
• D. $1+{\cos }^{2}C$

Answer 1

The correct option is B ${\cos }^{2}C$

It is given that $A+B+C=0\Rightarrow C=-(A+B)$

Now, $x^2+2xy\sin C+y^2={\sin }^{2}A+2\sin A\cos BsinC+{\cos }^{2}B={\sin }^{2}A+2\sin A\cos Bsin(-(A+B\left)\right)+{\cos }^{2}B={\sin }^{2}A-2\sin A\cos B[\sin A\cos B+\cos A\sin B]+{\cos }^{2}B={\sin }^{2}A-2\sin A\cos B\sin A\cos B-2\sin A\cos B\cos A\sin B+{\cos }^{2}B$

$={\sin }^{2}A-{\sin }^{2}A{\cos }^{2}B+{\cos }^{2}B-{\sin }^{2}A{\cos }^{2}B-2\sin A\cos B\cos A\sin B={\sin }^{2}A(1-co{s}^{2}B)+co{s}^{2}B(1-si{n}^{2}A)-2\sin A\cos B\cos A\sin B={\sin }^{2}Asi{n}^{2}B+co{s}^{2}Bco{s}^{2}A-2\sin A\cos B\cos A\sin B={\sin }^{2}Asi{n}^{2}B-\sin A\cos B\cos A\sin B+co{s}^{2}Bco{s}^{2}A-\sin A\cos B\cos A\sin B=\sin A\sin B[\sin A\sin B-\cos A\cos B]+cosAcosB[cosAcosB-sinAsinB]=\sin A\sin B[\sin A\sin B-\cos A\cos B]-cosAcosB[sinAsinB-cosA\cos B]=[\sin A\sin B-\cos A\cos B][\sin A\sin B-\cos A\cos B]=[-\cos (A+B\left)\right][-\cos (A+B\left)\right]$

which can be written as

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