Question

$\sin A+\sin B=a$ and $\cos A+\cos B=b$ then find the value of $\cos (A-B)$ is

• A. $\frac{a^2+b^2-1}{2}$
• B. $\frac{a^2+b^2-2}{2}$
• C. $a^2+b^2$
• D. none of these

Answer 1

The correct option is B $\frac{a^2+b^2-2}{2}$

Given,

$\sin A+\sin B=a$......(1) and $\cos A+\cos B=b$......(2).

Now squaring and adding (1) and (2) we get,

${\sin }^{2}A+{\cos }^{2}A+{\sin }^{2}B+{\cos }^{2}B+2\sin A.\sin B+2\cos A.\cos B=a^2+b^2$

or, $2\cos (A-B)=a^2+b^2-2$

or, $\cos (A-B)=\frac{a^2+b^2-2}{2}$.