Exercise 397

Let $a,b$ be real numbers such that $\sin a+\sin b=1$ and $\cos a+\cos b=1$. Find $\sin(a+b)$.

Solution

By sum-to-product, as in the problem, $t=\tan\frac{a+b}2=\dfrac{\sin a+\sin b}{\cos a+\cos b}=1$ (both sums are nonzero).
So $\frac{a+b}2=45^\circ+180^\circ k$ and $a+b=90^\circ+360^\circ k$ for an integer $k$. Equivalently, $\sin(a+b)=\dfrac{2t}{1+t^2}=\boxed1$.