Question

If sin A+ sin B=c and cosA+cosB =c, where c is not equal to 0, express sinB + cosB in terms of c.

Answer 1

\[\begin{align}
\sin(A)+\sin(B)&=c\implies \sin(A)=c-\sin(B)\\
\therefore \sin^2(A)&=(c-\sin(B))^2\\
&=c^2-2c\sin(B)+sin^2(B)
\end{align}\]similarly, from the second equation we can get:\[\cos^2(A)=c^2-2c\cos(B)+\cos^2(B)\]using \(\cos^2(A)+\sin^2(A)=1\) we can then get:\[\begin{align}
1&=2c^2-2c\sin(B)-2c\cos(B)+\sin^2(B)+\cos^2(B)\\
&=2c^2-2c(\sin(B)+\cos(B))+1\\
\therefore 2c(\sin(B)+\cos(B))&=2c^2\\
\therefore \sin(B)+\cos(B)&=c
\end{align}\]

Answer 2

this was notan easy derivation to spot.
maybe someone else has found a quicker/better derivation?

Answer 3

Here's and alternative:
Let
\[A=\theta+\phi, B=\theta-\phi\]

\[\sin(\theta+\phi)+\sin(\theta-\phi)=\cos(\theta)+\cos(\theta-\phi)=c\]
Expand by sum/difference formulae
Let\[P=\sin(\theta+\phi)+\sin(\theta-\phi)\]\[=\sin(\theta)\cos(\phi)+\cos(\theta)\sin(\phi)+\sin(\theta)\cos(\phi)-\cos(\theta)\sin(\phi)\]\[=2\sin(\theta)\cos(\phi)\]
\[Q=\cos(\theta+\phi)+\cos(\theta-\phi)\]\[=\cos(\theta)\cos(\phi)-\sin(\theta)\sin(\phi)+\cos(\theta)\cos(\phi)-\sin(\theta)\sin(\phi)\]\[=2\cos(\theta)\cos(\phi)\]

Equate
\[P=Q=c\]
\[\sin(\theta)\cos(\phi)=\cos(\theta)\cos(\phi)\]\[\sin(\theta)=\cos(\theta)\]\[\theta=\frac{\pi}{4} or \frac{5\pi}{4}\]
From which we show that cos(B)=sin(A) and so
sin(A)+cos(B)=sin(A)+sin(B)=c