Question

If sin α + sin β = a and cos α + cos β = b, show that

(i) $\sin \left(\mathrm{\alpha }+\mathrm{\beta }\right)=\frac{2ab}{a^2+b^2}$
(ii) $\cos \left(\mathrm{\alpha }+\mathrm{\beta }\right)=\frac{b^2-a^2}{b^2+a^2}$

Answer 1

(i)
$$\begin{gathered} a^2+b^2={{\left(\sin \alpha +\sin \beta \right)}^2+(\cos \alpha +\cos \beta )}^2 \\ \Rightarrow a^2+b^2={\sin }^2\alpha +{\sin }^2\beta +2\sin \alpha \sin \beta +{\cos }^2\alpha +{\cos }^2\beta +2\cos \alpha \cos \beta \\ \Rightarrow a^2+b^2={\sin }^2\alpha +{\cos }^2\alpha +{\sin }^2\beta +{\cos }^2\beta +2\left(\sin \alpha \sin \beta +\cos \alpha \cos \beta \right) \\ \Rightarrow a^2+b^2=2+2\cos (\alpha -\beta )...\left(1\right) \end{gathered}$$

Now,

$$\begin{gathered} b^2-a^2={(\cos \alpha +\cos \beta )}^2-{\left(\sin \alpha +\sin \beta \right)}^2 \\ \Rightarrow b^2-a^2={\cos }^2\alpha +{\cos }^2\beta -{\sin }^2\alpha -{\sin }^2\beta +2\cos \alpha \cos \beta -2\sin \alpha \sin \beta \\ \Rightarrow b^2-a^2=({\cos }^2\alpha -{\sin }^2\beta )+({\cos }^2\beta -{\sin }^2\alpha )-2\cos (\alpha +\beta ) \\ \Rightarrow b^2-a^2=2\cos (\alpha +\beta )\cos (\alpha -\beta )+2\cos (\alpha -\beta ) \\ \Rightarrow b^2-a^2=\cos (\alpha +\beta )\left(2+2\cos (\alpha -\beta )\right)...\left(2\right) \end{gathered}$$

From (1) and (2), we have

$$\begin{gathered} b^2-a^2=\cos (\alpha +\beta )\left(a^2+b^2\right) \\ \Rightarrow \frac{b^2-a^2}{a^2+b^2}=\cos (\alpha +\beta ) \end{gathered}$$

$$\begin{gathered} \Rightarrow \sin \left(\alpha +\beta \right)=\sqrt{1-{\cos }^2(\alpha +\beta )} \\ \Rightarrow \sin \left(\alpha +\beta \right)=\sqrt{1-{\left(\frac{b^2-a^2}{b^2+a^2}\right)}^2}=\sqrt{\frac{b^4+a^4-b^4-a^4+4a^2{b}^2}{{\left(b^2+a^2\right)}^2}} \\ \Rightarrow \sin \left(\alpha +\beta \right)=\frac{2ab}{a^2+b^2} \end{gathered}$$

(ii)

$$\begin{gathered} a^2+b^2={{\left(\sin \alpha +\sin \beta \right)}^2+(\cos \alpha +\cos \beta )}^2 \\ ={\sin }^2\alpha +{\sin }^2\beta +{\cos }^2\alpha +{\cos }^2\beta +2\sin \alpha \sin \beta +2\cos \alpha \cos \beta \\ =2+2\cos (\alpha -\beta ) \\ \Rightarrow b^2-a^2={(\cos \alpha +\cos \beta )}^2-{\left(\sin \alpha +\sin \beta \right)}^2 \\ \Rightarrow b^2-a^2={\cos }^2\alpha +{\cos }^2\beta -{\sin }^2\alpha -{\sin }^2\beta +2\cos \alpha \cos \beta -2\sin \alpha \sin \beta \\ \Rightarrow b^2-a^2=({\cos }^2\alpha -{\sin }^2\beta )+({\cos }^2\beta -{\sin }^2\alpha )-2\cos (\alpha +\beta ) \\ \Rightarrow b^2-a^2=2\cos (\alpha +\beta )\cos (\alpha -\beta )+2\cos (\alpha -\beta ) \\ \Rightarrow b^2-a^2=\cos (\alpha +\beta )\left(2+2\cos (\alpha -\beta )\right) \\ \Rightarrow b^2-a^2=\cos (\alpha +\beta )\left(a^2+b^2\right) \\ \frac{b^2-a^2}{a^2+b^2}=\cos (\alpha +\beta ) \end{gathered}$$