Aus Online Mathematik Brückenkurs 1

We can write the expression \displaystyle \text{sin}\left( x+y \right) in terms of \displaystyle \text{sin }x, \displaystyle \text{cos }x, \displaystyle \text{sin }y and \displaystyle \text{cos }y if we use the addition formula for sine,

\displaystyle \text{sin}\left( x+y \right)=\sin x\centerdot \cos y+\cos x\centerdot \sin y

In turn, it is possible to express the factors \displaystyle \text{cos }x and \displaystyle \text{cos }y in terms of \displaystyle \text{sin }x and \displaystyle \text{sin }y by using the Pythagorean identity,

\displaystyle \begin{align} & \cos x=\pm \sqrt{1-\text{sin}^{2}x}=\pm \sqrt{1-\left( {2}/{3}\; \right)^{2}}=\pm \frac{\sqrt{5}}{3} \\ & \cos y=\pm \sqrt{1-\text{sin}^{2}y}=\pm \sqrt{1-\left( {1}/{3}\; \right)^{2}}=\pm \frac{2\sqrt{2}}{3} \\ \end{align}

Because \displaystyle x and \displaystyle y are angles in the first quadrant, \displaystyle \text{cos }x and \displaystyle \text{cos }y are positive, so we in fact have

\displaystyle \cos x=\frac{\sqrt{5}}{3} and \displaystyle \cos y=\frac{2\sqrt{2}}{3}

Finally, we obtain

\displaystyle \sin \left( x+y \right)=\frac{2}{3}\centerdot \frac{2\sqrt{2}}{3}+\frac{\sqrt{5}}{3}\centerdot \frac{1}{3}=\frac{4\sqrt{2}+\sqrt{5}}{9}