Aus Online Mathematik Brückenkurs 1

Using the addition formula, we rewrite \displaystyle \text{sin}\left( x+y \right) as

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

If we use the same solution procedure as in exercise a, we use the Pythagorean identity

\displaystyle \cos ^{2}v+\sin ^{2}v=1 to express the unknown factors \displaystyle x\text{ } and \displaystyle y\text{ } in terms of \displaystyle \text{cos }x\text{ } and \displaystyle \text{cos }y,

\displaystyle \begin{align} & \sin x=\pm \sqrt{1-\text{cos}^{2}x}=\pm \sqrt{1-\left( \frac{2}{5} \right)^{2}}=\pm \sqrt{1-\frac{4}{25}}=\pm \frac{\sqrt{21}}{5}, \\ & \sin y=\pm \sqrt{1-\text{cos}^{2}y}=\pm \sqrt{1-\left( \frac{3}{5} \right)^{2}}=\pm \sqrt{1-\frac{9}{25}}=\pm \frac{4}{5} \\ \end{align}

The angles \displaystyle x\text{ } and \displaystyle y\text{ } lie in the first quadrant and both \displaystyle \text{sin }x\text{ } and \displaystyle \text{sin }y\text{ } are therefore positive, i.e.

\displaystyle \sin x=\frac{\sqrt{21}}{5} and \displaystyle \sin y=\frac{4}{5}

Thus, the answer is

\displaystyle \text{sin}\left( x+y \right)=\frac{\sqrt{21}}{5}\centerdot \frac{3}{5}+\frac{2}{5}\centerdot \frac{4}{5}=\frac{3\sqrt{21}+8}{25}