Aus Online Mathematik Brückenkurs 1

Using the addition formula, we rewrite \displaystyle \sin (x+y) as

\displaystyle \sin (x+y) = \sin x\cdot\cos y + \cos x\cdot\sin y\,\textrm{.}

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 \sin x and \displaystyle \sin y in terms of \displaystyle \cos x and \displaystyle \cos y,

\displaystyle \begin{align} \sin x &= \pm\sqrt{1-\cos^2\!x} = \pm\sqrt{1-\bigl(\tfrac{2}{5}\bigr)^2} = \pm\sqrt{1-\tfrac{4}{25}} = \pm\frac{\sqrt{21}}{5}\,,\\[5pt] \sin y &= \pm\sqrt{1-\cos^2\!y} = \pm\sqrt{1-\bigl(\tfrac{3}{5}\bigr)^2} = \pm\sqrt{1-\tfrac{9}{25}} = \pm\frac{4}{5}\,\textrm{.} \end{align}

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

\displaystyle \sin x = \frac{\sqrt{21}}{5}\qquad\text{and}\qquad\sin y = \frac{4}{5}\,\textrm{.}

Thus, the answer is

\displaystyle \sin (x+y) = \frac{\sqrt{21}}{5}\cdot\frac{3}{5} + \frac{2}{5}\cdot\frac{4}{5} = \frac{3\sqrt{21}+8}{25}\,\textrm{.}