Lösung 4.3:7a
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Version vom 15:07, 22. Okt. 2008
We can write the expression \displaystyle \sin (x+y) in terms of \displaystyle \sin x, \displaystyle \cos x, \displaystyle \sin y and \displaystyle \cos y if we use the addition formula for sine,
| \displaystyle \sin (x+y) = \sin x\cdot \cos y + \cos x\cdot \sin y\,\textrm{.} |
In turn, it is possible to express the factors \displaystyle \cos x and \displaystyle \cos y in terms of \displaystyle \sin x and \displaystyle \sin y by using the Pythagorean identity,
| \displaystyle \begin{align}
\cos x &= \pm \sqrt{1-\sin^2\!x} = \pm \sqrt{1-(2/3)^2} = \pm\frac{\sqrt{5}}{3}\,,\\[5pt] \cos y &= \pm \sqrt{1-\sin^2\!y} = \pm \sqrt{1-(1/3)^{2}} = \pm \frac{2\sqrt{2}}{3}\,\textrm{.} \end{align} |
Because x and y are angles in the first quadrant, \displaystyle \cos x and \displaystyle \cos y are positive, so we in fact have
| \displaystyle \cos x = \frac{\sqrt{5}}{3}\qquad\text{and}\qquad\cos y = \frac{2\sqrt{2}}{3}\,\textrm{.} |
Finally, we obtain
| \displaystyle \sin (x+y) = \frac{2}{3}\cdot \frac{2\sqrt{2}}{3} + \frac{\sqrt{5}}{3}\cdot \frac{1}{3} = \frac{4\sqrt{2} + \sqrt{5}}{9}\,\textrm{.} |
