Lösung 4.3:7a

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We can write the expression <math>\sin (x+y)</math> in terms of <math>\sin x</math>, <math>\cos x</math>, <math>\sin y</math> and <math>\cos y</math> if we use the addition formula for sine,
We can write the expression <math>\sin (x+y)</math> in terms of <math>\sin x</math>, <math>\cos x</math>, <math>\sin y</math> and <math>\cos y</math> if we use the addition formula for sine,
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{{Displayed math||<math>\sin (x+y) = \sin x\cdot \cos y + \cos x\cdot \sin y\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\sin (x+y) = \sin x\cdot \cos y + \cos x\cdot \sin y\,\textrm{.}</math>}}
In turn, it is possible to express the factors <math>\cos x</math> and <math>\cos y</math> in terms of <math>\sin x</math> and <math>\sin y</math> by using the Pythagorean identity,
In turn, it is possible to express the factors <math>\cos x</math> and <math>\cos y</math> in terms of <math>\sin x</math> and <math>\sin y</math> by using the Pythagorean identity,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\cos x &= \pm \sqrt{1-\sin^2\!x} = \pm \sqrt{1-(2/3)^2} = \pm\frac{\sqrt{5}}{3}\,,\\[5pt]
\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{.}
\cos y &= \pm \sqrt{1-\sin^2\!y} = \pm \sqrt{1-(1/3)^{2}} = \pm \frac{2\sqrt{2}}{3}\,\textrm{.}
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Because ''x'' and ''y'' are angles in the first quadrant, <math>\cos x</math> and <math>\cos y</math> are positive, so we in fact have
Because ''x'' and ''y'' are angles in the first quadrant, <math>\cos x</math> and <math>\cos y</math> are positive, so we in fact have
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{{Displayed math||<math>\cos x = \frac{\sqrt{5}}{3}\qquad\text{and}\qquad\cos y = \frac{2\sqrt{2}}{3}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\cos x = \frac{\sqrt{5}}{3}\qquad\text{and}\qquad\cos y = \frac{2\sqrt{2}}{3}\,\textrm{.}</math>}}
Finally, we obtain
Finally, we obtain
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{{Displayed math||<math>\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{.}</math>}}
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{{Abgesetzte Formel||<math>\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{.}</math>}}

Version vom 08:56, 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{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.3:7a“