Lösung 4.3:7b

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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Using the addition formula, we rewrite <math>\sin (x+y)</math> as
Using the addition formula, we rewrite <math>\sin (x+y)</math> as
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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>}}
If we use the same solution procedure as in exercise a, we use the Pythagorean identity <math>\cos^2\!v + \sin^2\!v = 1</math> to express the unknown factors <math>\sin x</math> and <math>\sin y</math> in terms of <math>\cos x</math> and <math>\cos y</math>,
If we use the same solution procedure as in exercise a, we use the Pythagorean identity <math>\cos^2\!v + \sin^2\!v = 1</math> to express the unknown factors <math>\sin x</math> and <math>\sin y</math> in terms of <math>\cos x</math> and <math>\cos y</math>,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\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 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{.}
\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{.}
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The angles ''x'' and ''y'' lie in the first quadrant and both <math>\sin x</math> and <math>\sin y</math> are therefore positive, i.e.
The angles ''x'' and ''y'' lie in the first quadrant and both <math>\sin x</math> and <math>\sin y</math> are therefore positive, i.e.
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{{Displayed math||<math>\sin x = \frac{\sqrt{21}}{5}\qquad\text{and}\qquad\sin y = \frac{4}{5}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\sin x = \frac{\sqrt{21}}{5}\qquad\text{and}\qquad\sin y = \frac{4}{5}\,\textrm{.}</math>}}
Thus, the answer is
Thus, the answer is
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{{Displayed math||<math>\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{.}</math>}}
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{{Abgesetzte Formel||<math>\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{.}</math>}}

Version vom 08:56, 22. Okt. 2008

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{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.3:7b“