Solution 4.3:7a

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m (Lösning 4.3:7a moved to Solution 4.3:7a: Robot: moved page)
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We can write the expression
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<center> [[Image:4_3_7a.gif]] </center>
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<math>\text{sin}\left( x+y \right)</math>
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in terms of
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<math>\text{sin }x</math>,
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<math>\text{cos }x</math>,
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<math>\text{sin }y</math>
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and
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<math>\text{cos }y</math>
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if we use the addition formula for sine,
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<math>\text{sin}\left( x+y \right)=\sin x\centerdot \cos y+\cos x\centerdot \sin y</math>
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In turn, it is possible to express the factors
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<math>\text{cos }x</math>
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and
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<math>\text{cos }y</math>
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in terms of
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<math>\text{sin }x</math>
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and
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<math>\text{sin }y</math>
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by using the Pythagorean identity,
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<math>\begin{align}
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& \cos x=\pm \sqrt{1-\text{sin}^{2}x}=\pm \sqrt{1-\left( {2}/{3}\; \right)^{2}}=\pm \frac{\sqrt{5}}{3} \\
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& \cos y=\pm \sqrt{1-\text{sin}^{2}y}=\pm \sqrt{1-\left( {1}/{3}\; \right)^{2}}=\pm \frac{2\sqrt{2}}{3} \\
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\end{align}</math>
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Because
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<math>x</math>
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and
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<math>y</math>
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are angles in the first quadrant,
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<math>\text{cos }x</math>
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and
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<math>\text{cos }y</math>
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are positive, so we in fact have
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<math>\cos x=\frac{\sqrt{5}}{3}</math>
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and
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<math>\cos y=\frac{2\sqrt{2}}{3}</math>
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Finally, we obtain
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<math>\sin \left( x+y \right)=\frac{2}{3}\centerdot \frac{2\sqrt{2}}{3}+\frac{\sqrt{5}}{3}\centerdot \frac{1}{3}=\frac{4\sqrt{2}+\sqrt{5}}{9}</math>

Revision as of 10:14, 30 September 2008

We can write the expression \displaystyle \text{sin}\left( x+y \right) in terms of \displaystyle \text{sin }x, \displaystyle \text{cos }x, \displaystyle \text{sin }y and \displaystyle \text{cos }y if we use the addition formula for sine,


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


In turn, it is possible to express the factors \displaystyle \text{cos }x and \displaystyle \text{cos }y in terms of \displaystyle \text{sin }x and \displaystyle \text{sin }y by using the Pythagorean identity,


\displaystyle \begin{align} & \cos x=\pm \sqrt{1-\text{sin}^{2}x}=\pm \sqrt{1-\left( {2}/{3}\; \right)^{2}}=\pm \frac{\sqrt{5}}{3} \\ & \cos y=\pm \sqrt{1-\text{sin}^{2}y}=\pm \sqrt{1-\left( {1}/{3}\; \right)^{2}}=\pm \frac{2\sqrt{2}}{3} \\ \end{align}


Because \displaystyle x and \displaystyle y are angles in the first quadrant, \displaystyle \text{cos }x and \displaystyle \text{cos }y are positive, so we in fact have


\displaystyle \cos x=\frac{\sqrt{5}}{3} and \displaystyle \cos y=\frac{2\sqrt{2}}{3}


Finally, we obtain


\displaystyle \sin \left( x+y \right)=\frac{2}{3}\centerdot \frac{2\sqrt{2}}{3}+\frac{\sqrt{5}}{3}\centerdot \frac{1}{3}=\frac{4\sqrt{2}+\sqrt{5}}{9}

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