Solution 4.3:7a

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We can write the expression
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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,
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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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{{Displayed math||<math>\sin (x+y) = \sin x\cdot \cos y + \cos x\cdot \sin y\,\textrm{.}</math>}}
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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 <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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\cos x &= \pm \sqrt{1-\sin^2\!x} = \pm \sqrt{1-(2/3)^2} = \pm\frac{\sqrt{5}}{3}\,,\\[5pt]
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\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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\end{align}</math>}}
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In turn, it is possible to express the factors
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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
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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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{{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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<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
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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<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>
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Current revision

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{.}
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