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Solution 4.3:7b

From Förberedande kurs i matematik 1

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m (Lösning 4.3:7b moved to Solution 4.3:7b: Robot: moved page)
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Using the addition formula, we rewrite
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<center> [[Image:4_3_7b.gif]] </center>
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<math>\text{sin}\left( x+y \right)</math>
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as
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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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If we use the same solution procedure as in exercise a, we use the Pythagorean identity
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<math>\cos ^{2}v+\sin ^{2}v=1</math>
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to express the unknown factors
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<math>x\text{ }</math>
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and
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<math>y\text{ }</math>
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in terms of
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<math>\text{cos }x\text{ }</math>
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and
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<math>\text{cos }y</math>,
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<math>\begin{align}
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& \sin x=\pm \sqrt{1-\text{cos}^{2}x}=\pm \sqrt{1-\left( \frac{2}{5} \right)^{2}}=\pm \sqrt{1-\frac{4}{25}}=\pm \frac{\sqrt{21}}{5}, \\
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& \sin y=\pm \sqrt{1-\text{cos}^{2}y}=\pm \sqrt{1-\left( \frac{3}{5} \right)^{2}}=\pm \sqrt{1-\frac{9}{25}}=\pm \frac{4}{5} \\
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\end{align}</math>
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The angles
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<math>x\text{ }</math>
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and
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<math>y\text{ }</math>
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lie in the first quadrant and both
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<math>\text{sin }x\text{ }</math>
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and
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<math>\text{sin }y\text{ }</math>
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are therefore positive, i.e.
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<math>\sin x=\frac{\sqrt{21}}{5}</math>
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and
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<math>\sin y=\frac{4}{5}</math>
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Thus, the answer is
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<math>\text{sin}\left( x+y \right)=\frac{\sqrt{21}}{5}\centerdot \frac{3}{5}+\frac{2}{5}\centerdot \frac{4}{5}=\frac{3\sqrt{21}+8}{25}</math>

Revision as of 10:29, 30 September 2008

Using the addition formula, we rewrite sinx+y  as


sinx+y=sinxcosy+cosxsiny 


If we use the same solution procedure as in exercise a, we use the Pythagorean identity

cos2v+sin2v=1 to express the unknown factors x and y in terms of cos x and cos y,

sinx=1−cos2x=1−522=1−425=521siny=1−cos2y=1−532=1−925=54 


The angles x and y lie in the first quadrant and both sin x and sin y are therefore positive, i.e.


sinx=521  and siny=54


Thus, the answer is


sinx+y=52153+5254=25321+8 

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