Solution 4.3:8b

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m (Lösning 4.3:8b moved to Solution 4.3:8b: Robot: moved page)
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<center> [[Image:4_3_8b.gif]] </center>
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Because
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<math>\tan v=\frac{\sin v}{\cos v}</math>, the left-hand side can be written using
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<math>\cos v</math>
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as the common denominator:
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<math>\frac{1}{\cos v}-\tan v=\frac{1}{\cos v}-\frac{\sin v}{\cos v}=\frac{\text{1-}\sin v}{\cos v}</math>
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Now, we observe that if we multiply top and bottom by with
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<math>\text{1}+\sin v</math>, the denominator will contain the denominator of the right-hand side as a factor and, in addition, the numerator can be simplified to give
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<math>\text{1}-\sin ^{2}v\text{ }=\cos ^{2}v</math>, using the conjugate rule:
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<math>\begin{align}
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& \frac{\text{1-}\sin v}{\cos v}=\frac{\text{1-}\sin v}{\cos v}\centerdot \frac{1+\sin v}{1+\sin v}=\frac{1-\sin ^{2}v}{\cos v\left( 1+\sin v \right)} \\
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& =\frac{\cos ^{2}v}{\cos v\left( 1+\sin v \right)}. \\
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\end{align}</math>
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Eliminating
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<math>\cos v</math>
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then gives the answer:
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<math>\frac{\cos ^{2}v}{\cos v\left( 1+\sin v \right)}=\frac{\cos v}{1+\sin v}</math>

Revision as of 10:53, 30 September 2008


Because \displaystyle \tan v=\frac{\sin v}{\cos v}, the left-hand side can be written using \displaystyle \cos v as the common denominator:


\displaystyle \frac{1}{\cos v}-\tan v=\frac{1}{\cos v}-\frac{\sin v}{\cos v}=\frac{\text{1-}\sin v}{\cos v}


Now, we observe that if we multiply top and bottom by with \displaystyle \text{1}+\sin v, the denominator will contain the denominator of the right-hand side as a factor and, in addition, the numerator can be simplified to give \displaystyle \text{1}-\sin ^{2}v\text{ }=\cos ^{2}v, using the conjugate rule:


\displaystyle \begin{align} & \frac{\text{1-}\sin v}{\cos v}=\frac{\text{1-}\sin v}{\cos v}\centerdot \frac{1+\sin v}{1+\sin v}=\frac{1-\sin ^{2}v}{\cos v\left( 1+\sin v \right)} \\ & =\frac{\cos ^{2}v}{\cos v\left( 1+\sin v \right)}. \\ \end{align}


Eliminating \displaystyle \cos v then gives the answer:


\displaystyle \frac{\cos ^{2}v}{\cos v\left( 1+\sin v \right)}=\frac{\cos v}{1+\sin v}

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