anonymous
  • anonymous
Find the value for cos(θ) if the following conditions hold: cos(2θ)=1/root2 and 180degrees<θ<270degrees
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
Refer to a solution using the Mathematica v9 computer program.
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anonymous
  • anonymous
\( \huge cos(2θ)= \cos^2 \theta -\sin^2 \theta = \frac{1}{\sqrt{2}}\) Rationalize \( \huge \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2} \) \( \huge \cos^2 \theta -\sin^2 \theta = \frac{\sqrt{2}}{2}\) \( \huge \cos^2 \theta = \frac{\sqrt{2}}{2} +\sin^2 \theta\) Now we use this identity \( \huge \sin^2 \theta = \frac{1-\cos(2\theta)}{2}\) \( \huge \cos^2 \theta = \frac{\sqrt{2}}{2} +\frac{1-cos(2\theta)}{2}\) Remember \( \huge \cos(2\theta) = \frac{\sqrt{2}}{2}\) \( \huge \cos^2 \theta = \frac{\sqrt{2}}{2} +\frac{1-\frac{\sqrt{2}}{2}}{2}\)\) \( \huge \cos^2 \theta = \frac{2+\sqrt{2}}{4}\) Now square each side \( \huge \sqrt{\cos^2 \theta} = \sqrt{\frac{2+\sqrt{2}}{4}}\) \( \huge \cos \theta = \frac{\sqrt{2+\sqrt{2}}}{2}\) And since we are in the third quadrant it will be \( \huge \cos \theta = -\frac{\sqrt{2+\sqrt{2}}}{2}\) WOW that was a lot of work :-)
anonymous
  • anonymous
You're taking cosine of 2Θ so multiply the interval by 2 \(\cos 2\theta = \frac{ 1 }{ \sqrt2 }=\frac{ \sqrt2 }{ 2 }\) , 360°<Θ<540° \[2\theta=405°\] \[\theta=202.5°\]

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