A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
Find the value for cos(θ) if the following conditions hold: cos(2θ)=1/root2 and 180degrees<θ<270degrees
anonymous
 one year ago
Find the value for cos(θ) if the following conditions hold: cos(2θ)=1/root2 and 180degrees<θ<270degrees

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Refer to a solution using the Mathematica v9 computer program.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\( \huge cos(2θ)= \cos^2 \theta \sin^2 \theta = \frac{1}{\sqrt{2}}\) 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 \( \huge \sin^2 \theta = \frac{1\cos(2\theta)}{2}\) \( \huge \cos^2 \theta = \frac{\sqrt{2}}{2} +\frac{1cos(2\theta)}{2}\) 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
 one year ago
Best ResponseYou've already chosen the best response.0You're taking cosine of 2Θ so multiply the interval by 2 \(\cos 2\theta = \frac{ 1 }{ \sqrt2 }=\frac{ \sqrt2 }{ 2 }\) , 360°<Θ<540° \[2\theta=405°\] \[\theta=202.5°\]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.