A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
If x=4tanθ, use trigonometric substitution to write √(16+x^2) as a trigonometric function of θ, where 0<θ<π/2
anonymous
 5 years ago
If x=4tanθ, use trigonometric substitution to write √(16+x^2) as a trigonometric function of θ, where 0<θ<π/2

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, first substitute \[4\tan \theta\] for x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you have\[\sqrt{16+16\tan ^{2}\theta}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am assuming that you would plug in 4tanθ to get: ` √(16+(4tanθ)^2) just not sure where to go from there

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can factor out a 16 and take it out of the radical \[\sqrt{16(1+\tan ^{2}\theta)} = 4\sqrt{(1+\tan ^{2}\theta)}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now you need to make a trig substitution. \[1+\tan ^{2}\theta=\sec ^{2}\theta\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK... so when you say take it out of the radical you mean...? sorry this is where I get confused

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since we had \[\sqrt{16(1+\tan ^{2}\theta)}\] we can evaluate the square root of 16

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and take it out of the square root

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so then the final step would be \[4\sqrt{\sec ^{2}\theta}=4\sec \theta\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0does that make sense?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sort of... I'm trying to figure out how you got 1+tan^2θ=sec^2θ to turn into 4√(sec^2θ)=4secθ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[4\sqrt{1+\tan ^{2}\theta}=4\sqrt{\sec ^{2}\theta}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I first just replaced 1+tan^2θ with sec^2θ

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then, we can take the square root of sec^2θ

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok that makes sense thanks!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0say I had one with cot.... √(9+(3cot^2θ) √(9+9cotθ)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or I mean √(9+9cot^2θ)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok then I think....1√(1+cot^2θ) how do I know what to substitute then?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Or I mean 9√(1+cot^2θ)
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.