A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

Given cos(theta)=4/9 and csc(theta)<0, find sin(theta) and tan(theta).

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can you help me and i will help you too

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @heatherchaloupek2004 What do you need help with hon?

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    cos(theta)=4/9, then \[\cos ^{2}(\theta)=16/81\] \[\sin(\theta)=\sqrt{1-\cos^{2}(\theta)}\] \[\sin(\theta)=\sqrt{1-16/81}=\sqrt{65}/9\] \[\tan(\theta)=\sin(\theta)/\cos(\theta)=\sqrt{65}/4\]

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @SwapnilP so, how do you get to sqrt(65)/9?

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Here it is: \[We have \cos(\theta)=4/9 and \sin(\theta)=\sqrt{65}/9\] so \[\tan(\theta)=\sin(\theta)/\cos(\theta)=\[(\sqrt{65}/9)/(4/9)=(\sqrt{65}/9)*(9/4)=\sqrt{65}/4\]

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\sqrt{1-16/81}=\sqrt{(81-16)/81}=\sqrt{65/81}=\sqrt{65}/\sqrt{81}=\sqrt{65}/9\]

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    LOh I gent it'!

  8. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.