A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

karatechopper

  • 4 years ago

GT please help in Trig

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

    Look at this link: http://en.wikipedia.org/wiki/Sine Particularly section "Relation to Unit Circle". In trignometry, sine and cosine are two basic functions. Their values are defined as ratios of sides of right triangle (formed inside the unit circle) based on various angle measures.

  2. karatechopper
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok..then..

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

    Really, first study that part and ask questions. If you understand this basic point, rest is all bunch of re-arranging and playing with it.

  4. karatechopper
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok

  5. karatechopper
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well i do have one question, but its not only a sine question its a cos,sin, and tan question. may i ask?

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

    |dw:1327282151207:dw|

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

    Ask any question any time.

  8. karatechopper
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok well how do you know if you have to solve for sin, cos, or tan? i see that some questions dont tell you..

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

    In that above triangle sin(x) = a/c cos(x) = b/c In unit circle, c = 1 (radius of that unit circle).

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

    tan is nothing but ratio of "sin and cos" tan = sin/cos

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

    What function you solve for depends on the triangle's angles and measures in question. In my triangle above, x is the specific angle in the left bottom. If x was the right top angle, then sin and cos will "flip".

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

    In general: sin = opposite side of angle / hypotenuse cos = adjacent side of angle / hypotenuse

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

    |dw:1327282536012:dw|

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

    sin(x) = a/c cos(x) = b/c tan(x) = sin(x)/cos(x) = a/c / b/c = a/b sin(y) = b/c cos(y) = a/c tan(y) = sin(y)/cos(y) = b/c / a/c = b/a

  15. karatechopper
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ?

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

    what?

  17. 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.