Physical Science Cos, Sin, Tan

Discussion in 'Physical Science' started by tablet, Nov 22, 2004.

  1. tablet

    tablet Premium Member

    What are the use of these? I'm more interest in plotting graphics using these but I have never fully understand them. Flash is a very interesting program that allows me to play with these function and it has alot of uses in graphics programming.

    I'm no mathematician. But would like to fully understand those three function. If you have to use number and equations in your explaination, go ahead! I'm all ears.
  2. helenheaven

    helenheaven Premium Member

    Trigonometry is a nightmare from high school... I never want to go there again!

    Actually trig can be quite fun...but I prefer algebra, some maths can be quite intriguing
  3. tablet

    tablet Premium Member

    Hel:Trigonometry is a nightmare from high school... I never want to go there again!

    Really? Interesting. IF you're not a math person and you try to learn all of it it becomes very frustrating, but if you only learn the part that interests you then you'll have a good start. :saint2:

    I didn't really like math until I got into programming but before that I really hate math. I still do! I guess one thing lead to another.
  4. pineappleupsidedown

    pineappleupsidedown Premium Member

    they are the relationship between any two sides of a triangle, more specially a right triangle.

    For example. if you have a triangle whose sides are both one, and the hypotenus is 1.414...(you can figure out the hypotenus be taking each side,squaring it, adding them together and taking the square root of the answer. 1^2+1^2=hypotenus squared, or the square root of 2, which is 1.414)


    Cos is the relationship of the side touching the angle, divided by the hypotenus or Adjecent/Hypotenus

    Sin is the relationship of the side not touching the angle, divided by the hypotenus, or Opposite/Hypotenus

    Tan is the relationship of the side not touching the angle, divided by the side touching the angle, or Opposite over adjecent.


    so the Cos of this triangle would be 1/1.414, or .707. to find the angle, you take the inverse cos (looks like cos^-1) of .707. this will give you the angle. the angle is 45. to do the reverse, you take the cos(45) and you get .707

    Did you get that?

  5. junior_smith

    junior_smith Premium Member

    yeah its all trig, sign is for when you have the oppisite and adjacent lengths, cosisgn for when you have oppostie and hyptonose and tangent for adjacent and hyptonose if i remeber my pneumonic ddvices lool
  6. pineappleupsidedown

    pineappleupsidedown Premium Member

    its actually what i had b4 junior, Sine= O/H, Cosine= A/H, and Tangent= O/A. just trying to keep tablet from getting confused.

  7. tablet

    tablet Premium Member

    I just found an online plotting program for these function. Check it out!

    Where's helen? she went offline so not to witness this madness again.

    I think if you can give me example of its application in the real world it might help. I think Todd know what I mean when I say that... ;)
  8. pineappleupsidedown

    pineappleupsidedown Premium Member


    alright, lets say you have to build a handicap ramp, and it cannot be steeper than 20 degrees. you have to make the ramp go up 24 inches. How long will the ramp have to be?


  9. tablet

    tablet Premium Member

    I like the diagram. A thousand thanks! I will absorb this knowledge bit by bit.

  10. amantine

    amantine Premium Member

    You know what's even more fun? Sine and cosine in complex numbers. It may seen more difficult, but it is actually a lot simpler.

    We'll need this formulas:

    e^(i*a) = cos(a) + i*sin(a)
    cos(-a) = cos(a)
    sin(-a) = -sin(a)

    First the definition of the cosine for a real number a:

    2*cos(a) = cos(a) + i*sin(a) + cos(a) - i*sin(a)
    2*cos(a) = cos(a) + i*sin(a) + cos(-a) + i*sin(-a)
    2*cos(a) = e^(i*a) + e^(-i*a)
    cos(a) = (e^(i*a) + e^(-i*a))/2

    A complex number means replacing a with a+b*i and that gives:

    cos(a+b*i) = (e^(i*a-b) + e^(-i*a+b))/2

    The same method gives

    sin(a+b*i) = (e^(i*a-b) - e^(-i*a+b))/(2*i)

    If you want to check that these are correct, you can search for cosine and sine at MathWorld.
  11. pineappleupsidedown

    pineappleupsidedown Premium Member

    lol, that is about what im doing in math right now, and that still looked confusing.