1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Deriving trig formulas

  1. Aug 29, 2008 #1
    hello. i need help deriving trig formulas.

    the first problem i most derive is

    51. sin^2 x + cos^2 x = 1


    i can use these identities:

    a. sin(-x) = -sinx
    b. cos(-x) = cos(x)
    c. cos(x+y) = cosxcosy - sinxsiny
    d. sin(x+y) = sinxcosy + cosxsiny

    thanks for any help
     
  2. jcsd
  3. Aug 29, 2008 #2
    hello. i need help deriving trig formulas.

    the first problem i most derive is

    51. sin^2 x + cos^2 x = 1


    i can use these identities:

    a. sin(-x) = -sinx
    b. cos(-x) = cos(x)
    c. cos(x+y) = cosxcosy - sinxsiny
    d. sin(x+y) = sinxcosy + cosxsiny

    thanks for any help
     
  4. Aug 29, 2008 #3
    i think i posted in the wrong area...sorry...my bad
     
  5. Aug 29, 2008 #4
    well, the identity

    [tex] sin^2x+cos^2x=1[/tex] is the fundamental identity in trig. It can be derived directly from the unit trig circle.

    So, what u basically do is this. You draw the oriented trig. circle, which means that your radius is 1. then your sinx will lay on the vertical line ( along the y-axis, if you place the trig circle in the origin of the coordinate system) and your cosx will lie along the x-axis.

    Now you need to use pythagorean theorem, so you'll have :[tex] x^2+y^2=1[/tex] but remember that x is actually the sin and y cos. and radius is 1.
     
  6. Aug 29, 2008 #5

    danago

    User Avatar
    Gold Member

    Try starting with cos(x + -x)
     
  7. Aug 29, 2008 #6

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    Look at your equation C. What happens when you set y = -x?
     
  8. Aug 30, 2008 #7

    Gib Z

    User Avatar
    Homework Helper

    It's ok, at least you know now.

    I don't actually believe its possible to derive that equation 51 from those listed identities. It's much easier to derive using the definition of the trig functions on the unit circle anyway - much more fundamental. Identities a and b are quite basic and don't exactly help, and c & d are more complex identities than eqn 51 itself.
     
  9. Aug 30, 2008 #8
    There is indeed some unclarity about what one is supposed to use in the proof. If we define two functions, both identically zero, [tex]f_1=0[/tex] and [tex]f_2=0[/tex], then these functions satisfy the properties

    [tex]
    f_1(-x)=-f_1(x)
    [/tex]

    [tex]
    f_2(-x) = f_2(x)
    [/tex]

    [tex]
    f_2(x + y) = f_2(x)f_2(y) - f_1(x)f_1(y)
    [/tex]

    [tex]
    f_1(x+y) = f_2(x) f_1(y) + f_1(x) f_2(y)
    [/tex]

    and it is clear that nobody will prove the identity

    [tex]
    f_1^2(x) + f_2^2(x) = 1
    [/tex]

    now.
     
  10. Aug 30, 2008 #9

    atyy

    User Avatar
    Science Advisor

    How about using your third identity, with y=-x:

    LHS: cos(x+(-x)) = cos(0) = 1

    RHS: cos(x)cos(-x) - sin(x)sin(-x)
    =cos(x)cos(x) -sin(x)[-sin(x)] (Here we use your first two identities)
    =cos2x+sin 2x

    So the identities you gave are not enough. We also needed cos(0)=1.
     
    Last edited: Aug 30, 2008
  11. Aug 30, 2008 #10

    Defennder

    User Avatar
    Homework Helper

    ? I don't see what this has got to do with the original post. Your f1 and f2 are constant zero functions, so how does this relate to sin and cos?
     
  12. Aug 30, 2008 #11

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    His point is that f1 and f2 satisfy all the given equations except f12+ f22= 1 so it is impossible to use the given equations to prove that.
     
  13. Aug 30, 2008 #12

    Defennder

    User Avatar
    Homework Helper

    Ok, I see.
     
  14. Aug 30, 2008 #13

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi atyy! Nice! :smile:
    No … we can say sin(0) = sin(-0) = -sin(0), so sin(0) = 0.

    Then cos(x) = cos(0 + x) = cos(0)cos(x) + 0 sin(x), so cos(0) = 1. :smile:
     
  15. Aug 30, 2008 #14
    atyy, very nice remark. I didn't notice that such small addition to the assumptions was enough. btw. the original post was quite hmhm.. "homework like". Could it be that that was a little bit too explicit answer?

    Carefully!

    [tex]
    \cos(x) = \cos(0)\cos(x)\quad\implies\quad \cos(0)=1\;\textrm{or}\;\cos(x)=0
    [/tex]
     
  16. Aug 30, 2008 #15

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    No … [tex]\implies\quad \cos(0)=1\;\textrm{ , or}\;\cos(x)=0 \text{ FOR ALL x}[/tex] …

    and if cosx = 0 for all x, then sinx = 0 also. :cry:
     
  17. Aug 30, 2008 #16
    I see. So we don't need to assume that cos(0)=1, but it suffices to assume that cos(x)!=0 with some x, and then cos(0)=1 follows.
     
  18. Aug 30, 2008 #17

    atyy

    User Avatar
    Science Advisor


    Wow, jostpuur and tiny-tim! That's the nicest thing I've learnt in some time.:rofl:
     
  19. Aug 30, 2008 #18
  20. Aug 30, 2008 #19

    symbolipoint

    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    That first one , [tex] sin^2x+cos^2x=1[tex]
    is merely the Pythagorean Theorem applied to the unit circle.
     
  21. Aug 31, 2008 #20
    i made progress, but hit a dead end again...

    sin^2x + cos^2x = 1
    sin^2x + (cosx)(cosx) = 1
    sin^2x + cos(-x)(cosx) = 1
    sin^2x = 1

    can any one bring it home...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Deriving trig formulas
Loading...