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Proof: Identity formula of sin(x)^2 + cos(x)^2 = 1 for *degrees*

  1. Oct 27, 2013 #1
    1. The problem statement, all variables and given/known data

    We are given two sets of functions: sin(x) and cos(x); S(x) and C(x). In the former, x is measured in radians, in the latter x is measured in degrees.
    It is possible to convert between the two using the following relations:
    sin(x) = S(mx), cos(x) = C(mx) where m=180/pi
    S(x) = sin(nx), C(x) = cos(nx), where n = pi/180

    Given the identity formula sin(x)^2 + cos(x)^2 = 1, what is a similar identity relating S(x) and C(x)?

    2. Relevant equations
    sin(x) = S(mx), cos(x) = C(mx) where m=180/pi
    S(x) = sin(nx), C(x) = cos(nx), where n = pi/180
    sin(x)^2 + cos(x)^2 = 1

    3. The attempt at a solution

    I tried to show sin(nx)^2 + cos(nx)^2 = 1, which then implies S(x)^2 +C(x)^2 = 1

    (d/dx)(sin(nx)^2 + cos(nx)^2) = (d/dx)1
    2*sin(nx)*cos(nx)*(n) -2*sin(nx)*cos(nx)*(n) = 0 for all n contained in ℝ,
    so S(x)^2 +C(x)^2 = 1

    Is this right in any way? The topic being covered is derivatives so I thought I need to use them.
    Can I simply make the assumption that sin(nx)^2 + cos(nx)^2 = 1? How should I justify this? Is there any other sort of manipulation I can do to, perhaps starting with sin(x)^2 + cos(x)^2 = 1 and then subbing in sin(x) = S(mx) and cos(x) = C(mx), and then getting the "m"s out of the brackets in some way?
     
  2. jcsd
  3. Oct 27, 2013 #2

    haruspex

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    I see no need for calculus here. You know sin(x)^2 + cos(x)^2 = 1 is true for all x. Given S(y^2) +C(y)^2 for some y, what might be a useful x to consider?
     
  4. Oct 27, 2013 #3
    So can I simply say sin(x)^2 + cos(x)^2 = 1 is true for all x, so sin(nx)^2 + cos(nx)^2 = 1?
    I'm a bit confused about why I would put y into S(y)^2 + C(y)^2... why wouldn't I put y into the sin and cos identity?
    ie: sin(y)^2 + cos(y)^2 = 1 for all y, take y = nx, therefore S(x)^2 + C(x)^2 = 1
     
  5. Oct 27, 2013 #4

    haruspex

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    Yes. I introduced y to make it less confusing, but it isn't necessary.
     
  6. Oct 27, 2013 #5
    Got it, thanks! I was thinking it must be more complex than it really is.
     
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