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Persimmon
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Homework Statement
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)?
Homework 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
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?