Date: 01/07/97 at 20:03:43 From: Stephen A Barder Subject: Trig Identities I'm a junior in high school and got this problem for homework today: Verify the following identity: sin(x+y+z) = sinxcosycosz + cosxsinycosz + cosxcosysinz - sinxsinysinz I know how to evaluate stuff like sin(x+y), but I don't even know where to start with three variables. Your help would be really appreciated. Tim Barder
Date: 01/08/97 at 00:54:54 From: Doctor Pete Subject: Re: Trig Identities Hi, It's really not as difficult as you may think: If you know that Sin[w+x] = Sin[w]Cos[x] + Cos[w]Sin[x] then it seems perfectly natural to ask what this would be if w = y+z. This gives: Sin[(y+z)+x] = Sin[y+z]Cos[x] + Cos[y+z]Sin[x] Now, I leave it to you to further expand this expression, which is fairly straightfoward. In a similar vein, note that: Cos[x] = Cos[x] Cos[2x] = Cos[x]^2 - Sin[x]^2 = Cos[x]^2 - (1-Cos[x]^2) = 2 Cos[x]^2 - 1 Cos[3x] = Cos[x + 2x] = Cos[x]Cos[2x] - Sin[x]Sin[2x] = Cos[x](2 Cos[x]^2 - 1) - Sin[x](2 Sin[x]Cos[x]) = 2 Cos[x]^3 - Cos[x] - 2 Sin[x]^2 Cos[x] = 2 Cos[x]^3 - Cos[x] - 2 (1-Cos[x]^2)Cos[x] = 4 Cos[x]^3 - 3 Cos[x] Cos[4x] = Cos[2x + 2x] = 2 Cos[2x]^2 - 1 = 2 (2 Cos[x]^2 - 1)^2 - 1 = 2 (4 Cos[x]^4 - 4 Cos[x]^2 + 1) - 1 = 8 Cos[x]^4 - 8 Cos[x]^2 + 1 Apparently, we can always express Cos[kx] for any positive integer k as some combination of powers of Cos[x]. Challenge 1 (easy): Find Cos[5x], Cos[6x], Cos[7x] in terms of powers of Cos[x]. Challenge 2 (medium): Find Sin[kx] in terms of powers of Sin[x] for integer values of k from 1 to 7. Challenge 3 (hard): Note that we can view Cos[kx] and Sin[kx] as polynomials in terms of Cos[x] and Sin[x], respectively. From what you found in parts 1, 2, is there a pattern in the coefficients of these polynomials? (Don't forget terms with coefficient 0.) Can you guess the pattern? Challenge 4 (hard): If you guessed a pattern in part 3, prove that this pattern works; that is, find the polynomial C[k,x] in Cos[x] such that C[k,x] = Cos[kx], and the polynomial S[k,x] in Sin[x] such that S[k,x] = Sin[kx]. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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