Trigonometric IdentityDate: 02/12/98 at 16:02:07 From: Toby Smith Subject: Trigonometric Identity I have to prove that sin 7x = 7(sin x) - 56(sin x)^3 + 112(sin x)^5 - (sin x)^7 But I have reached the point where I have: sin 7x = 7(sin x) - 56(sin x)^3 + 112(sin x)^5 - 64(sin x)^7 I have graphed both of these on a Sharp EL-9300 Graphics Calculator, and I think they are equal. How can this be? But, more importantly, how can mine be changed to make it identical to the one I am after? Regards, Toby Smith Date: 02/13/98 at 07:47:26 From: Doctor Anthony Subject: Re: Trigonometric Identity You don't mention which method you used to find sin(7x). One method is to evaluate sin(6x + x) = sin(6x)cos(x) + cos(6x)sin(x) and work on down with similar expressions for sin(6x), sin(5x) and so on. This method is laborious and likely to end up with mistakes in the heavy algebra. A better method is using DeMoivre's theorem. cos(7x) + i.sin(7x) = (cos(x) + i.sin(x))^7 Now expand the right hand side using the binomial theorem and take the imaginary part to get sin(7x) Using c to represent cos(x) and s to represent sin(x) (c + i.s)^7 = c^7 + 7.c^6.(i.s) + 21.c^5.(i.s)^2 + 35.c^4.(i.s)^3 + 35.c^3.(i.s)^4 + 21.c^2.(i.s)^5 + 7.c.(i.s)^6 + (i.s)^7 Take out the odd powers on the right as these will be the imaginary part. Remember that i^3 = -i and i^4 = +1 so i^5 = i, i^7 = -i We then get sin(7x) = 7cos^6(x).sin(x) - 35cos^4(x).sin^3(x) + 21cos^2(x).sin^5(x) - sin^7(x) The last stage is to convert the various cos terms to sin terms using the identity cos^2(x) = 1 - sin^2(x) cos^4(x) = 1 - 2sin^2(x) + sin^4(x) cos^6(x) = 1 - 3sin^2(x) + 3sin^4(x) - sin^6(x) So putting in these expressions, we get sin(7x)= 7sin(x)[1-3sin^2(x)+3sin^4(x)-sin^6(x)] - 35sin^3(x)[1-2sin^2(x) + sin^4(x)] + 21sin^5(x)[1-sin^2(x)] - sin^7(x) = 7sin(x) - 56sin^3(x) + 112sin^5(x) - 64sin^7(x) The second version that you give is the correct one. The two versions will not be equal except for small values of sin(x), where for example .01^7 = 10^(-14) -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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