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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

849.0. "sin(nx) = f(sin(x),cos(x)) ???" by VIDEO::OSMAN (type video::user$7:[osman]eric.vt240) Thu Mar 31 1988 19:52

In a recent topic, Mr. Gilbert pointed out the following:

	sin 2x = 2 sin x cos x
	sin 3x = 3 sin x - 4 (sin x)^3
	sin 4x = cos x ( 4 sin x - 8 (sin x)^3 )
	sin 5x = 16 (sin x)^5 - 20 (sin x)^3 + 5 sin x


Can anyone figure out the pattern here and state the closed form for:

	sin (nx) = f(sin(x),cos(x))

Alternately, how about

	sin(nx) = f(sin((n-1)x))       [recursive formula]



/Eric

T.RTitleUserPersonal
Name
DateLines
849.1How about RTFBMEIS::WOLFFI feel the need, the need for speedThu Mar 31 1988 19:587
    I think this is a case for looking into a math formula collection,
    like "Handbuch der Mathematik".
    
    I look it up, and see if I can find something.
    
    	Julian.
    
849.2 :-) 8BALL::HALLYBYou have the right to remain silent.Thu Mar 31 1988 20:417
>    I think this is a case for looking into a math formula collection,
>    like "Handbuch der Mathematik".

    I think Eric was looking for something more direct, like Gilbert
    showed, only without putting the bucket of water on the floor.
    
    Oops, excuse me.  Wrong note.
849.3If you really want ot play with these...AKQJ10::YARBROUGHWhy is computing so labor intensive?Thu Mar 31 1988 20:446
MAPLE handles this kind of expression quite well for fixed n; try

> expand(sin(10*X), X);

for example. It may give some insights into the general form, which MAPLE 
is able to compute quite quickly.
849.4Here you have it...AKQJ10::YARBROUGHWhy is computing so labor intensive?Thu Mar 31 1988 20:528
Aha, found it in Bronshtein & Semendyayev. Spoiler follows the <FF>.


Sin(nx) = n*cos(x)^(n-1)*sin(x) - C(n,3)cos(x)^(n-3)*sin(x)^3 + 
	C(n,5)cos(x)^(n-5)*sin(x)^5 - ...

Cos(nx) = cos(x)^n - C(n,2)cos(x)^(n-2)*sin(x)^2 + 
	C(n,4)cos(x)^(n-4)*sin(x)^4 - ...
849.5Yupe, that's the book I meant.MEIS::WOLFFI feel the need, the need for speedFri Apr 01 1988 00:1914
    Lynn,
    
    	same book I looked in... but after I typed it in now I might
    as well post it.

sin(nx) =


     n-1                /n\     n-3      3       /n\	 n-5	  5
n cos     (x) sin(x) - (   ) cos   (x)sin (x) + (   ) cos   (x)sin (x) -...
                        \3/                      \5/

        Julian.
    
849.6CADM::ROTHIf you plant ice you'll harvest windMon Apr 04 1988 16:004
    Such identities are much more transparent in terms of the complex
    exponential, exp(n*i*t) = cos(n*t) + i*sin(n*t).

    - Jim