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Thread: Alternative forms of cosine

  1. #1
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    Alternative forms of cosine

    cos(pi/2) is equal to zero, so If we simplify cos( (pi/2) - (2*pi*x) ) will it be cos(pi/2) - cos(2*pi*x) OR cos(pi/2) + cos(2*pi*x). Also why did they change cos to sine? It's all illustrated in the image.
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    Re: Alternative forms of cosine

    Quote Originally Posted by asilvester635 View Post
    cos(pi/2) is equal to zero, so If we simplify cos( (pi/2) - (2*pi*x) ) will it be cos(pi/2) - cos(2*pi*x) OR cos(pi/2) + cos(2*pi*x). Also why did they change cos to sine? It's all illustrated in the image.
    First, $\cos(x-y)=\sin(x)\sin(y)+\cos(x)\cos(y)$
    $ \begin{align*}\cos((\pi/2)- (2\pi x))&=\sin(\pi/2)\sin(2\pi x) \\&=(1)\sin(2\pi x) \end{align*}$
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    Re: Alternative forms of cosine

    Quote Originally Posted by asilvester635 View Post
    cos(pi/2) is equal to zero, so If we simplify cos( (pi/2) - (2*pi*x) ) will it be cos(pi/2) - cos(2*pi*x) OR cos(pi/2) + cos(2*pi*x). Also why did they change cos to sine? It's all illustrated in the image.
    First, rid your mind of the idea that f(a+ b) is very often f(a)+ f(b)! That's part of the condition that a function be linear (the other is that f(ax)= af(x) for constant a) and linear functions, while very simple and very nice, are not very common. Certainly the sine and cosine functions are NOT linear (their graphs are not straight lines). That cos(a+ b) is NOT equal to cos(a)+ cos(b) should be clear by taking a= b= 0: cos(0+ 0)= cos(0)= 1 while cos(0)+ cos(0)= 1+ 1= 2.

    One of the things you should have learned in Trigonometry is the "sum laws"- sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b) and cos(a+ b)= cos(a)cos(b)- sin(a)sin(b).

    Notice that, if a= b= 0, those rules give sin(0+ 0)= sin(0)= sin(0)cos(0)+ cos(0)sin(0)= 0(1)+ 1(0)= 0 and cos(0+ 0)= cos(0)= cos(0)cos(0)- sin(0)sin(0)= 1(1)- 0(0)= 1.
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