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Say I have a cos function, $y = 1.2cos(0.503x) + 5$ And say I want to replicate it using sine, so expressing the above function in sine so that it gives "the same wave".

How would I do this? I know I have to find a phase shift, but I don't get the steps on how to do it. Also, please no use of Wolframalpha or the like, because I'll have to do these types of problems in class without them.

Thank you in advance.

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  • $\begingroup$ do you just want $1.2\cos(0.503x + \pi/2) + 5 = 1.2\sin(0.503(x + \pi/(2*.503)) + 5?$ $\endgroup$
    – abel
    Jun 7, 2015 at 2:34
  • $\begingroup$ Not sure what you mean... $\endgroup$
    – user164403
    Jun 7, 2015 at 2:36
  • $\begingroup$ Yes, that is what I want. $\endgroup$
    – user164403
    Jun 7, 2015 at 2:40

1 Answer 1

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Hint:

$\cos(x)$ is $\sin(x)$ shifted by $\pi/2$

$$\therefore \sin(x) = \cos(x - \pi/2)$$

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  • $\begingroup$ I see. So heres what I got from that: Since period is 2p/|k| = 2pi / 0.5 which is approx 4pi, and there is a (pi/2) shift between cos and sin, (pi/2) is a fourth of the period 4pi that I have. Therefore, I'd be shifting approx. 12.5/4, and my equation would be 1.2sin(0.503(x+12.5/4)) + 5? $\endgroup$
    – user164403
    Jun 7, 2015 at 2:31
  • $\begingroup$ @user164403: Try graphing it on WolframAlpha to verify. $\endgroup$
    – user174622
    Jun 7, 2015 at 2:34
  • $\begingroup$ They line up and do in fact work. Thank you for your help. $\endgroup$
    – user164403
    Jun 7, 2015 at 2:36
  • $\begingroup$ @user164403 Then you're done. $\endgroup$
    – user174622
    Jun 7, 2015 at 2:41

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