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Replicating a cosine graph with sine, given transformations?

This is an extension question of my previous post.

Say I have an equation like $y = 7 \cos(0.96(x-3)) + 11$.

How would I find the sine equivalent that lines exactly with it? I thought that $\sin$ and $\cos$ differ only by a phase shift of $\displaystyle -\frac{\pi}{2}$, when do I need to use reflections?

Thanks

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2 Answers 2

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$A * \sin( B * x - C ) + D$

A = amplitude

B = Period, usually given as pi/B

C = phase shift (or horizontal offset if you prefer)

D = vertical offset

Since you are only changing the horizontal to correctly duplicate sine in cosine or vice versa, the ONLY thing that should change is the phase shift, which is $\pm \pi/2$ to cause it to align with the opposite function.

addendum A & B are assumed to be 1 if not otherwise given C & D are assumed to be 0 if not otherwise given

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  • $\begingroup$ apologies, the bracketing was wrong, fixed now $\endgroup$
    – Nox Ferocia
    Jun 24, 2022 at 20:26
  • $\begingroup$ That works. It can easily be confirmed algebraically, or one can play with the graphs here. OP could easily convert their formula to this form perform the phase shift, and if they really must have the answer in their original form they can convert it back. Personally I prefer it the way you have shown it. $\endgroup$
    – David K
    Jun 24, 2022 at 22:42
  • $\begingroup$ I think OP may be confusing (X -3) as being (X-C) when the whole thing should be X since it is being multiplied by 0.96. OPs completed correction should look like $7sin(0.96(x-3)+pi/2)$ and NOT $7sin(0.96(x-3+pi/2))$ $\endgroup$
    – Nox Ferocia
    Jun 26, 2022 at 19:03
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what about $$\begin{align} y & = 7 \cos(0.96(x-3)) + 11 \\ &= 7 \sin(\pi/2 + 0.96(x-3)) + 11\\ & = 7\sin(0.96(x - 3 + \pi/(2 \times 0.96)) + 11\\ &= 7 \sin(0.96(x- 1.3637) + 11? \end{align}$$

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  • $\begingroup$ Hmm, I don't get your steps. How did you do this? My logic was to add 6.5/4 since that is a "quarter of the period" here and pi/2 is a quarter of the standard 2pi period. $\endgroup$
    – user164403
    Jun 7, 2015 at 17:49
  • $\begingroup$ @user164403. $\sin(t+\pi/2) = \cos t$ is an identity, meaning is true for all $t.$ that is what i used. $\endgroup$
    – abel
    Jun 7, 2015 at 17:55
  • $\begingroup$ What would you do if the question asked for the amplitude to be negative for sin? $\endgroup$
    – user164403
    Jun 7, 2015 at 17:56
  • $\begingroup$ @user164403, i will add a $\pi$ to the argument of $\sin$ and negate the amplitude. $\endgroup$
    – abel
    Jun 7, 2015 at 17:57
  • $\begingroup$ Hmm, I tried doing that. I got -7sin(0.96(x+0.27)) + 11, but it gives me a much more shifted graph. What did I do wrong? I did -7sin(0.96(x-3) + pi)) + 11, and brought the pi back into the brackets getting phase shift of 0.27. $\endgroup$
    – user164403
    Jun 7, 2015 at 18:03

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