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?



what about $$\begin{align} y & = 7 \cos(0.96(x-3)) + 11 \\ &= 7 \sin(\pi/2 + 0.96(x-3)) + 11\\ & = 7\sin(0.96(x - 3 + \pi/(2 \times 0.96)) + 11\\ &= 7 \sin(0.96(x- 1.3637) + 11? \end{align}$$

  • $\begingroup$ 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 '15 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 '15 at 17:55
  • $\begingroup$ What would you do if the question asked for the amplitude to be negative for sin? $\endgroup$ – user164403 Jun 7 '15 at 17:56
  • $\begingroup$ @user164403, i will add a $\pi$ to the argument of $\sin$ and negate the amplitude. $\endgroup$ – abel Jun 7 '15 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 '15 at 18:03

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