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I've been asked to write two equation(one sine and one cosine) for the following graph

enter image description here

I'm understand axis of symmetry is $y=-10$, Period is$\frac{\pi}{30}$ and amplitude is $6$, are these values correct ? how will get and equation of sine and cosine from the graph.

Any help is appreciated, also any resource to learn this topic further will be helpful,

Thank you, Arif

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Wave-form in standard form

$$ (y-k)= A \sin \dfrac{ 2 \pi (x-h)}{\lambda} = A \cos{ [\frac{\pi}{2} -\dfrac{ 2 \pi (x-h)}{\lambda}]} $$

where we get $(h,k)$ as average values of sine wave inflection point ( below where you marked $15$) with maximum positive slope using the given crest and trough of the sine-wave for $ (x-,y-)$ coordinates to determine shifts/translations of a rigid sine curve.

$$k=\frac{-4-16}{2} = -10,\, A=6, \, h= \frac{6-24}{2} = -9, \lambda=60 \,$$

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  • $\begingroup$ please can you elaborate on as how we get the values of h and k. $\endgroup$ – Arif Feb 24 '18 at 14:13
  • $\begingroup$ Apologies, an error occured by oversight. Corrected it, explained in the answer. $\endgroup$ – Narasimham Feb 24 '18 at 14:56
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Period is $30$ and amplitude $6$ thus the function is in the form

$$f(x)=6 \cos \left(\frac{2\pi}{30}x+\phi\right)+C$$

where

  • C<10 is the constant for the vertical shifting
  • $\phi\approx -\frac{\pi}6$ is the constant for the horizontal shifting
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I tried using $Bx+C=0$, our sine function starts at $-24$ and $B=\frac{\pi}{30}$ substituting $x=-24, ~B=\frac{\pi}{30}$ in $Bx+C=0$ we get $C=\frac{24\pi}{30}$

our equation become $6\sin(\frac{\pi}{30}x+\frac{24\pi}{30})-10$. Is this correct ?

Thanks, Arif

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