I feel annoyed because I can't rearrange something like this... ^^;;e

Anyway, how do I rearrange this:

$\displaystyle y=A sin (Bx+C)+D$

into....

$\displaystyle y=A sin B (x-C)+D$

Can you please list the steps for me?

Thanks.

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- Nov 29th 2009, 01:35 AMmaybealwaysRearranging a general sine function
I feel annoyed because I can't rearrange something like this... ^^;;e

Anyway, how do I rearrange this:

$\displaystyle y=A sin (Bx+C)+D$

into....

$\displaystyle y=A sin B (x-C)+D$

Can you please list the steps for me?

Thanks. - Nov 29th 2009, 02:43 AMe^(i*pi)
This will only work if $\displaystyle B \neq 0$

Consider the function $\displaystyle Bx+k$. We can take out a factor of B (because it's non-zero to give)

$\displaystyle B\left(x+\frac{k}{B}\right)$

Let $\displaystyle C = \frac{k}{B}$ to give $\displaystyle B(x+C)$

Therefore putting this new expression in place of the old one gives $\displaystyle y=Asin[B(x+C)]$ - Nov 29th 2009, 03:06 AMmaybealways
Thank you, that helps a lot.

But for the equation you arrived at, $\displaystyle y=Asin[B(x+C)]$, why is the sign in the bracket an addition sign? Shouldn't it be a subtraction sign?

So, if I wanted to rearrange this: $\displaystyle y=3.123 sin (0.2178x-2.731)+18.43$ into the $\displaystyle

y=A sin B (x-C)+D

$, what would I need to do? - Nov 29th 2009, 04:37 AMHallsofIvy
?? You said before that you wanted to take the "B" outside the sine function.

Apparently, you just want to write 0.2178x- 2.731 as B(x- C). You can do that by "factoring out" 0.2178. 0.2178x- 2.731= 0.2178(x- 2.731/0.2718)= 0.2178(x- 10.04783) so "3.123 sin (0.2178x-2.731)+18.43= 3.123 sin(0.2178(x- 10.0483)+ 18.43.

That is "A sin B(x- C)+ D" with A= 3.123, B= 0.2178, C= 10.0483, and D= 18.43.