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.
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.
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 $\displaystyle y=Asin[B(x+C)]$
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?
?? 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.