Question

How do I convert: y = 6 sqrt(2) sin(x)-6 sqrt(2) cos(x)

into the form y = A sin(Bx-C) that has the same graph as the above?

Wolfram|Alpha tells me I'm supposed to get: y = -12sin(pi/4 - x)

But I want to know how I get there.

Answer 1

im waiting for some one to respond to see how they figure this out!

Answer 2

Me too..

Answer 3

Original Equation for better readability: \[y = 6 \sqrt(2) \sin(x) - 6 \sqrt(2) \cos(x)\]

And the equation in the \[y=A \sin(Bx-C)\] form:
\[y = -12\sin(\pi/4 - x)\]


....How do I get from original to final equation posted?

Answer 4

You want to put a sum of sine and cosine into one sine. You can consider the following:\[r \sin(x + \alpha) = r \cos \alpha \sin x + r \sin a \cos x\]

Answer 5

For the two expressions to be identical, you have to equate the coefficients on either side and solve for r and alpha.

Answer 6

Does this make sense?

Answer 7

So\[r \cos \alpha = 6 \sqrt{2}, r \sin \alpha = -6 \sqrt{2}\]

Answer 8

Divide sine by cosine to give\[\frac{r \sin \alpha}{r \sin \alpha}=\tan \alpha = -1 \rightarrow \alpha = \tan^{-1}(-1) = -\frac{\pi}{4}\]

Answer 9

Substitute this into either one of the sine or cosine expressions to get r:\[r \cos \alpha = r \cos (-\pi/4)=r \frac{1}{\sqrt{2}}=6 \sqrt{2} \rightarrow r = 12\]

Answer 10

what is happenin here

Answer 11

\[6\sqrt{2}\sin x -6\sqrt{2} \cos x = 12\sin(x - \frac{\pi}{4})\]

Answer 12

oh sweet, I think I can pick it up from here. Thanks so much!

Answer 13

No worries.

Answer 14

dam your goo at it

Answer 15

Your Wolframalpha answer is the same - \[12\sin(x - \pi/4) = 12 \sin (-(\pi/4 -x))=-12\sin (\pi/4-x)\]

Answer 16

thanks

Answer 17

oh, btw, on the step where you're dividing sin by cos to get tan, why did you divide 2 sins? Is that just a typo or am I missing something?

Answer 18

yeah, typo...

Answer 19

no prob, thanks a ton man, it makes sense now.

Answer 20

you're welcome