## anonymous 5 years ago 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.

1. anonymous

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

2. anonymous

Me too..

3. anonymous

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?

4. anonymous

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

5. anonymous

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

6. anonymous

Does this make sense?

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

what is happenin here

11. anonymous

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

12. anonymous

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

13. anonymous

No worries.

14. anonymous

15. anonymous

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

16. anonymous

thanks

17. anonymous

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?

18. anonymous

yeah, typo...

19. anonymous

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

20. anonymous

you're welcome