TuringTest
  • TuringTest
Quicky: Write\[\sin x-\cos x\]in the form\[A\sin(x+c)\]I totally forgot how to do this and can't find a reference.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
TuringTest
  • TuringTest
@satellite73 I know you can do this, please lend a hand.
anonymous
  • anonymous
yes we can do it
phi
  • phi
use sin(a+b)= sin(a)cos(b) + sin(b)cos(a)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

TuringTest
  • TuringTest
I just forgot how I used to know once...
phi
  • phi
or in your case use sin(a-b)= sin(a)cos(b) - sin(b)cos(a)
anonymous
  • anonymous
\[a\cos(\theta)+b\sin(\theta)=\sqrt{a^2+b^2}\sin(x+\tan^{-1}\frac{a}{b})\] if i am not mistaken
anonymous
  • anonymous
we can derive the identity by using the addition angle formula
TuringTest
  • TuringTest
so C is 3pi/4 ?
anonymous
  • anonymous
in this case you will have a = -1, b = 1, \[\sqrt{2}\sin(x+\tan^{-1}(-1))=\sqrt{2}\sin(x-\frac{\pi}{4})\]
TuringTest
  • TuringTest
...or - pi/4 okay thank you!
anonymous
  • anonymous
yw now i will see if i can remember how to derive the identity, it is not hard
.Sam.
  • .Sam.
to derive not hard at all
TuringTest
  • TuringTest
I'd like to see it, but I may not stick around for the show getting ready for class
TuringTest
  • TuringTest
what's up with the bike theme?
TuringTest
  • TuringTest
satellite wrote\[a\cos(\theta)+b\sin(\theta)=\sqrt{a^2+b^2}\sin(x+\tan^{-1}\frac{a}{b})\]I know it doesn't matter in this case, but shouldn't it be\[a\sin(\theta)+b\cos(\theta)=\sqrt{a^2+b^2}\sin(x+\tan^{-1}\frac{a}{b})\]? (sin and cos switched, so the coefficient of sin is in the numerator when you take arctan)
anonymous
  • anonymous
i am fairly sure i am right, that you should have \[\tan^{-1}(\frac{a}{b})\] where a is the coefficient of the cosine term it would be clearer if we derived the fromula
TuringTest
  • TuringTest
okay, then I guess I will make the derivation a priority
anonymous
  • anonymous
give me a second and i will start
TuringTest
  • TuringTest
no rush, I may not be able to stick around but I'll read it when I return if nothing else
anonymous
  • anonymous
ok i just need to refresh my memory.
anonymous
  • anonymous
ok it is not hard
anonymous
  • anonymous
put \[a\sin(x)+b\cos(x)=\frac{ra}{r}\sin(x)+\frac{br}{r}\cos(x)=r\left(\frac{a}{r}\sin(x)+\frac{b}{r}\cos(x)\right)\] where \[r=\sqrt{a^2+b^2}\]
anonymous
  • anonymous
then note that we can find \[\alpha\] with \[\sin(\alpha)=\frac{b}{\sqrt{a^2+b^2}}, \cos(\alpha)=\frac{a}{\sqrt{a^2+b^2}}\] and now you will see why it really is the "a" on top and the "b" on the bottom
phi
  • phi
Are you going to finish the derivation, or leave it to the reader?
TuringTest
  • TuringTest
gotchya... just to be clear, you meant that is why in this case the "b" is on top (you put b as the coef of cos this time)
anonymous
  • anonymous
\[a\sin(x)+b\cos(x)=r\left(\frac{a}{r}\sin(x)+\frac{b}{r}\cos(x)\right)\] \[=r\left(\cos(\alpha)\sin(x)+\sin(\alpha)\cos(x)\right)\] \[=r\sin(x+\alpha)\]
TuringTest
  • TuringTest
so\[\alpha=\tan^{-1}\frac ba\]
anonymous
  • anonymous
ok so lets see what we have, because i could have been wrong if it is \[a\sin(x)+b\cos(x)\] then you want \[\alpha\] with \[\cos(\alpha)=\frac{a}{r}\] \[\sin(\alpha)=\frac{b}{r}\] or \[\tan(\alpha)=\frac{b}{a}\] so \[\alpha =\tan^{-1}(\frac{b}{a})\] whew!
TuringTest
  • TuringTest
got it, thanks sat! engraving this into my brain right now....

Looking for something else?

Not the answer you are looking for? Search for more explanations.