Question

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.

Answer 1

@satellite73 I know you can do this, please lend a hand.

Answer 2

yes we can do it

Answer 3

use sin(a+b)= sin(a)cos(b) + sin(b)cos(a)

Answer 4

I just forgot how
I used to know once...

Answer 5

or in your case
use sin(a-b)= sin(a)cos(b) - sin(b)cos(a)

Answer 6

\[a\cos(\theta)+b\sin(\theta)=\sqrt{a^2+b^2}\sin(x+\tan^{-1}\frac{a}{b})\] if i am not mistaken

Answer 7

we can derive the identity by using the addition angle formula

Answer 8

so C is 3pi/4 ?

Answer 9

in this case you will have a = -1, b = 1,
\[\sqrt{2}\sin(x+\tan^{-1}(-1))=\sqrt{2}\sin(x-\frac{\pi}{4})\]

Answer 10

...or - pi/4
okay thank you!

Answer 11

yw now i will see if i can remember how to derive the identity, it is not hard

Answer 12

to derive not hard at all

Answer 13

I'd like to see it, but I may not stick around for the show
getting ready for class

Answer 14

what's up with the bike theme?

Answer 15

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)

Answer 16

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

Answer 17

okay, then I guess I will make the derivation a priority

Answer 18

give me a second and i will start

Answer 19

no rush, I may not be able to stick around
but I'll read it when I return if nothing else

Answer 20

ok i just need to refresh my memory.

Answer 21

ok it is not hard

Answer 22

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
\[r=\sqrt{a^2+b^2}\]

Answer 23

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

Answer 24

Are you going to finish the derivation, or leave it to the reader?

Answer 25

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)

Answer 26

\[a\sin(x)+b\cos(x)=r\left(\frac{a}{r}\sin(x)+\frac{b}{r}\cos(x)\right)\]
\[=r\left(\cos(\alpha)\sin(x)+\sin(\alpha)\cos(x)\right)\]
\[=r\sin(x+\alpha)\]

Answer 27

so\[\alpha=\tan^{-1}\frac ba\]

Answer 28

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!

Answer 29

got it, thanks sat!
engraving this into my brain right now....