math q

[TecHNooB]

Forgot how to do this

How do you turn..

A cos (x) + B sin(x)

into

C sin (x + invtan(D)) or C cos (x + invtan(D))

 

[OhSnap688]

Oh man Trig. I hated Trig. I came in here thinking I could help but I can't sorry . Google it! lol

 

[rocadelpunk]

I don't remember either. Might want to do some googling or wolfram alphaing of linear combinations of sin & cos

I suppose you could derive everything from triangle though : P

 

[MrDudeMan]

Are you sure x isn't a part of D? The way you have it written implies that is the case.

 

[darkxshade]

Originally posted by: TecHNooB
Forgot how to do this

How do you turn..

A cos (x) + B sin(x)

into

C sin (x + invtan(D)) or C cos (x + invtan(D))

A cos (x) + B sin(x)

cast magic


C sin (x + invtan(D)) or C cos (x + invtan(D))

Don't forget to say hocus pocus

 

[artikk]

probably some obscure trig identity combo I don't remember from hs.

 

[DrPizza]

Set Acosx + Bsinx = Csin(x+arctanD)
=C(sinx * cos(arctanD) + cosx * sin(arctanD) ) <sin of the sum of two angles identity>
draw pretty right triangle, label one leg D, the other leg 1, such that the tangent of one of the angles is D. Pythagorean theoremize the triangle to get the hypotenuse = sqrt(1+D²)
From there, the cosD = 1/sqrt(1+D²) and sinD = D/sqrt(1+D²)

Fill these into the 2nd line above:
=C[sinx / sqrt(1+D²) + Dcosx / sqrt(1+D²)
shuffle things around a bit, and pair up the cosx term on the left to the cosx term on the right;
the sinx term on the left, and the sinx term on the right.

Thus, A must = CD/sqrt(1+D²) and B = C/sqrt(1+D²)

Hope that helps; I seem to recall seeing something like this about 20 years ago, but haven't seen it since. Hopefully my derivation isn't off; I don't have a pen or pencil around to do it on paper first.
After that point, I assume you simply solve for C and D (2 equations, 2 unknowns) - should be simple.

And, hopefully there are no mistakes above; I didn't even stop to ponder any domain issues or anything of the sort. And, I didn't bother solving for C and D, so once I saw those solutions, it might point to a simpler derivation.

 

[eLiu]

http://en.wikipedia.org/wiki/Trigonometric_identity

scroll down to 'linear combinations'

 

[TecHNooB]

Found the proof using keyword 'linear combinations' and some other junk

http://www.mathcentre.ac.uk/re...rcostheta-alphaetc.pdf

 

[rocadelpunk]

Originally posted by: TecHNooB
Found the proof using keyword 'linear combinations' and some other junk

http://www.mathcentre.ac.uk/re...rcostheta-alphaetc.pdf

nice link!

 

[BlackTigers]

I hate trig identities. =/

Urghhhh.

 

[eLiu]

Originally posted by: TecHNooB
Found the proof using keyword 'linear combinations' and some other junk

http://www.mathcentre.ac.uk/re...rcostheta-alphaetc.pdf

oops, I didn't read the OP fully and thought you only wanted the specifics of what C and D are, not necessarily the proof. my bad, lol.

but looks like you found it anyway

 

[Sea Moose]

Originally posted by: darkxshade

Originally posted by: TecHNooB
Forgot how to do this

How do you turn..

A cos (x) + B sin(x)

into

C sin (x + invtan(D)) or C cos (x + invtan(D))

A cos (x) + B sin(x)

cast magic


C sin (x + invtan(D)) or C cos (x + invtan(D))

Don't forget to say hocus pocus

This