Trig Question

[Ikerous]

I'm working on some physics hw and i really need to know how to solve this equation

c = asin(x) + bcos(x)

What does x equal?

I don't remember learning this trig formula and none of the google resulst were in english ><

 

[Angry Lawyer]

X is (x+x)/2

-Angry Lawyer

 

[Ikerous]

I ended up.. doing something.. and got x = arcsin[(c-a^2-b^2)/ab] / 2

Is that right? Or reducable?

 

[Dan]

It's easy:

c = bsin(x) + asin(x)
c = (b+a)sin(x)
c/(b+a) = sin(x)
arcsin (c/(b+a)) = x


x = arcsin (c/(b+a))
dunno if that is equivalent to what you wrote, but it's a lot simpler

 

[Ikerous]

x = arcsin (c/(b+a))

I have no idea how you derived that, but i love you <3

Care to explain?

 

[Dan]

I edited it to show the process

 

[Ikerous]

Sob -_- i wrote the initial question wrong

c = asin(x) + bcos(x)

Thats definitely what i meant..

 

[Shamrock]

Sob -_- i wrote the initial question wrong

c = asin(x) + bcos(x)

Thats definitely what i meant..

You should edit your first post. :thumbs:

 

[Ikerous]

You should edit your first post. :thumbs:

Good point :|

 

[Shamrock]

Good point :|

Can't you go by the same process Dan did?

c = asin(x) + bcos(x)
c = (asin + bcos)(x)
c/(asin + bcos) = (x)


Of course I have yet to learn this. I'm not sure if the 'sin' and 'cos' are variables like such.

 

[tripleplay369]

c = asin(x) + bcos(x)
c = (asin + bcos)(x)

You definately can't do that..

 

[Shamrock]

You definately can't do that..

Why not? Just wondering.

 

[the_lone_wolf]

because asin(x) + bcos(x) doesn't equal (asin _bcos)(x)

it's like saying x/2 = x/4, it's mathmatically incorrect

 

[Dan]

Ok, it took me a while to remember this. I had to look through my old calc notes. It comes up a lot in 2nd order differential equations with stuff like vibrational damping when you want to rewrite the solution to an initial value problem in a more compact way.

You use the identity that sin(x+y) = sin(x)cos+sincos(x)

For visualization, imagine a right angle triangle with sides a and b perpendicular such that the hypotenuse, R = sqrt(a^2+b^2) and the angle t = arctan(b/a)

now b = Rsin(t) and a = Rcos(t)
sub this into the original equation and you get
c = asin(x) + bcos(x)
c = Rcos(t)sin(x) + Rsin(t)cos(x)
and using the trig identity from above:
c = Rsin(x + t)

Once you have this you can solve for x
arcsin(c/R)-t = x
In original terms this equals:
arcsin(c/(sqrt(a^2+b^2))) - arctan(b/a) = x

I never would have figured that one out on my own if the solution hadn't already been taught to me though.

 

[Ikerous]

Wow, nice dan

After my last post i left and got high with a buddy and tried the problem again and used sin(x) = (c - bcos(x))/a and sin^2 = 1 - cos^2 and set them equal and solved with the quadric formula and got an answer. Ill check to see if it's equal to yours. thanks

The original problem was a normal 'two functions, two unknowns' problem, trying to find X and Y in

A + Bcos(120) + Ccos(x) + Dcos(Y) = 0

Bsin(120) + Csin(x) + Dsin(Y) = 0

I used that method to get an answer (Those a, b, and c's are different though. They're just really long constants i assigned variables to so it was easier to write)
The answer is this huge formula though, i cant imagine simplifying it

But wow, if i just use that formula at the beginning its all easier, nice ><
Wow, i might not even need that ><

i ended up with

y = arcsin(c/(sqrt(a^2+b^2))) - arctan(b/a)

a=Bsin(120)
b=A + Bcos(120)
c=(C^2 - (D^2 + a^2 + b^2))/2D

Thanks for the formula ^_^

 

[JellyWorld]

Why not? Just wondering.

sin and cos are functions, not variables.

 

[Beerdude26]

Holy ****ing shit, rocket science

 

[Lou]

I'm working on some physics hw and i really need to know how to solve this equation

c = asin(x) + bcos(x)

What does x equal?

I don't remember learning this trig formula and none of the google resulst were in english ><

square both sides, then use trig identities

Ikerous: You can't solve these problems on your own and you want to teach your friend (aka: yourself) Calculus 3?

 

[Shamrock]

sin and cos are functions, not variables.

Thank you for the specific response.

 

[Dan]

square both sides, then use trig identities

Ikerous: You can't solve these problems on your own and you want to teach your friend (aka: yourself) Calculus 3?

Squaring both sides would not be helpful. You'd end up with:

c^2 = a^2 * sin^2(x) + 2ab * sin(x) * cos(x) + b^2 * cos^2(x)

I don't think that's any easier to work with

 

[Lou]

where

Next time use Wikipedia

Dan you are right. What you did is very clever. I don't think I would have thought of it on my own. Also, my first thought would have been to let a = sin and b = cos without the R, but now I can see why you threw in the R in there (because a and b could be greater/lower than 1/-1.