Recent content by solve

Ok. acos(x)+bsin(x)=Rsin(x+t) Why is there R in the above equation? Yes, you can expand it, but that doesn't explain what kind of reasoning keeps R in f(x)=Rsin(x+t)

Thanks.

Does that happen to be R? If yes, I'd like to know why and how R got there.

Hi,tiny-tim. See, sin(x+t)=sin(x)cos(t)+ cos(x)sin(t)= acos(x)+ bsin(x) Ok, a=cos(t) and b=sin(t), but how or why does R end up there?

Homework Statement acos(x)+bsin(x)=Rsin(x+t) Homework Equations The Attempt at a Solution Is there any way to show how R is "placed" in acos(x)+bsin(x)=Rsin(x+t) algebraically? I mean I could, probably, do acos(x)+bsin(x)=sin(t)cos(x)+ cos(t)sinx(x), but still somehow need R in it. Does R...

Thanks, Ratch.

Thanks, Everyone. Solved this one. Have some other questions related to this equation, but will start a new thread. Thanks.

Also, is it true that cos(pi/2- x)=cos(x- pi/2)= sinx ? Thanks, Ratch.

I don't think I understand you correctly. For instance, y=sin(x+ pi/4) leads y=sinx by a phase of pi/4. y=sin(x- pi/4) lags behind y=sinx with a phase difference of pi/4. Would that be correct? They have to do with the questions about the preference of use between -f(x) and f(-x) i.e...

Hi, Ratch. The logic here: -7cos(5x-3)= -7cos5(x-3/5), so the phase is -3/5. The logic: -cot(3x-4)= cot(-3x+4)= cot3(-x+4), so the phase looks to be 4/3, but the phase of -cot(3x-4) looks like -4/3, except I am not sure how negative cot influences the phase. Not equations, but just...

cosθ=0.825 and sinθ=-0.717 ? And sin and cos are both negative in third quadrant.

I get θ just by Rsinθ/Rcosθ= tanθ I don't quite get how I can chose θ so that \displaystyle \sin(\theta)=-\frac{4}{5} and \displaystyle \cos(\theta)=-\frac{3}{5}\,. Thanks.

Hi, people. This question indeed comes from a trig equation: 3cosx+ 4sinx=5. I am offered to solve it by using either f(x)=Rsin(x+theta) or f(x)= Rcos(x+phi). Very little explanation as to the background relationship of an equation of the form acosx+ bsinx=c and Rsin(x+theta). Nor do I have...

Now this perfectly stratifies my question. Thanks, Infinitum.

What up, Infinitum! Just can't get around to learning this LaTex. One of these days, for real. pi/2+ and -2kpi= -pi/2+ and-2kpi