Exploring Higher-Degree Polynomials: Questions and Concepts

[learningphysics]

How come when I graph: $$f(x)=\sqrt{3}\cos{x}\sin{x}$$ on my graphing calculator it looks like a earthquake scale on my HP 50g and then a regular cosine/sine function on my Ti84+?

On my hp50g, the maximums fluctuate ranging from .52 to 6.8?? What is happening here?

Edit: Okay, I just changed the window screen and they show a smooth graph now! I changed from like -64/64 horizontal to -2pi, 2pi...

Edit 2: Wait, both of them have different maximums of .52 to 6.8... okay this is weird.

How am I suppose to solve the problem when it asks for the amplitude and the amplitude fluctuates? Did I insert the function into the graphing calc wrong? Am I suppose to put parentheses somewhere?

Be careful about entering the functions especially for sin cos... use parenthese when needed. I assume the function is $$f(x)=\sqrt{3}(\cos{x})(\sin{x})$$ and not $$f(x)=\sqrt{3}\cos{(x\sin{x})}$$ right?

 

[lLovePhysics]

Be careful about entering the functions especially for sin cos... use parenthese when needed. I assume the function is $$f(x)=\sqrt{3}(\cos{x})(\sin{x})$$ and not $$f(x)=\sqrt{3}\cos{(x\sin{x})}$$ right?

Oh sorry, the function is suppose to be: $$f(x)=\sqrt{3}cosx+sinx$$

EDIT: Oh.. My.. God.. I was looking at the x- coordinate instead of the y! Yeah the y is 2 for all of the maximums. However, how come when the horiz window is large the maximums fluctuate??

Also, how do you calculate the amplitude of the function by hand? I'm confused on that.

 

[learningphysics]

Oh sorry, the function is suppose to be: $$f(x)=\sqrt{3}cosx+sinx$$

EDIT: Oh.. My.. God.. I was looking at the x- coordinate instead of the y! Yeah the y is 2 for all of the maximums. However, how come when the horiz window is large the maximums fluctuate??

Also, how do you calculate the amplitude of the function by hand? I'm confused on that.

Any function of the form Acosx + Bsinx has amplitude $$\sqrt{A^2+B^2}$$

 

[lLovePhysics]

Any function of the form Acosx + Bsinx has amplitude $$\sqrt{A^2+B^2}$$

Is there a rule/proof for this? Any keywords I can search??

 

[learningphysics]

Is there a rule/proof for this? Any keywords I can search??

http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/trig/phase.html

Check out the part that says superimposing sines and cosines..

The idea is to write Acosx+ Bsinx in the form Csin(x+y)

You know that:
Csin(x + y)
=C(sinx)(cosy) + C(cosx)(siny) (sum rule for sines)
=C(siny)(cosx) + C(cosy)(sinx) (just rearranging terms here)

So you set this equal to your original function and equate coefficients...
Csiny = A
Ccosy = B

And from here we get
C^2($$sin^2y + cos^2y$$) = A^2 + B^2
C^2 = A^2 + B^2

 

[lLovePhysics]

Whew, back again! Thanks for the explanation.

How do you find the domain of $$sin^{2}\theta$$

 

[Gib Z]

I'm sure someone will come up with some nice mumbo jumbo about the square function being a continuous bijection of the sine function unto the reals, if that is what is correct to say (someone tell me) but basically, there aren't any values that $$\sin^2 \theta$$ isn't defined for, so its good.

 

[lLovePhysics]

Why can you use this method for this problem?:

If $$\log_{10}m=\frac{1}{2}$$ then $$\log_{10}10m^{2}$$=

I just solved for m, using the first equation and plugged it into the 2nd to solve for the answer but it is wrong... Can you not do that but instead you need to use the log properties? Why is this?

 

[Dick]

Why can you use this method for this problem?:

If $$\log_{10}m=\frac{1}{2}$$ then $$\log_{10}10m^{2}$$=

I just solved for m, using the first equation and plugged it into the 2nd to solve for the answer but it is wrong... Can you not do that but instead you need to use the log properties? Why is this?

That's not even right, and exactly what problem are you talking about?

 

[lLovePhysics]

Nvm, my mistake :D

Is $$csc(2\theta)=\frac{1}{sin(2\theta)}$$?

 

[Dick]

Nvm, my mistake :D

Is $$csc(2\theta)=\frac{1}{sin(2\theta)}$$?

Yes?

 

[lLovePhysics]

Also, how come exponential equations are inverses of logarithms?

For example,

#1 $$b^{x}=y$$

if you switch the x and y terms (definition of inverse)

then: $$b^{y}=x$$

which is equal to: $$log_{b}x=y$$

I just don't get why the exponential eq. are inverses of logs. You can express eq. #1 as $$log_{b}y=x$$ and that is not the inverse of eq. #1 right? So... do they only become inverses after you switch the x and y and turn it into a log?

 

[HallsofIvy]

Also, how come exponential equations are inverses of logarithms?

For example,

#1 $$b^{x}=y$$

if you switch the x and y terms (definition of inverse)

then: $$b^{y}=x$$

which is equal to: $$log_{b}x=y$$

I just don't get why the exponential eq. are inverses of logs. You can express eq. #1 as $$log_{b}y=x$$ and that is not the inverse of eq. #1 right? So... do they only become inverses after you switch the x and y and turn it into a log?

Obviously properties like this depend upon what the functions are- that is, upon how they are defined.
What are your definitions of b^x and log_b(x)?

I know a way of defining b^x rigorously and then log_b(x) is definid as its inverse.
I also know a way of definging log_b(x) directly and then b^x is defined as its inverse.

 

[lLovePhysics]

I kind of see what you mean.

Here's an excerpt from my book about logs and exponential functions:

"The basic properties of exponents and logarithms and the fact that the exponential function and the logarithmic function are inverses lead to many interesting problems.

 

[lLovePhysics]

So inverse functions are reflected off y=x line while odd functions are reflected off the origin right? How do visualize something reflected off the origin? How is it different to a reflection off y=x?

 

[lLovePhysics]

Also, I dont' really get inverse trig functions, when you calculate arcsin(2/3) on your calculator, it only gives you the value that is within the "conventional domain" of -pi/2<=x<=pi/2 right?

However, when you solve for the angles of a trig solution like in the problem: $$3\sin^{2}\theta+10\sin \theta-8=0$$, you need to look for the reference angles? That is, the angles outside the conventional domain? (2nd quad, in this case)

Can someone come up with a general rule for this? Why do reference angles qualify as a solution when it is outside of the arcsin domain?