# Single trigonometric functions...

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• Oct 31st 2006, 03:49 PM
TomCat
Single trigonometric functions...
If I were to turn f(x) = 3 cos(2x) + 4 sin(2x) into a single function... Where would I want to start? Would I just use the double angle formulas and go from there?

Thanks,

-Tom
• Oct 31st 2006, 05:43 PM
topsquark
Quote:

Originally Posted by TomCat
If I were to turn f(x) = 3 cos(2x) + 4 sin(2x) into a single function... Where would I want to start? Would I just use the double angle formulas and go from there?

Thanks,

-Tom

You mean a single trig function to represent that? I don't think it can be done. However, to answer your comment about the double angle formulas this is what you would get:

f(x) = 3cos(2x) + 4sin(2x) = 3(cos^2(x) - sin^2(x)) + 4*2sin(x)cos(x)

$\displaystyle f(x) = 3cos^2(x) + 8sin(x)cos(x) - 3sin^2(x)$

As it happens you can factor this:
f(x) = [3cos(x) - sin(x)][cos(x) - 3sin(x)]

But that doesn't give you the form you want.

If you mean can you get it just in terms of sine or cosine, THAT we can do:

$\displaystyle f(x) = 3cos(2x) + 4sin(2x) = 3(1 - 2sin^2(x)) + 8sin(x)cos(x)$

$\displaystyle f(x) = 3 - 6sin^2(x) + 8 sin(x) \sqrt{1 - sin^2(x)}$

-Dan
• Oct 31st 2006, 06:01 PM
ThePerfectHacker
But,
$\displaystyle \cos x\not = \sqrt{1-\sin^2 x}$

Maybe in physics okay, but not heir.
• Oct 31st 2006, 06:22 PM
classicstrings
I thought cos^2(x) + sin^2(x) = 1

Rearranging gives cos(x) = root(1 - sin^2(x))?
• Oct 31st 2006, 06:22 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
But,
$\displaystyle \cos x\not = \sqrt{1-\sin^2 x}$

Maybe in physics okay, but not heir.

(Sigh) Okay.

$\displaystyle f(x) = 3 - 6sin^2(x) \pm 8sin(x) \sqrt{1 - sin^2(x)}$ with the sign of the $\displaystyle \pm$ chosen depending on which quadrant x is in.

-Dan
• Oct 31st 2006, 06:33 PM
ThePerfectHacker
Quote:

Originally Posted by classicstrings
I thought cos^2(x) + sin^2(x) = 1

Rearranging gives cos(x) = root(1 - sin^2(x))?

These are not equivalent statements.
---
Basically, look at the graphs show below.

~~~
Topsquark, you should define a function,
$\displaystyle \mbox{TPH}(x)$ that is $\displaystyle 1$ if $\displaystyle x\in [-\pi/2 +2\pi k,\pi/2+2\pi k]$ otherwise $\displaystyle -1$
Then you can say,
$\displaystyle \cos x=\mbox{TPH}(x)\sqrt{1-\sin^2 x}$

I actually think my function can be useful. Look at the Heaviside function.

Can you express my function as a fourier series or will it lead to Gibbs phenomena?
• Oct 31st 2006, 07:28 PM
classicstrings
How do I type sin^2 on my calculator? When I type sin it automatically gives me sin( .
• Oct 31st 2006, 11:47 PM
earboth
Quote:

Originally Posted by classicstrings
How do I type sin^2 on my calculator? When I type sin it automatically gives me sin( .

Hi,

you have to use brackets
sin^2(x) has to be typed: (sin(x))^2

EB
• Nov 1st 2006, 03:00 AM
TomCat
Okay... I need to get just one type (sin or cos).
• Nov 1st 2006, 04:30 AM
CaptainBlack
Quote:

Originally Posted by TomCat
If I were to turn f(x) = 3 cos(2x) + 4 sin(2x) into a single function... Where would I want to start? Would I just use the double angle formulas and go from there?

Thanks,

-Tom

The idea with this type of question is to reduce

$\displaystyle f(x)=3 \cos(2x) + 4 \sin(2x)$

to some thing like:

$\displaystyle f(x)=r [\sin(\theta) \cos(2x) + \cos(\theta) \sin(2x)]$

Then to use the addition formula for $\displaystyle \sin$ to reduce this to:

$\displaystyle f(x)=r \sin(\theta+2x)$

For this to work we need:

$\displaystyle r \sin(\theta)=3$ and $\displaystyle r \cos(\theta)=4$,

so $\displaystyle \theta = \arctan(3/4)$ and $\displaystyle r=\sqrt{3^2+4^2}=5$.

RonL
• Nov 1st 2006, 12:47 PM
TomCat
I think I get what you're saying. I was looking over this in class with a friend, and she was just confusing me more. I don't think she was doing it right either.

I would just do something similar for f(x)= Acos(bx) + Bsin(bx) correct?

Thank you!

-Tom
• Nov 1st 2006, 01:48 PM
CaptainBlack
Quote:

Originally Posted by TomCat
I think I get what you're saying. I was looking over this in class with a friend, and she was just confusing me more. I don't think she was doing it right either.

I would just do something similar for f(x)= Acos(bx) + Bsin(bx) correct?

Thank you!

-Tom

Yes, exactly.

RonL
• Nov 2nd 2006, 10:02 AM
TomCat
Hmm... That was over the heads of everyone I showed that to. Are there any other ways to do it that are more Pre-Calcish?
• Nov 2nd 2006, 11:01 AM
CaptainBlack
Quote:

Originally Posted by TomCat
Hmm... That was over the heads of everyone I showed that to. Are there any other ways to do it that are more Pre-Calcish?

It was pure basic trigonometry, which should be pre-calculus (not that
I would know we don't have the stereotyped math sequence that you
seem to have in the US).

The only things you need to know to do it is the sin addition formula and
the trig form of Pythagoras's theorem, which are:

$\displaystyle \sin(A+B)=\sin(A)\cos(B) + \cos(A)\sin(B)$

and:

$\displaystyle \sin^2(\theta)+\cos^2(\theta)=1$

and the rest is just basic algebra.

RonL
• Nov 2nd 2006, 11:11 AM
topsquark
Quote:

Originally Posted by CaptainBlack
It was pure basic trigonometry, which should be pre-calculus (not that
I would know we don't have the stereotyped math sequence that you
seem to have in the US).

The only things you need to know to do it is the sin addition formula and
the trig form of Pythagoras's theorem, which are:

$\displaystyle \sin(A+B)=\sin(A)\cos(B) + \cos(A)\sin(B)$

and:

$\displaystyle \sin^2(\theta)+\cos^2(\theta)=1$

and the rest is just basic algebra.

RonL

You might possibly call it "Advanced Trigonometry" but it's definately Pre-Calculus level.

-Dan
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