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