New answers tagged trigonometry
2
Cos[a + b] Sinc[a + b] == Sinc[2 (a + b)] /. Sinc[x_] -> Sin[x]/x // Simplify
Cos[x/2] Sinc[x/2] + Cos[y/2] Sinc[y/2] == Sinc[x] + Sinc[y]
/.Sinc[x_] -> Sin[x]/x // Simplify
FullSimplify[ Cos[x/2] Sinc[x/2] + Cos[y/2] Sinc[y/2] == Sinc[x] + Sinc[y],
TransformationFunctions -> {Automatic, Reduce[#, {x, y}, Reals] &}]
2
Sometimes, a preliminary application of FunctionExpand[] works wonders:
Cos[a + b] Sinc[a + b] == Sinc[2 (a + b)] // FunctionExpand // FullSimplify
True
2
You can try proceeding this way:
rule = Cos[x_] Sinc[x_] -> Sinc[2 x]
Cos[x/2] Sinc[x/2] + Cos[y/2] Sinc[y/2] == Sinc[x] + Sinc[y] /. rule
Cos[a + b] Sinc[a + b] == Sinc[2 (a + b)] /. rule
Which both give
True
as expected. Also note that your assumptions make no difference as
Cos[x/2] Sinc[x/2] == Sinc[x] // FullSimplify
gives True without any ...
2
this is the kind of "hack" that I often throw together to illustrate math concepts for my students. The code is terrible but the result looks okay. Perhaps there is something here you can use...
The code...
Manipulate[
If[angle < 0,
endpt = {Cos[Abs[angle] Degree], - Sin[Abs[angle] Degree]} +
0.1 Abs[angle] Degree {Cos[
Abs[angle] ...
1
To give people ideas:
angle[a_] := Module[{p}, Show[
p = PolarPlot[1000 + 10 x, {x, 0, a}, PlotStyle -> Thick],
Graphics[{Thick, Extract[p, Position[p, _Hue]][[1]],
Line[{{0, 0}, 1.9 Extract[p, Position[p, _Line]][[1, 1, -1]]}]}]]];
angle[(2 360 + 45) Degree]
Remember to watch for negative angles.
6
Here's something I've used before:
rotCircle[angle_, ctr_, base1_, base2_, directives___] :=
With[{step = 0.05 Sign[angle], spiral = 0.01},
{directives,
Arrow[Table[ctr + (1.1 + spiral s) (Cos[s] base1 + Sin[s] base2),
{s, Append[Range[0, angle, step], angle]}]]}
]
Graphics[rotCircle[10., {0, 0}, {1, 0}, {0, 1}, Red, Thick], PlotRange -> ...
3
Does this give you a start?
DynamicModule[{p = {1.5, 0}},
Deploy@LocatorPane[Dynamic[p],
Dynamic@Graphics[
{Brown, Thick, Line[{{0, 0}, p/Norm[p]*1.5}],
Circle[{0, 0}, 1, {Mod[ArcTan @@ p, 2 Pi], 2 Pi}]},
PlotRange -> {{-2, 2}, {-2, 2}}, Axes -> True,
AxesOrigin -> {0, 0}], Appearance -> None
]
]
3
Maybe you could expand the Vin to Fourier series to "normalize" it.
For example there are three of them kind:
VinSet = {10 Cos[1000 t - π/2], 9 Cos[400 t + π/4], Cos[t + 3.45]};
coeffSet = FourierCoefficient[# /. Times[ω_?NumericQ t] :> t, t, 1] & /@ VinSet
$\left\{-5 i,\frac{9 \sqrt[4]{-1}}{2}, -0.476409-0.151771 i\right \}$
{2 Abs[#], ...
1
As for your second question, here's one way that is very similar to what we'd do by hand. Use the exponential form and then identify the phase and magnitude.
Clear[A, p, t]
Vin = 10 Exp[-Pi/2]*Exp[1000 t I];
{A, p} = Replace[Vin, A_ Exp[p_] -> {A Exp[Re[p]], Im[p]}]
A now holds the amplitude and p holds the phase. If you have the trigonometric form you ...
2
If you specify the angle as a real number (rather than an exact integer), it does not do the transformation to Sin. For instance
Vin = 10 Cos[1000 t - Pi/2.0]
and
Vin = 10 Cos[1000 t - 90.0 Degree]
both do what you ask.
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