# Tag Info

0

@user3419717 as the comments suggest this question seems (i) like homework (ii) does not really require Mathematica as once you observe to obvious solution considerations of periodicity produce all the others (iii) if your intention is to use Mathematica to solve this problem then some code attempts (consistent with rules in (i) is expected. I could not ...

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Solve[Sin[2 x] == Cos[36/180] && 0 < x < 2 Pi, x] 180/Pi %[[All, 1, 2]] // Simplify

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Usually such issue is caused by Automatic Range and PlotRange->All would be a fix. It's not the case though. Szabolcs has noticed: That region of the contour plot is white because the expression insolation[phi, 23.5 Cos[t]] evaluates to non-real complex numbers at those t, phi points. I tried this by right-clicking the plot, and using GetCoordinates ...

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After patching your data and fixing/adjusting code (removed unneeded Total, upped samples): theta = data[[All, 1]]; w = data[[All, 2]]; num = Dimensions[theta][[1]]; n = 150; m = Table[ Flatten[Table[{Sin[i1*theta[[i2]]], 0}, {i1, 1, n}]] + Flatten[Table[{0, Cos[i1*theta[[i2]]]}, {i1, 1, n}]], {i2, 1, num}]; x = LeastSquares[m, w]; f[t_] := ...

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You say your data is a list of $(X,Y)$ so it looks like this: ListLinePlot[data, Frame -> True, FrameLabel -> {"X", "Y"}] Far from been a random coil where the definition of persistent length makes sense. Let see, we want to calculate the average $\cos(\theta) = \hat{v_1} \cdot \hat{v_1}$, as a function of the contour distance, so I use Dot[v1, ...

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The issue we encounter here is an apparent incompleteness of the recent updates in the system, we should remember that Solve has been updated in the recent versions of Mathematica and although documentation pages say "last modified in 8", one can distinguish various different issues between ver.8 and ver.9, it's just a state of art. In ver. 8 we get: ...

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Given: expr = Cos[2 B g t] Sin[u]^2 - I Sin[2 B g t] Sin[u]^2 ... one approach is: FullSimplify[expr, ExcludedForms -> {Cos[_], Sin[_]}] (Cos[2 B g t] - I Sin[2 B g t]) Sin[u]^2 Another approach worth exploring is to use a custom ComplexityFunction, as per: FF[ee_] := 1000 Count[ee, _Exp, {0, Infinity}] + LeafCount[ee] FullSimplify[expr, ...

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