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0

Coincidently, I recently made the following routine for the exact task: Clear[TrigShrink]; TrigShrink[exp_, trgt_, lag_: \[Phi]] := Module[{sign, xExp, xRes, xTrig, cCos, cSin, tan, xlag, clag}, xExp = exp // TrigExpand // Collect[#, {Sin[trgt], Cos[trgt]}] &; xRes = xExp /. {Sin[trgt] -> 0, Cos[trgt] -> 0}; xTrig = xExp - xRes; cCos = ...


0

Not sure this is what you need : A = 1/6; B = -Sqrt[3]/24; ω = 8 Sqrt[3]; r = Sqrt[A^2 + B^2] ClearAll[t]; sol = Solve[ A Cos[ω t] + B Sin[ω t] == r Cos[ω t - ϕ], ϕ]; t = 0; ϕ /. sol[[1]] (* ConditionalExpression[-ArcCos[4/Sqrt[19]] - 2 π C[1], C[1] ∈ Integers] *)


4

Just as @Artes, I started with the expression theta^2 - 1/4. FullSimplify cannot reduce this expression to 0, but with some help it can: TrigToExp[theta^2 - 1/4 ] // FullSimplify // Together // Factor This gives a complicated expression with to linear factors in the numerator. Simplify these factors: FullSimplify /@ % (* 0 *)


6

This is an interesting question demonstrating how related Mathematica trigonometric functions work. Besides recommended purely mathematical approach there are many symbolic tools in the system which one may exploit to get a symbolic result. At first one can think about acting with TrigToExp on the expression and then trying another ways, however then you ...


2

It's about plotting things and extracting the information you need. Here's one way: give the answer a name and then evaluate it over a range of the integers C[1]. Extract from this the x-values and then calculate the corresponding y-values. Then plot the curves and the points, and combine with Show: ans = Solve[Cos[x] == 1/2, x]; xValsRule = (ans /. C[1] ...


4

Since it's triangle functions, I'll try Fourier transformation: FourierCoefficient[#, x, n] & /@ (A Sin[x] + B Cos[x] == 0) Reduce[ForAll[{n}, %], {A, B}] A == 0 && B == 0


2

in inverse kinematics, this sort of thing is done all the time. There are general analytical formulas used in solving the inverse kinematics, and one of this is the one asked here. The solution is always to use atan2. So the solution for this problem, as given in the text book, is, given $a\cos(\theta)+b\sin(\theta)=c$ then $$ \theta = {\rm ...


4

One approach (that works for this simple problem) is to make the substitutions $\cos(x)\rightarrow u \in \mathbb{R}$ and $\sin(x)\rightarrow v \in \mathbb{R}$ with the additional constraint that $u^2+v^2=1$. This approach transforms the transcedental equation to a polynomial (actually, linear) one, and MMA has no problem providing the answer: ...



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