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I am expanding on my comment. You want to find $x,y$ such that: $$X\cot X + Y\cot Y =0, \ X=d\times x,\ Y=d\times y, \quad \text{and}\quad Y=f(X).$$ $d$ can be seen as a scaling parameter, for simplicity I write the equations here with $d=1$. The problem becomes: $$x\cot x + y \cot y=0\quad\text{and}\quad y=f(x)$$ These are two equations, that individually ...

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\$Version (* "12.1.1 for Mac OS X x86 (64-bit) (June 19, 2020)" *) Clear["Global*"] expr1 = (1/2)*α* Integrate[Sin[2*Pi*(fi - fk + δf)*t], {t, 0, T}] + Ν1 (* Ν1 + (α Sin[π T (fi - fk + δf)]^2)/(2 π (fi - fk + δf)) *) If you don't like this form, force a different form using a custom ComplexityFunction expr2 = Simplify[expr1, ...

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Clear["Global*"] eqn = Sin[2 a] Cos[3 a] == Tan[4 a]; The functions are periodic so there are many periodic solutions. Solve[eqn, a] // Short[#, 4] & Restrict the range of a to some region of interest sol = Solve[{eqn, 0 <= a <= 2 Pi}, a] The approximate numeric values are sol // N (* {{a -> 0.}, {a -> 1.5708}, {a -> 3....

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Does this do what you want? expr=(1/(2 (fj - fk) π)) Cos[(fj - fk) π T + (fj - fk) π (T + δt)] Sin[(fj - fk) π T - (fj - fk) π (T + δt)]; expr /. (f : Cos | Sin)[a_] :> f[Simplify[a]] This applies Simplify only to the arguments of Sin and Cos Alternatively, Simplify[expr,Trig->False] (This prevents Mathematica from applying Trig ...

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FindIntegerNullVector is your friend. Notice that this method allows you to obtain an even simpler expression. A. Find relations between sets of different functions v={ArcCsc[18/Sqrt[50+17 Sqrt]],ArcTan[Sqrt],ArcCot[3+Sqrt]}; v={ArcCot[Sqrt],ArcCot[2 Sqrt],ArcCot[3+Sqrt]}; v={ArcCot[2 Sqrt],ArcCot[Sqrt],ArcTan[Sqrt]}; v=...

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