Can we solve for $delta$ in $fracπ2 - fracRrβ = arcsinbiggl(frac(R+r)sin(δ)rbiggr)+fracRrδ$?

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Solve for $delta$:



$$fracpi2 - fracRrβ = arcsinleft(frac(R+r)sin deltarright)+fracRrδ$$




My problem is that I can't comprehend how to get both $delta$s out of a trig operation at the same time. If I apply 'sine' to both sides, I understand the 'arcsine' will be canceled, but does this operation apply 'sine' to $fracRrdelta$ on the right, making it $sinleft(fracRrdeltaright)$ ???



Is this problem a question of a trigonometric derivative?




My second question is perhaps a more important one. I do not have a textbook pertaining to trigonometry, much less trigonometric derivatives. The math book I currently own is about matrices and calculus, but only having to do with 'regular' derivatives; no trigonometry derivatives, obviously.



What is a good textbook for trigonometry (regular trig-algebra)?
What is a good textbook for trigonometry derivatives? Is there a textbook that contains both of these subjects?



Thanks is advance.







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  • You need a numerical method. Remember that $x=cos(x)$ does not show analytical solutions. If you give test values for $r,R,beta$, I could shos you.
    – Claude Leibovici
    yesterday










  • You should ask for textbook recommendations in a separate question. That said, do a search first. I'm pretty sure that people have asked for such recommendations a few times.
    – Blue
    yesterday














up vote
0
down vote

favorite













Solve for $delta$:



$$fracpi2 - fracRrβ = arcsinleft(frac(R+r)sin deltarright)+fracRrδ$$




My problem is that I can't comprehend how to get both $delta$s out of a trig operation at the same time. If I apply 'sine' to both sides, I understand the 'arcsine' will be canceled, but does this operation apply 'sine' to $fracRrdelta$ on the right, making it $sinleft(fracRrdeltaright)$ ???



Is this problem a question of a trigonometric derivative?




My second question is perhaps a more important one. I do not have a textbook pertaining to trigonometry, much less trigonometric derivatives. The math book I currently own is about matrices and calculus, but only having to do with 'regular' derivatives; no trigonometry derivatives, obviously.



What is a good textbook for trigonometry (regular trig-algebra)?
What is a good textbook for trigonometry derivatives? Is there a textbook that contains both of these subjects?



Thanks is advance.







share|cite|improve this question





















  • You need a numerical method. Remember that $x=cos(x)$ does not show analytical solutions. If you give test values for $r,R,beta$, I could shos you.
    – Claude Leibovici
    yesterday










  • You should ask for textbook recommendations in a separate question. That said, do a search first. I'm pretty sure that people have asked for such recommendations a few times.
    – Blue
    yesterday












up vote
0
down vote

favorite









up vote
0
down vote

favorite












Solve for $delta$:



$$fracpi2 - fracRrβ = arcsinleft(frac(R+r)sin deltarright)+fracRrδ$$




My problem is that I can't comprehend how to get both $delta$s out of a trig operation at the same time. If I apply 'sine' to both sides, I understand the 'arcsine' will be canceled, but does this operation apply 'sine' to $fracRrdelta$ on the right, making it $sinleft(fracRrdeltaright)$ ???



Is this problem a question of a trigonometric derivative?




My second question is perhaps a more important one. I do not have a textbook pertaining to trigonometry, much less trigonometric derivatives. The math book I currently own is about matrices and calculus, but only having to do with 'regular' derivatives; no trigonometry derivatives, obviously.



What is a good textbook for trigonometry (regular trig-algebra)?
What is a good textbook for trigonometry derivatives? Is there a textbook that contains both of these subjects?



Thanks is advance.







share|cite|improve this question














Solve for $delta$:



$$fracpi2 - fracRrβ = arcsinleft(frac(R+r)sin deltarright)+fracRrδ$$




My problem is that I can't comprehend how to get both $delta$s out of a trig operation at the same time. If I apply 'sine' to both sides, I understand the 'arcsine' will be canceled, but does this operation apply 'sine' to $fracRrdelta$ on the right, making it $sinleft(fracRrdeltaright)$ ???



Is this problem a question of a trigonometric derivative?




My second question is perhaps a more important one. I do not have a textbook pertaining to trigonometry, much less trigonometric derivatives. The math book I currently own is about matrices and calculus, but only having to do with 'regular' derivatives; no trigonometry derivatives, obviously.



What is a good textbook for trigonometry (regular trig-algebra)?
What is a good textbook for trigonometry derivatives? Is there a textbook that contains both of these subjects?



Thanks is advance.









share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited yesterday









Blue

43.6k868141




43.6k868141









asked yesterday









user101434

44




44











  • You need a numerical method. Remember that $x=cos(x)$ does not show analytical solutions. If you give test values for $r,R,beta$, I could shos you.
    – Claude Leibovici
    yesterday










  • You should ask for textbook recommendations in a separate question. That said, do a search first. I'm pretty sure that people have asked for such recommendations a few times.
    – Blue
    yesterday
















  • You need a numerical method. Remember that $x=cos(x)$ does not show analytical solutions. If you give test values for $r,R,beta$, I could shos you.
    – Claude Leibovici
    yesterday










  • You should ask for textbook recommendations in a separate question. That said, do a search first. I'm pretty sure that people have asked for such recommendations a few times.
    – Blue
    yesterday















You need a numerical method. Remember that $x=cos(x)$ does not show analytical solutions. If you give test values for $r,R,beta$, I could shos you.
– Claude Leibovici
yesterday




You need a numerical method. Remember that $x=cos(x)$ does not show analytical solutions. If you give test values for $r,R,beta$, I could shos you.
– Claude Leibovici
yesterday












You should ask for textbook recommendations in a separate question. That said, do a search first. I'm pretty sure that people have asked for such recommendations a few times.
– Blue
yesterday




You should ask for textbook recommendations in a separate question. That said, do a search first. I'm pretty sure that people have asked for such recommendations a few times.
– Blue
yesterday










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Letting $Q = R/r$ and rearranging a bit gives



$$
fracpi2 - Qbeta - Qdelta = arcsin bigl( (Q + 1) sin delta bigr)
$$



then taking $sin(cdot)$ of both sides and using $sin (pi/2 - x) = cos x$ results in



$$
cos bigl( Q(beta + delta) bigr) = (Q + 1) sin delta.
$$



If $Q = 1$ then the equation can be solved exactly. Otherwise, you can rewrite the equation as



$$
delta = f(delta) = arcsin left( fraccos bigl( Q(beta + delta) bigr)Q + 1 right)
$$



and repeatedly apply $f$ to some initial guess for a solution $delta$. It turns out that this fixed-point iteration converges in this particular case.






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    1 Answer
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    active

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    1 Answer
    1






    active

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    active

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    active

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    up vote
    2
    down vote













    Letting $Q = R/r$ and rearranging a bit gives



    $$
    fracpi2 - Qbeta - Qdelta = arcsin bigl( (Q + 1) sin delta bigr)
    $$



    then taking $sin(cdot)$ of both sides and using $sin (pi/2 - x) = cos x$ results in



    $$
    cos bigl( Q(beta + delta) bigr) = (Q + 1) sin delta.
    $$



    If $Q = 1$ then the equation can be solved exactly. Otherwise, you can rewrite the equation as



    $$
    delta = f(delta) = arcsin left( fraccos bigl( Q(beta + delta) bigr)Q + 1 right)
    $$



    and repeatedly apply $f$ to some initial guess for a solution $delta$. It turns out that this fixed-point iteration converges in this particular case.






    share|cite|improve this answer

























      up vote
      2
      down vote













      Letting $Q = R/r$ and rearranging a bit gives



      $$
      fracpi2 - Qbeta - Qdelta = arcsin bigl( (Q + 1) sin delta bigr)
      $$



      then taking $sin(cdot)$ of both sides and using $sin (pi/2 - x) = cos x$ results in



      $$
      cos bigl( Q(beta + delta) bigr) = (Q + 1) sin delta.
      $$



      If $Q = 1$ then the equation can be solved exactly. Otherwise, you can rewrite the equation as



      $$
      delta = f(delta) = arcsin left( fraccos bigl( Q(beta + delta) bigr)Q + 1 right)
      $$



      and repeatedly apply $f$ to some initial guess for a solution $delta$. It turns out that this fixed-point iteration converges in this particular case.






      share|cite|improve this answer























        up vote
        2
        down vote










        up vote
        2
        down vote









        Letting $Q = R/r$ and rearranging a bit gives



        $$
        fracpi2 - Qbeta - Qdelta = arcsin bigl( (Q + 1) sin delta bigr)
        $$



        then taking $sin(cdot)$ of both sides and using $sin (pi/2 - x) = cos x$ results in



        $$
        cos bigl( Q(beta + delta) bigr) = (Q + 1) sin delta.
        $$



        If $Q = 1$ then the equation can be solved exactly. Otherwise, you can rewrite the equation as



        $$
        delta = f(delta) = arcsin left( fraccos bigl( Q(beta + delta) bigr)Q + 1 right)
        $$



        and repeatedly apply $f$ to some initial guess for a solution $delta$. It turns out that this fixed-point iteration converges in this particular case.






        share|cite|improve this answer













        Letting $Q = R/r$ and rearranging a bit gives



        $$
        fracpi2 - Qbeta - Qdelta = arcsin bigl( (Q + 1) sin delta bigr)
        $$



        then taking $sin(cdot)$ of both sides and using $sin (pi/2 - x) = cos x$ results in



        $$
        cos bigl( Q(beta + delta) bigr) = (Q + 1) sin delta.
        $$



        If $Q = 1$ then the equation can be solved exactly. Otherwise, you can rewrite the equation as



        $$
        delta = f(delta) = arcsin left( fraccos bigl( Q(beta + delta) bigr)Q + 1 right)
        $$



        and repeatedly apply $f$ to some initial guess for a solution $delta$. It turns out that this fixed-point iteration converges in this particular case.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered yesterday









        D. G.

        464




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