Is there always exactly one solution to $a cosleft(frac{x}{2}right)- b sinleft(frac{x}{2}right) = 0$ in the...












2












$begingroup$


I have a probably simple question.
If I have an equation like
begin{align}
a cos left(frac{x}{2}right) - b sin left(frac{x}{2}right) &= 0\
frac ab cos left(frac{x}{2}right) - sin left(frac{x}{2}right) &= 0
end{align}
with $a, b > 0$. Is it true that there is always precisely one solution in the interval $0 leq x leq 2 pi$.



I think it is obvious, but can I use this in a proof without proving it?






trigonometry







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  • 1




    $begingroup$
    Never assume something is obvious if you cannot state why it is obvious.
    $endgroup$
    – Wesley Strik
    Jan 19 at 15:19
















2












$begingroup$


I have a probably simple question.
If I have an equation like
begin{align}
a cos left(frac{x}{2}right) - b sin left(frac{x}{2}right) &= 0\
frac ab cos left(frac{x}{2}right) - sin left(frac{x}{2}right) &= 0
end{align}
with $a, b > 0$. Is it true that there is always precisely one solution in the interval $0 leq x leq 2 pi$.



I think it is obvious, but can I use this in a proof without proving it?






trigonometry







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Never assume something is obvious if you cannot state why it is obvious.
    $endgroup$
    – Wesley Strik
    Jan 19 at 15:19














2












2








2





$begingroup$


I have a probably simple question.
If I have an equation like
begin{align}
a cos left(frac{x}{2}right) - b sin left(frac{x}{2}right) &= 0\
frac ab cos left(frac{x}{2}right) - sin left(frac{x}{2}right) &= 0
end{align}
with $a, b > 0$. Is it true that there is always precisely one solution in the interval $0 leq x leq 2 pi$.



I think it is obvious, but can I use this in a proof without proving it?






trigonometry







share|cite|improve this question











$endgroup$




I have a probably simple question.
If I have an equation like
begin{align}
a cos left(frac{x}{2}right) - b sin left(frac{x}{2}right) &= 0\
frac ab cos left(frac{x}{2}right) - sin left(frac{x}{2}right) &= 0
end{align}
with $a, b > 0$. Is it true that there is always precisely one solution in the interval $0 leq x leq 2 pi$.



I think it is obvious, but can I use this in a proof without proving it?






trigonometry




trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 19 at 16:13









Wesley Strik

2,113423




2,113423










asked Jan 19 at 14:58









garax91garax91

405




405








  • 1




    $begingroup$
    Never assume something is obvious if you cannot state why it is obvious.
    $endgroup$
    – Wesley Strik
    Jan 19 at 15:19














  • 1




    $begingroup$
    Never assume something is obvious if you cannot state why it is obvious.
    $endgroup$
    – Wesley Strik
    Jan 19 at 15:19








1




1




$begingroup$
Never assume something is obvious if you cannot state why it is obvious.
$endgroup$
– Wesley Strik
Jan 19 at 15:19




$begingroup$
Never assume something is obvious if you cannot state why it is obvious.
$endgroup$
– Wesley Strik
Jan 19 at 15:19










2 Answers
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$begingroup$

It is $$tan(frac{x}{2})=frac{a}{b}$$






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    $begingroup$

    Simply note the definition of tangent: $tan(x) = dfrac{sin(x)}{cos(x)}$.



    $$acosleft(frac{x}{2}right)-bsinleft(frac{x}{2}right) = 0 iff acosleft(frac{x}{2}right) = bsinleft(frac{x}{2}right) iff \ frac{a}{b} = frac{sinleft(frac{x}{2}right)}{cosleft(frac{x}{2}right)} iff frac{a}{b} = tanleft(frac{x}{2}right)$$



    Since the range of $tan(x)$ includes all real numbers, there is a solution in the interval $x in [0, 2pi]$.






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      2 Answers
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      2 Answers
      2






      active

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      active

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      active

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      4












      $begingroup$

      It is $$tan(frac{x}{2})=frac{a}{b}$$






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        It is $$tan(frac{x}{2})=frac{a}{b}$$






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          It is $$tan(frac{x}{2})=frac{a}{b}$$






          share|cite|improve this answer









          $endgroup$



          It is $$tan(frac{x}{2})=frac{a}{b}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 19 at 15:00









          Dr. Sonnhard GraubnerDr. Sonnhard Graubner

          76.7k42866




          76.7k42866























              2












              $begingroup$

              Simply note the definition of tangent: $tan(x) = dfrac{sin(x)}{cos(x)}$.



              $$acosleft(frac{x}{2}right)-bsinleft(frac{x}{2}right) = 0 iff acosleft(frac{x}{2}right) = bsinleft(frac{x}{2}right) iff \ frac{a}{b} = frac{sinleft(frac{x}{2}right)}{cosleft(frac{x}{2}right)} iff frac{a}{b} = tanleft(frac{x}{2}right)$$



              Since the range of $tan(x)$ includes all real numbers, there is a solution in the interval $x in [0, 2pi]$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Simply note the definition of tangent: $tan(x) = dfrac{sin(x)}{cos(x)}$.



                $$acosleft(frac{x}{2}right)-bsinleft(frac{x}{2}right) = 0 iff acosleft(frac{x}{2}right) = bsinleft(frac{x}{2}right) iff \ frac{a}{b} = frac{sinleft(frac{x}{2}right)}{cosleft(frac{x}{2}right)} iff frac{a}{b} = tanleft(frac{x}{2}right)$$



                Since the range of $tan(x)$ includes all real numbers, there is a solution in the interval $x in [0, 2pi]$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Simply note the definition of tangent: $tan(x) = dfrac{sin(x)}{cos(x)}$.



                  $$acosleft(frac{x}{2}right)-bsinleft(frac{x}{2}right) = 0 iff acosleft(frac{x}{2}right) = bsinleft(frac{x}{2}right) iff \ frac{a}{b} = frac{sinleft(frac{x}{2}right)}{cosleft(frac{x}{2}right)} iff frac{a}{b} = tanleft(frac{x}{2}right)$$



                  Since the range of $tan(x)$ includes all real numbers, there is a solution in the interval $x in [0, 2pi]$.






                  share|cite|improve this answer









                  $endgroup$



                  Simply note the definition of tangent: $tan(x) = dfrac{sin(x)}{cos(x)}$.



                  $$acosleft(frac{x}{2}right)-bsinleft(frac{x}{2}right) = 0 iff acosleft(frac{x}{2}right) = bsinleft(frac{x}{2}right) iff \ frac{a}{b} = frac{sinleft(frac{x}{2}right)}{cosleft(frac{x}{2}right)} iff frac{a}{b} = tanleft(frac{x}{2}right)$$



                  Since the range of $tan(x)$ includes all real numbers, there is a solution in the interval $x in [0, 2pi]$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 19 at 15:15









                  KM101KM101

                  6,0351525




                  6,0351525






























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