If $ sin alpha = frac 45 $ and $ cos beta = frac{5}{13} $, prove that $ cos frac{alpha-beta}{2} =...












2












$begingroup$


I can solve it easily if I assume that $ 0 < alpha, beta < frac{pi}{2}$



But there is no mention of the quadrants in which $ alpha $ and $ beta $ lie in.



Is the question wrong ?






trigonometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    You know $cos alpha$ and $sin beta$ up to a sign. Look at the formulas you have for $cos frac{alpha-beta}{2}$ and see how they change when you vary the signs of those two quantities. Best case, you see that flipping each sign does nothing, worst case, you have four cases to consider, one for each possible combination of signs.
    $endgroup$
    – Aaron
    Jan 13 '17 at 10:45






  • 1




    $begingroup$
    I did just that, and I got 4 different answers, only one of which matches $ frac{8}{sqrt{65}} $
    $endgroup$
    – Nathuram
    Jan 13 '17 at 10:46








  • 1




    $begingroup$
    You're right: you can't prove that equality.
    $endgroup$
    – egreg
    Jan 13 '17 at 10:48






  • 1




    $begingroup$
    In that case, without verifying your work, I would say yes, the question is wrong, or at least incomplete, if it doesn't say something else that would give you a way of specifying those signs.
    $endgroup$
    – Aaron
    Jan 13 '17 at 10:49
















2












$begingroup$


I can solve it easily if I assume that $ 0 < alpha, beta < frac{pi}{2}$



But there is no mention of the quadrants in which $ alpha $ and $ beta $ lie in.



Is the question wrong ?






trigonometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    You know $cos alpha$ and $sin beta$ up to a sign. Look at the formulas you have for $cos frac{alpha-beta}{2}$ and see how they change when you vary the signs of those two quantities. Best case, you see that flipping each sign does nothing, worst case, you have four cases to consider, one for each possible combination of signs.
    $endgroup$
    – Aaron
    Jan 13 '17 at 10:45






  • 1




    $begingroup$
    I did just that, and I got 4 different answers, only one of which matches $ frac{8}{sqrt{65}} $
    $endgroup$
    – Nathuram
    Jan 13 '17 at 10:46








  • 1




    $begingroup$
    You're right: you can't prove that equality.
    $endgroup$
    – egreg
    Jan 13 '17 at 10:48






  • 1




    $begingroup$
    In that case, without verifying your work, I would say yes, the question is wrong, or at least incomplete, if it doesn't say something else that would give you a way of specifying those signs.
    $endgroup$
    – Aaron
    Jan 13 '17 at 10:49














2












2








2


1



$begingroup$


I can solve it easily if I assume that $ 0 < alpha, beta < frac{pi}{2}$



But there is no mention of the quadrants in which $ alpha $ and $ beta $ lie in.



Is the question wrong ?






trigonometry







share|cite|improve this question









$endgroup$




I can solve it easily if I assume that $ 0 < alpha, beta < frac{pi}{2}$



But there is no mention of the quadrants in which $ alpha $ and $ beta $ lie in.



Is the question wrong ?






trigonometry




trigonometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 13 '17 at 10:15









NathuramNathuram

54211




54211












  • $begingroup$
    You know $cos alpha$ and $sin beta$ up to a sign. Look at the formulas you have for $cos frac{alpha-beta}{2}$ and see how they change when you vary the signs of those two quantities. Best case, you see that flipping each sign does nothing, worst case, you have four cases to consider, one for each possible combination of signs.
    $endgroup$
    – Aaron
    Jan 13 '17 at 10:45






  • 1




    $begingroup$
    I did just that, and I got 4 different answers, only one of which matches $ frac{8}{sqrt{65}} $
    $endgroup$
    – Nathuram
    Jan 13 '17 at 10:46








  • 1




    $begingroup$
    You're right: you can't prove that equality.
    $endgroup$
    – egreg
    Jan 13 '17 at 10:48






  • 1




    $begingroup$
    In that case, without verifying your work, I would say yes, the question is wrong, or at least incomplete, if it doesn't say something else that would give you a way of specifying those signs.
    $endgroup$
    – Aaron
    Jan 13 '17 at 10:49


















  • $begingroup$
    You know $cos alpha$ and $sin beta$ up to a sign. Look at the formulas you have for $cos frac{alpha-beta}{2}$ and see how they change when you vary the signs of those two quantities. Best case, you see that flipping each sign does nothing, worst case, you have four cases to consider, one for each possible combination of signs.
    $endgroup$
    – Aaron
    Jan 13 '17 at 10:45






  • 1




    $begingroup$
    I did just that, and I got 4 different answers, only one of which matches $ frac{8}{sqrt{65}} $
    $endgroup$
    – Nathuram
    Jan 13 '17 at 10:46








  • 1




    $begingroup$
    You're right: you can't prove that equality.
    $endgroup$
    – egreg
    Jan 13 '17 at 10:48






  • 1




    $begingroup$
    In that case, without verifying your work, I would say yes, the question is wrong, or at least incomplete, if it doesn't say something else that would give you a way of specifying those signs.
    $endgroup$
    – Aaron
    Jan 13 '17 at 10:49
















$begingroup$
You know $cos alpha$ and $sin beta$ up to a sign. Look at the formulas you have for $cos frac{alpha-beta}{2}$ and see how they change when you vary the signs of those two quantities. Best case, you see that flipping each sign does nothing, worst case, you have four cases to consider, one for each possible combination of signs.
$endgroup$
– Aaron
Jan 13 '17 at 10:45




$begingroup$
You know $cos alpha$ and $sin beta$ up to a sign. Look at the formulas you have for $cos frac{alpha-beta}{2}$ and see how they change when you vary the signs of those two quantities. Best case, you see that flipping each sign does nothing, worst case, you have four cases to consider, one for each possible combination of signs.
$endgroup$
– Aaron
Jan 13 '17 at 10:45




1




1




$begingroup$
I did just that, and I got 4 different answers, only one of which matches $ frac{8}{sqrt{65}} $
$endgroup$
– Nathuram
Jan 13 '17 at 10:46






$begingroup$
I did just that, and I got 4 different answers, only one of which matches $ frac{8}{sqrt{65}} $
$endgroup$
– Nathuram
Jan 13 '17 at 10:46






1




1




$begingroup$
You're right: you can't prove that equality.
$endgroup$
– egreg
Jan 13 '17 at 10:48




$begingroup$
You're right: you can't prove that equality.
$endgroup$
– egreg
Jan 13 '17 at 10:48




1




1




$begingroup$
In that case, without verifying your work, I would say yes, the question is wrong, or at least incomplete, if it doesn't say something else that would give you a way of specifying those signs.
$endgroup$
– Aaron
Jan 13 '17 at 10:49




$begingroup$
In that case, without verifying your work, I would say yes, the question is wrong, or at least incomplete, if it doesn't say something else that would give you a way of specifying those signs.
$endgroup$
– Aaron
Jan 13 '17 at 10:49










2 Answers
2






active

oldest

votes


















0












$begingroup$

Hint:



$sin alpha>0$ then $0<alpha<pi$ and $cos beta>0$ then $-frac{pi}2<beta<frac{pi}2$ and $-frac{pi}2<-beta<frac{pi}2$



Thus $$-frac{pi}2<alpha-beta<frac{3pi}2$$
and
$$-frac{pi}4<frac{alpha-beta}2<frac{3pi}4$$



1) $sin^2alpha+cos^2alpha=1$, then $cosalpha=pmsqrt{1-sin^2alpha}$ and $sin beta=pmsqrt{1-cos^2beta}$



2) $cos (alpha-beta)=cosalphacosbeta+sinalphasinbeta$



3)$|cos frac{alpha-beta}2|=sqrt{frac{1+cos(alpha-beta)}2}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @JonathanY. since $0<alpha,beta<pi/2$ we have that $-pi/2<(alpha-beta)/2<pi/2$ hence the $cos(alpha-beta)/2$ is nonnegative.
    $endgroup$
    – Math-fun
    Jan 13 '17 at 10:31












  • $begingroup$
    @JonathanY. my bad I did not read the statement properly.
    $endgroup$
    – Math-fun
    Jan 13 '17 at 10:34










  • $begingroup$
    Taking into account recent edits, if $cos(alpha-beta)$ could be uniquely determined, then so would $cosfrac{alpha-beta}{2}$. But can step 2 be completed like you wish? Also, consider that comments by OP suggest that he's tried this already. (In short, I'm asking if you followed through on your hints and have a complete solution in mind.)
    $endgroup$
    – Jonathan Y.
    Jan 13 '17 at 10:54



















0












$begingroup$

HINT:



begin{align}
cos(alpha/2-beta/2)&=sqrt{frac{1+cos(alpha-beta)}{2}}=sqrt{(1+cosalphacosbeta+sinalphasinbeta)/2}\
end{align}



$$ cosalpha =frac35, cosbeta = frac{5}{13},, sinalpha=frac45,, sinbeta =frac{12}{13}, $$



plug in to get $ 8/sqrt{65}$



For other quadrants calculate other three values also which are possible with inverse trig functions.



$$sqrt{(1pmcosalphacosbetapmsinalphasinbeta)/2}$$






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Hint:



    $sin alpha>0$ then $0<alpha<pi$ and $cos beta>0$ then $-frac{pi}2<beta<frac{pi}2$ and $-frac{pi}2<-beta<frac{pi}2$



    Thus $$-frac{pi}2<alpha-beta<frac{3pi}2$$
    and
    $$-frac{pi}4<frac{alpha-beta}2<frac{3pi}4$$



    1) $sin^2alpha+cos^2alpha=1$, then $cosalpha=pmsqrt{1-sin^2alpha}$ and $sin beta=pmsqrt{1-cos^2beta}$



    2) $cos (alpha-beta)=cosalphacosbeta+sinalphasinbeta$



    3)$|cos frac{alpha-beta}2|=sqrt{frac{1+cos(alpha-beta)}2}$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @JonathanY. since $0<alpha,beta<pi/2$ we have that $-pi/2<(alpha-beta)/2<pi/2$ hence the $cos(alpha-beta)/2$ is nonnegative.
      $endgroup$
      – Math-fun
      Jan 13 '17 at 10:31












    • $begingroup$
      @JonathanY. my bad I did not read the statement properly.
      $endgroup$
      – Math-fun
      Jan 13 '17 at 10:34










    • $begingroup$
      Taking into account recent edits, if $cos(alpha-beta)$ could be uniquely determined, then so would $cosfrac{alpha-beta}{2}$. But can step 2 be completed like you wish? Also, consider that comments by OP suggest that he's tried this already. (In short, I'm asking if you followed through on your hints and have a complete solution in mind.)
      $endgroup$
      – Jonathan Y.
      Jan 13 '17 at 10:54
















    0












    $begingroup$

    Hint:



    $sin alpha>0$ then $0<alpha<pi$ and $cos beta>0$ then $-frac{pi}2<beta<frac{pi}2$ and $-frac{pi}2<-beta<frac{pi}2$



    Thus $$-frac{pi}2<alpha-beta<frac{3pi}2$$
    and
    $$-frac{pi}4<frac{alpha-beta}2<frac{3pi}4$$



    1) $sin^2alpha+cos^2alpha=1$, then $cosalpha=pmsqrt{1-sin^2alpha}$ and $sin beta=pmsqrt{1-cos^2beta}$



    2) $cos (alpha-beta)=cosalphacosbeta+sinalphasinbeta$



    3)$|cos frac{alpha-beta}2|=sqrt{frac{1+cos(alpha-beta)}2}$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @JonathanY. since $0<alpha,beta<pi/2$ we have that $-pi/2<(alpha-beta)/2<pi/2$ hence the $cos(alpha-beta)/2$ is nonnegative.
      $endgroup$
      – Math-fun
      Jan 13 '17 at 10:31












    • $begingroup$
      @JonathanY. my bad I did not read the statement properly.
      $endgroup$
      – Math-fun
      Jan 13 '17 at 10:34










    • $begingroup$
      Taking into account recent edits, if $cos(alpha-beta)$ could be uniquely determined, then so would $cosfrac{alpha-beta}{2}$. But can step 2 be completed like you wish? Also, consider that comments by OP suggest that he's tried this already. (In short, I'm asking if you followed through on your hints and have a complete solution in mind.)
      $endgroup$
      – Jonathan Y.
      Jan 13 '17 at 10:54














    0












    0








    0





    $begingroup$

    Hint:



    $sin alpha>0$ then $0<alpha<pi$ and $cos beta>0$ then $-frac{pi}2<beta<frac{pi}2$ and $-frac{pi}2<-beta<frac{pi}2$



    Thus $$-frac{pi}2<alpha-beta<frac{3pi}2$$
    and
    $$-frac{pi}4<frac{alpha-beta}2<frac{3pi}4$$



    1) $sin^2alpha+cos^2alpha=1$, then $cosalpha=pmsqrt{1-sin^2alpha}$ and $sin beta=pmsqrt{1-cos^2beta}$



    2) $cos (alpha-beta)=cosalphacosbeta+sinalphasinbeta$



    3)$|cos frac{alpha-beta}2|=sqrt{frac{1+cos(alpha-beta)}2}$






    share|cite|improve this answer











    $endgroup$



    Hint:



    $sin alpha>0$ then $0<alpha<pi$ and $cos beta>0$ then $-frac{pi}2<beta<frac{pi}2$ and $-frac{pi}2<-beta<frac{pi}2$



    Thus $$-frac{pi}2<alpha-beta<frac{3pi}2$$
    and
    $$-frac{pi}4<frac{alpha-beta}2<frac{3pi}4$$



    1) $sin^2alpha+cos^2alpha=1$, then $cosalpha=pmsqrt{1-sin^2alpha}$ and $sin beta=pmsqrt{1-cos^2beta}$



    2) $cos (alpha-beta)=cosalphacosbeta+sinalphasinbeta$



    3)$|cos frac{alpha-beta}2|=sqrt{frac{1+cos(alpha-beta)}2}$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 13 '17 at 10:43

























    answered Jan 13 '17 at 10:24









    Roman83Roman83

    14.4k31956




    14.4k31956












    • $begingroup$
      @JonathanY. since $0<alpha,beta<pi/2$ we have that $-pi/2<(alpha-beta)/2<pi/2$ hence the $cos(alpha-beta)/2$ is nonnegative.
      $endgroup$
      – Math-fun
      Jan 13 '17 at 10:31












    • $begingroup$
      @JonathanY. my bad I did not read the statement properly.
      $endgroup$
      – Math-fun
      Jan 13 '17 at 10:34










    • $begingroup$
      Taking into account recent edits, if $cos(alpha-beta)$ could be uniquely determined, then so would $cosfrac{alpha-beta}{2}$. But can step 2 be completed like you wish? Also, consider that comments by OP suggest that he's tried this already. (In short, I'm asking if you followed through on your hints and have a complete solution in mind.)
      $endgroup$
      – Jonathan Y.
      Jan 13 '17 at 10:54


















    • $begingroup$
      @JonathanY. since $0<alpha,beta<pi/2$ we have that $-pi/2<(alpha-beta)/2<pi/2$ hence the $cos(alpha-beta)/2$ is nonnegative.
      $endgroup$
      – Math-fun
      Jan 13 '17 at 10:31












    • $begingroup$
      @JonathanY. my bad I did not read the statement properly.
      $endgroup$
      – Math-fun
      Jan 13 '17 at 10:34










    • $begingroup$
      Taking into account recent edits, if $cos(alpha-beta)$ could be uniquely determined, then so would $cosfrac{alpha-beta}{2}$. But can step 2 be completed like you wish? Also, consider that comments by OP suggest that he's tried this already. (In short, I'm asking if you followed through on your hints and have a complete solution in mind.)
      $endgroup$
      – Jonathan Y.
      Jan 13 '17 at 10:54
















    $begingroup$
    @JonathanY. since $0<alpha,beta<pi/2$ we have that $-pi/2<(alpha-beta)/2<pi/2$ hence the $cos(alpha-beta)/2$ is nonnegative.
    $endgroup$
    – Math-fun
    Jan 13 '17 at 10:31






    $begingroup$
    @JonathanY. since $0<alpha,beta<pi/2$ we have that $-pi/2<(alpha-beta)/2<pi/2$ hence the $cos(alpha-beta)/2$ is nonnegative.
    $endgroup$
    – Math-fun
    Jan 13 '17 at 10:31














    $begingroup$
    @JonathanY. my bad I did not read the statement properly.
    $endgroup$
    – Math-fun
    Jan 13 '17 at 10:34




    $begingroup$
    @JonathanY. my bad I did not read the statement properly.
    $endgroup$
    – Math-fun
    Jan 13 '17 at 10:34












    $begingroup$
    Taking into account recent edits, if $cos(alpha-beta)$ could be uniquely determined, then so would $cosfrac{alpha-beta}{2}$. But can step 2 be completed like you wish? Also, consider that comments by OP suggest that he's tried this already. (In short, I'm asking if you followed through on your hints and have a complete solution in mind.)
    $endgroup$
    – Jonathan Y.
    Jan 13 '17 at 10:54




    $begingroup$
    Taking into account recent edits, if $cos(alpha-beta)$ could be uniquely determined, then so would $cosfrac{alpha-beta}{2}$. But can step 2 be completed like you wish? Also, consider that comments by OP suggest that he's tried this already. (In short, I'm asking if you followed through on your hints and have a complete solution in mind.)
    $endgroup$
    – Jonathan Y.
    Jan 13 '17 at 10:54











    0












    $begingroup$

    HINT:



    begin{align}
    cos(alpha/2-beta/2)&=sqrt{frac{1+cos(alpha-beta)}{2}}=sqrt{(1+cosalphacosbeta+sinalphasinbeta)/2}\
    end{align}



    $$ cosalpha =frac35, cosbeta = frac{5}{13},, sinalpha=frac45,, sinbeta =frac{12}{13}, $$



    plug in to get $ 8/sqrt{65}$



    For other quadrants calculate other three values also which are possible with inverse trig functions.



    $$sqrt{(1pmcosalphacosbetapmsinalphasinbeta)/2}$$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      HINT:



      begin{align}
      cos(alpha/2-beta/2)&=sqrt{frac{1+cos(alpha-beta)}{2}}=sqrt{(1+cosalphacosbeta+sinalphasinbeta)/2}\
      end{align}



      $$ cosalpha =frac35, cosbeta = frac{5}{13},, sinalpha=frac45,, sinbeta =frac{12}{13}, $$



      plug in to get $ 8/sqrt{65}$



      For other quadrants calculate other three values also which are possible with inverse trig functions.



      $$sqrt{(1pmcosalphacosbetapmsinalphasinbeta)/2}$$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        HINT:



        begin{align}
        cos(alpha/2-beta/2)&=sqrt{frac{1+cos(alpha-beta)}{2}}=sqrt{(1+cosalphacosbeta+sinalphasinbeta)/2}\
        end{align}



        $$ cosalpha =frac35, cosbeta = frac{5}{13},, sinalpha=frac45,, sinbeta =frac{12}{13}, $$



        plug in to get $ 8/sqrt{65}$



        For other quadrants calculate other three values also which are possible with inverse trig functions.



        $$sqrt{(1pmcosalphacosbetapmsinalphasinbeta)/2}$$






        share|cite|improve this answer











        $endgroup$



        HINT:



        begin{align}
        cos(alpha/2-beta/2)&=sqrt{frac{1+cos(alpha-beta)}{2}}=sqrt{(1+cosalphacosbeta+sinalphasinbeta)/2}\
        end{align}



        $$ cosalpha =frac35, cosbeta = frac{5}{13},, sinalpha=frac45,, sinbeta =frac{12}{13}, $$



        plug in to get $ 8/sqrt{65}$



        For other quadrants calculate other three values also which are possible with inverse trig functions.



        $$sqrt{(1pmcosalphacosbetapmsinalphasinbeta)/2}$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 13 '17 at 11:14

























        answered Jan 13 '17 at 11:02









        NarasimhamNarasimham

        20.8k62158




        20.8k62158






























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            Antonio Litta Visconti Arese

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            Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
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