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

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Multi tool use












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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            menu    Wanted List (rus) Main Projects    FAMI Dumping project    Pirate cart's ID numbers list    My carts collection    Unpublished NES cheats    >All titles     Numbered titles     A titles     B titles     C titles     D titles     E titles     F titles     G titles     H titles     I titles     J titles     K titles     L titles     M titles     N titles     O titles     P titles     Q titles     R titles     S titles     T titles     U titles     V titles     W titles     X titles     Y titles     Z titles     Unique cheats     Buried secrets     Fixed generic cheats     Generic cheats     Secret passwords     Password generators     List of games with no cheats found    Unpublished Game Boy cheats    Unpublished Game Boy Color cheats Misc Stuff    Pirate Market    BONZA    FORTUNA    Decrypted SmyNES ROMs    NES inside (rus)    FCEUMM project Blog    cah4e3.wordpress.com Mail me:    cah4e3 at gmail dot c...
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            SpongeBob's Truth or Square (Wii)

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            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! SpongeBob's Truth or Square (Wii) From The Cutting Room Floor Jump to: navigation, search SpongeBob's Truth or Square Developer : Heavy Iron Studios Publisher : THQ Platforms : Wii Released in US : October 26, 2009 Released in EU : December 2009 This game has hidden development-related text. This game has unused graphics. This game has unused models. This game has unused sounds. This game has unused text. This page is rather stubbly and could use some expansion. Are you a bad enough dude to rescue this article? SpongeBob's Truth or Square is a game based off the tenth anniversary TV special of the same name. The game was made by Heavy Iron Studios, the same company behind SpongeBob SquarePants: Battle for Bikini Bottom and The SpongeBob Square...
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