If $x$ and $y$ are acute, and $sin y = 3 cos (x+y) sin x$⁡, then find the maximum value of $tan y$












2












$begingroup$



Given $x,y$ are acute angles such that
$$sin y = 3 cos(x+y)sin x$$
Find the maximum value of $tan ⁡y$.




Attempt:
We have



$$begin{aligned} 3(cos x cos y - sin x sin y) sin x & = sin y \ 3 cos x sin x - 3 sin^2 x tan y & = tan y \ 3 cos x sin x & = tan y(1 +3 sin^2 x) \ tan y & = dfrac{3 sin x cos x} {1+3 sin^2 x} end{aligned}$$



Now, how about the next step? Or maybe I did some mistakes?






trigonometry optimization maxima-minima a.m.-g.m.-inequality







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$endgroup$








  • 1




    $begingroup$
    I think now you could use $1 = sin^2 x + cos^2 x$, then divide the top and the bottom by $cos^2 x$ and transform the equation to a fraction of $tan x$.
    $endgroup$
    – xbh
    Jan 17 at 6:49
















2












$begingroup$



Given $x,y$ are acute angles such that
$$sin y = 3 cos(x+y)sin x$$
Find the maximum value of $tan ⁡y$.




Attempt:
We have



$$begin{aligned} 3(cos x cos y - sin x sin y) sin x & = sin y \ 3 cos x sin x - 3 sin^2 x tan y & = tan y \ 3 cos x sin x & = tan y(1 +3 sin^2 x) \ tan y & = dfrac{3 sin x cos x} {1+3 sin^2 x} end{aligned}$$



Now, how about the next step? Or maybe I did some mistakes?






trigonometry optimization maxima-minima a.m.-g.m.-inequality







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I think now you could use $1 = sin^2 x + cos^2 x$, then divide the top and the bottom by $cos^2 x$ and transform the equation to a fraction of $tan x$.
    $endgroup$
    – xbh
    Jan 17 at 6:49














2












2








2





$begingroup$



Given $x,y$ are acute angles such that
$$sin y = 3 cos(x+y)sin x$$
Find the maximum value of $tan ⁡y$.




Attempt:
We have



$$begin{aligned} 3(cos x cos y - sin x sin y) sin x & = sin y \ 3 cos x sin x - 3 sin^2 x tan y & = tan y \ 3 cos x sin x & = tan y(1 +3 sin^2 x) \ tan y & = dfrac{3 sin x cos x} {1+3 sin^2 x} end{aligned}$$



Now, how about the next step? Or maybe I did some mistakes?






trigonometry optimization maxima-minima a.m.-g.m.-inequality







share|cite|improve this question











$endgroup$





Given $x,y$ are acute angles such that
$$sin y = 3 cos(x+y)sin x$$
Find the maximum value of $tan ⁡y$.




Attempt:
We have



$$begin{aligned} 3(cos x cos y - sin x sin y) sin x & = sin y \ 3 cos x sin x - 3 sin^2 x tan y & = tan y \ 3 cos x sin x & = tan y(1 +3 sin^2 x) \ tan y & = dfrac{3 sin x cos x} {1+3 sin^2 x} end{aligned}$$



Now, how about the next step? Or maybe I did some mistakes?






trigonometry optimization maxima-minima a.m.-g.m.-inequality




trigonometry optimization maxima-minima a.m.-g.m.-inequality






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share|cite|improve this question













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edited Jan 17 at 7:00









Michael Rozenberg

110k1896201




110k1896201










asked Jan 17 at 6:41









Shane Dizzy SukardyShane Dizzy Sukardy

60819




60819








  • 1




    $begingroup$
    I think now you could use $1 = sin^2 x + cos^2 x$, then divide the top and the bottom by $cos^2 x$ and transform the equation to a fraction of $tan x$.
    $endgroup$
    – xbh
    Jan 17 at 6:49














  • 1




    $begingroup$
    I think now you could use $1 = sin^2 x + cos^2 x$, then divide the top and the bottom by $cos^2 x$ and transform the equation to a fraction of $tan x$.
    $endgroup$
    – xbh
    Jan 17 at 6:49








1




1




$begingroup$
I think now you could use $1 = sin^2 x + cos^2 x$, then divide the top and the bottom by $cos^2 x$ and transform the equation to a fraction of $tan x$.
$endgroup$
– xbh
Jan 17 at 6:49




$begingroup$
I think now you could use $1 = sin^2 x + cos^2 x$, then divide the top and the bottom by $cos^2 x$ and transform the equation to a fraction of $tan x$.
$endgroup$
– xbh
Jan 17 at 6:49










3 Answers
3






active

oldest

votes


















2












$begingroup$

By AM-GM $$tan{y}=frac{3sin{x}cos{x}}{1+3sin^2x}=frac{3sin{x}cos{x}}{cos^2x+4sin^2x}leqfrac{3sin{x}cos{x}}{2sqrt{cos^2xcdot4sin^2x}}=frac{3}{4}.$$
The equality occurs for $cos{x}=2sin{x},$ which says that we got a maximal value.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much, Sir.
    $endgroup$
    – Shane Dizzy Sukardy
    Jan 17 at 7:09










  • $begingroup$
    @Shane Dizzy Sukardy You are welcome!
    $endgroup$
    – Michael Rozenberg
    Jan 17 at 7:10



















1












$begingroup$

Put $t = tan x$, then $tan y = dfrac{3t}{1+4t^2}le dfrac{3t}{4t} = dfrac{3}{4}$, which is the max of $tan y$ with equality occurs when $t = dfrac{1}{2}$ or $2sin x = cos x$...the rest is simple...






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    $$tan y=dfrac{3tan x}{1+4tan^2x}$$



    $$iff(4tan y)tan ^2x-3tan x+tan y=0$$ which is a Quadratic Equation $tan x$



    As $tan x$ is real, the discriminant must be $ge0$



    i.e., $$3^2-4(4tan y)ge0tan yifftan^2yledfrac9{16}iff-dfrac34letan yledfrac34$$






    share|cite|improve this answer









    $endgroup$














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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      By AM-GM $$tan{y}=frac{3sin{x}cos{x}}{1+3sin^2x}=frac{3sin{x}cos{x}}{cos^2x+4sin^2x}leqfrac{3sin{x}cos{x}}{2sqrt{cos^2xcdot4sin^2x}}=frac{3}{4}.$$
      The equality occurs for $cos{x}=2sin{x},$ which says that we got a maximal value.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thank you very much, Sir.
        $endgroup$
        – Shane Dizzy Sukardy
        Jan 17 at 7:09










      • $begingroup$
        @Shane Dizzy Sukardy You are welcome!
        $endgroup$
        – Michael Rozenberg
        Jan 17 at 7:10
















      2












      $begingroup$

      By AM-GM $$tan{y}=frac{3sin{x}cos{x}}{1+3sin^2x}=frac{3sin{x}cos{x}}{cos^2x+4sin^2x}leqfrac{3sin{x}cos{x}}{2sqrt{cos^2xcdot4sin^2x}}=frac{3}{4}.$$
      The equality occurs for $cos{x}=2sin{x},$ which says that we got a maximal value.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thank you very much, Sir.
        $endgroup$
        – Shane Dizzy Sukardy
        Jan 17 at 7:09










      • $begingroup$
        @Shane Dizzy Sukardy You are welcome!
        $endgroup$
        – Michael Rozenberg
        Jan 17 at 7:10














      2












      2








      2





      $begingroup$

      By AM-GM $$tan{y}=frac{3sin{x}cos{x}}{1+3sin^2x}=frac{3sin{x}cos{x}}{cos^2x+4sin^2x}leqfrac{3sin{x}cos{x}}{2sqrt{cos^2xcdot4sin^2x}}=frac{3}{4}.$$
      The equality occurs for $cos{x}=2sin{x},$ which says that we got a maximal value.






      share|cite|improve this answer









      $endgroup$



      By AM-GM $$tan{y}=frac{3sin{x}cos{x}}{1+3sin^2x}=frac{3sin{x}cos{x}}{cos^2x+4sin^2x}leqfrac{3sin{x}cos{x}}{2sqrt{cos^2xcdot4sin^2x}}=frac{3}{4}.$$
      The equality occurs for $cos{x}=2sin{x},$ which says that we got a maximal value.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 17 at 6:56









      Michael RozenbergMichael Rozenberg

      110k1896201




      110k1896201












      • $begingroup$
        Thank you very much, Sir.
        $endgroup$
        – Shane Dizzy Sukardy
        Jan 17 at 7:09










      • $begingroup$
        @Shane Dizzy Sukardy You are welcome!
        $endgroup$
        – Michael Rozenberg
        Jan 17 at 7:10


















      • $begingroup$
        Thank you very much, Sir.
        $endgroup$
        – Shane Dizzy Sukardy
        Jan 17 at 7:09










      • $begingroup$
        @Shane Dizzy Sukardy You are welcome!
        $endgroup$
        – Michael Rozenberg
        Jan 17 at 7:10
















      $begingroup$
      Thank you very much, Sir.
      $endgroup$
      – Shane Dizzy Sukardy
      Jan 17 at 7:09




      $begingroup$
      Thank you very much, Sir.
      $endgroup$
      – Shane Dizzy Sukardy
      Jan 17 at 7:09












      $begingroup$
      @Shane Dizzy Sukardy You are welcome!
      $endgroup$
      – Michael Rozenberg
      Jan 17 at 7:10




      $begingroup$
      @Shane Dizzy Sukardy You are welcome!
      $endgroup$
      – Michael Rozenberg
      Jan 17 at 7:10











      1












      $begingroup$

      Put $t = tan x$, then $tan y = dfrac{3t}{1+4t^2}le dfrac{3t}{4t} = dfrac{3}{4}$, which is the max of $tan y$ with equality occurs when $t = dfrac{1}{2}$ or $2sin x = cos x$...the rest is simple...






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Put $t = tan x$, then $tan y = dfrac{3t}{1+4t^2}le dfrac{3t}{4t} = dfrac{3}{4}$, which is the max of $tan y$ with equality occurs when $t = dfrac{1}{2}$ or $2sin x = cos x$...the rest is simple...






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Put $t = tan x$, then $tan y = dfrac{3t}{1+4t^2}le dfrac{3t}{4t} = dfrac{3}{4}$, which is the max of $tan y$ with equality occurs when $t = dfrac{1}{2}$ or $2sin x = cos x$...the rest is simple...






          share|cite|improve this answer









          $endgroup$



          Put $t = tan x$, then $tan y = dfrac{3t}{1+4t^2}le dfrac{3t}{4t} = dfrac{3}{4}$, which is the max of $tan y$ with equality occurs when $t = dfrac{1}{2}$ or $2sin x = cos x$...the rest is simple...







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 17 at 6:54









          DeepSeaDeepSea

          71.5k54488




          71.5k54488























              0












              $begingroup$

              $$tan y=dfrac{3tan x}{1+4tan^2x}$$



              $$iff(4tan y)tan ^2x-3tan x+tan y=0$$ which is a Quadratic Equation $tan x$



              As $tan x$ is real, the discriminant must be $ge0$



              i.e., $$3^2-4(4tan y)ge0tan yifftan^2yledfrac9{16}iff-dfrac34letan yledfrac34$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $$tan y=dfrac{3tan x}{1+4tan^2x}$$



                $$iff(4tan y)tan ^2x-3tan x+tan y=0$$ which is a Quadratic Equation $tan x$



                As $tan x$ is real, the discriminant must be $ge0$



                i.e., $$3^2-4(4tan y)ge0tan yifftan^2yledfrac9{16}iff-dfrac34letan yledfrac34$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $$tan y=dfrac{3tan x}{1+4tan^2x}$$



                  $$iff(4tan y)tan ^2x-3tan x+tan y=0$$ which is a Quadratic Equation $tan x$



                  As $tan x$ is real, the discriminant must be $ge0$



                  i.e., $$3^2-4(4tan y)ge0tan yifftan^2yledfrac9{16}iff-dfrac34letan yledfrac34$$






                  share|cite|improve this answer









                  $endgroup$



                  $$tan y=dfrac{3tan x}{1+4tan^2x}$$



                  $$iff(4tan y)tan ^2x-3tan x+tan y=0$$ which is a Quadratic Equation $tan x$



                  As $tan x$ is real, the discriminant must be $ge0$



                  i.e., $$3^2-4(4tan y)ge0tan yifftan^2yledfrac9{16}iff-dfrac34letan yledfrac34$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 17 at 14:15









                  lab bhattacharjeelab bhattacharjee

                  228k15159279




                  228k15159279






























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