Calculate $sinfrac{pi}{16}$ from given trigonometric identities












1












$begingroup$


There is the choice between four trigonometric identities
$$sin(4phi)=8cos^{3}(phi)sin(phi)-4cos(phi)sin(phi)$$
$$cos(4phi)=8cos^{4}(phi) -8cos^{2}(phi)+1$$
$$sin^{4}(phi)=1/8(cos(4phi) -4cos(2phi)+3)$$
$$cos^{4}(phi)=1/8(cos(4phi)+4cos(2phi)+3)$$
to calculate $sinfrac{pi}{16}$. When using the first I got stuck trying to eliminate cos expressions and bring $sin(4phi)$ and $sin(phi)$ on different sides of the equation, as in
$$sin(4phi)=sin (2phi)(2-4sin^{2}(phi))$$
What way to go ?






trigonometry







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$endgroup$












  • $begingroup$
    Fun related reading: J. Matthew H. Peters, Some trigonometric ratios in surd form, Mathematical Gazette 66 #438 (December 1982), 296-299 AND Ron Knott's web page Exact Trigonometric Function Values.
    $endgroup$
    – Dave L. Renfro
    Dec 11 '18 at 17:49










  • $begingroup$
    Just use three times the bisection formulas $sinfrac{x}{2}=sqrt{frac{1-cos x}{2}}$, $cosfrac{x}{2}=sqrt{frac{1+cos x}{2}}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 11 '18 at 18:52
















1












$begingroup$


There is the choice between four trigonometric identities
$$sin(4phi)=8cos^{3}(phi)sin(phi)-4cos(phi)sin(phi)$$
$$cos(4phi)=8cos^{4}(phi) -8cos^{2}(phi)+1$$
$$sin^{4}(phi)=1/8(cos(4phi) -4cos(2phi)+3)$$
$$cos^{4}(phi)=1/8(cos(4phi)+4cos(2phi)+3)$$
to calculate $sinfrac{pi}{16}$. When using the first I got stuck trying to eliminate cos expressions and bring $sin(4phi)$ and $sin(phi)$ on different sides of the equation, as in
$$sin(4phi)=sin (2phi)(2-4sin^{2}(phi))$$
What way to go ?






trigonometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    Fun related reading: J. Matthew H. Peters, Some trigonometric ratios in surd form, Mathematical Gazette 66 #438 (December 1982), 296-299 AND Ron Knott's web page Exact Trigonometric Function Values.
    $endgroup$
    – Dave L. Renfro
    Dec 11 '18 at 17:49










  • $begingroup$
    Just use three times the bisection formulas $sinfrac{x}{2}=sqrt{frac{1-cos x}{2}}$, $cosfrac{x}{2}=sqrt{frac{1+cos x}{2}}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 11 '18 at 18:52














1












1








1





$begingroup$


There is the choice between four trigonometric identities
$$sin(4phi)=8cos^{3}(phi)sin(phi)-4cos(phi)sin(phi)$$
$$cos(4phi)=8cos^{4}(phi) -8cos^{2}(phi)+1$$
$$sin^{4}(phi)=1/8(cos(4phi) -4cos(2phi)+3)$$
$$cos^{4}(phi)=1/8(cos(4phi)+4cos(2phi)+3)$$
to calculate $sinfrac{pi}{16}$. When using the first I got stuck trying to eliminate cos expressions and bring $sin(4phi)$ and $sin(phi)$ on different sides of the equation, as in
$$sin(4phi)=sin (2phi)(2-4sin^{2}(phi))$$
What way to go ?






trigonometry







share|cite|improve this question









$endgroup$




There is the choice between four trigonometric identities
$$sin(4phi)=8cos^{3}(phi)sin(phi)-4cos(phi)sin(phi)$$
$$cos(4phi)=8cos^{4}(phi) -8cos^{2}(phi)+1$$
$$sin^{4}(phi)=1/8(cos(4phi) -4cos(2phi)+3)$$
$$cos^{4}(phi)=1/8(cos(4phi)+4cos(2phi)+3)$$
to calculate $sinfrac{pi}{16}$. When using the first I got stuck trying to eliminate cos expressions and bring $sin(4phi)$ and $sin(phi)$ on different sides of the equation, as in
$$sin(4phi)=sin (2phi)(2-4sin^{2}(phi))$$
What way to go ?






trigonometry




trigonometry






share|cite|improve this question













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asked Dec 11 '18 at 17:25









B. HauffaB. Hauffa

62




62












  • $begingroup$
    Fun related reading: J. Matthew H. Peters, Some trigonometric ratios in surd form, Mathematical Gazette 66 #438 (December 1982), 296-299 AND Ron Knott's web page Exact Trigonometric Function Values.
    $endgroup$
    – Dave L. Renfro
    Dec 11 '18 at 17:49










  • $begingroup$
    Just use three times the bisection formulas $sinfrac{x}{2}=sqrt{frac{1-cos x}{2}}$, $cosfrac{x}{2}=sqrt{frac{1+cos x}{2}}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 11 '18 at 18:52


















  • $begingroup$
    Fun related reading: J. Matthew H. Peters, Some trigonometric ratios in surd form, Mathematical Gazette 66 #438 (December 1982), 296-299 AND Ron Knott's web page Exact Trigonometric Function Values.
    $endgroup$
    – Dave L. Renfro
    Dec 11 '18 at 17:49










  • $begingroup$
    Just use three times the bisection formulas $sinfrac{x}{2}=sqrt{frac{1-cos x}{2}}$, $cosfrac{x}{2}=sqrt{frac{1+cos x}{2}}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 11 '18 at 18:52
















$begingroup$
Fun related reading: J. Matthew H. Peters, Some trigonometric ratios in surd form, Mathematical Gazette 66 #438 (December 1982), 296-299 AND Ron Knott's web page Exact Trigonometric Function Values.
$endgroup$
– Dave L. Renfro
Dec 11 '18 at 17:49




$begingroup$
Fun related reading: J. Matthew H. Peters, Some trigonometric ratios in surd form, Mathematical Gazette 66 #438 (December 1982), 296-299 AND Ron Knott's web page Exact Trigonometric Function Values.
$endgroup$
– Dave L. Renfro
Dec 11 '18 at 17:49












$begingroup$
Just use three times the bisection formulas $sinfrac{x}{2}=sqrt{frac{1-cos x}{2}}$, $cosfrac{x}{2}=sqrt{frac{1+cos x}{2}}$.
$endgroup$
– Jack D'Aurizio
Dec 11 '18 at 18:52




$begingroup$
Just use three times the bisection formulas $sinfrac{x}{2}=sqrt{frac{1-cos x}{2}}$, $cosfrac{x}{2}=sqrt{frac{1+cos x}{2}}$.
$endgroup$
– Jack D'Aurizio
Dec 11 '18 at 18:52










2 Answers
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$begingroup$

Hint: Note that $cos(pi/4) = dfrac{1}{sqrt{2}}$, Use the second one of your list and put $y = cos^2(pi/16)$ to get a quadratic equation in $y$, and then use the identity $sin(pi/16) = sqrt{1-cos^2(pi/16)}= sqrt{1-y}$ to finish.






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$endgroup$





















    0












    $begingroup$

    Use the second identity:



    $$ 8cos^4left(frac{pi}{16}right) - 8cos^2left(frac{pi}{16}right) + 1 = cosleft(frac{pi}{4}right) = frac{sqrt2}{2} $$



    Let $x = cos^2left(frac{pi}{16}right)$ then



    $$ 8x^2-8x = frac{sqrt{2}-2}{2} $$



    Completing the square:
    $$ 4x^2 - 4x + 1 = frac{sqrt{2}-2}{4}+1 = frac{2+sqrt{2}}{4} $$



    $$ (2x-1)^2 = frac{2+sqrt{2}}{4} $$



    Note that $2x-1 = 2cos^2left(frac{pi}{16}right)-1 = cos left(frac{pi}{8}right) > 0$ so we take the positive root



    $$ 2x-1 = frac{sqrt{2+sqrt{2}}}{2} $$



    $$ x = frac{2+sqrt{2+sqrt{2}}}{4} $$



    Then we have



    $$ sinleft(frac{pi}{16}right) = sqrt{1-x} = frac{sqrt{2-sqrt{2+sqrt{2}}}}{2} $$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






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      2 Answers
      2






      active

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      active

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      active

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      1












      $begingroup$

      Hint: Note that $cos(pi/4) = dfrac{1}{sqrt{2}}$, Use the second one of your list and put $y = cos^2(pi/16)$ to get a quadratic equation in $y$, and then use the identity $sin(pi/16) = sqrt{1-cos^2(pi/16)}= sqrt{1-y}$ to finish.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        Hint: Note that $cos(pi/4) = dfrac{1}{sqrt{2}}$, Use the second one of your list and put $y = cos^2(pi/16)$ to get a quadratic equation in $y$, and then use the identity $sin(pi/16) = sqrt{1-cos^2(pi/16)}= sqrt{1-y}$ to finish.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          Hint: Note that $cos(pi/4) = dfrac{1}{sqrt{2}}$, Use the second one of your list and put $y = cos^2(pi/16)$ to get a quadratic equation in $y$, and then use the identity $sin(pi/16) = sqrt{1-cos^2(pi/16)}= sqrt{1-y}$ to finish.






          share|cite|improve this answer











          $endgroup$



          Hint: Note that $cos(pi/4) = dfrac{1}{sqrt{2}}$, Use the second one of your list and put $y = cos^2(pi/16)$ to get a quadratic equation in $y$, and then use the identity $sin(pi/16) = sqrt{1-cos^2(pi/16)}= sqrt{1-y}$ to finish.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 11 '18 at 17:36

























          answered Dec 11 '18 at 17:31









          DeepSeaDeepSea

          71.2k54487




          71.2k54487























              0












              $begingroup$

              Use the second identity:



              $$ 8cos^4left(frac{pi}{16}right) - 8cos^2left(frac{pi}{16}right) + 1 = cosleft(frac{pi}{4}right) = frac{sqrt2}{2} $$



              Let $x = cos^2left(frac{pi}{16}right)$ then



              $$ 8x^2-8x = frac{sqrt{2}-2}{2} $$



              Completing the square:
              $$ 4x^2 - 4x + 1 = frac{sqrt{2}-2}{4}+1 = frac{2+sqrt{2}}{4} $$



              $$ (2x-1)^2 = frac{2+sqrt{2}}{4} $$



              Note that $2x-1 = 2cos^2left(frac{pi}{16}right)-1 = cos left(frac{pi}{8}right) > 0$ so we take the positive root



              $$ 2x-1 = frac{sqrt{2+sqrt{2}}}{2} $$



              $$ x = frac{2+sqrt{2+sqrt{2}}}{4} $$



              Then we have



              $$ sinleft(frac{pi}{16}right) = sqrt{1-x} = frac{sqrt{2-sqrt{2+sqrt{2}}}}{2} $$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Use the second identity:



                $$ 8cos^4left(frac{pi}{16}right) - 8cos^2left(frac{pi}{16}right) + 1 = cosleft(frac{pi}{4}right) = frac{sqrt2}{2} $$



                Let $x = cos^2left(frac{pi}{16}right)$ then



                $$ 8x^2-8x = frac{sqrt{2}-2}{2} $$



                Completing the square:
                $$ 4x^2 - 4x + 1 = frac{sqrt{2}-2}{4}+1 = frac{2+sqrt{2}}{4} $$



                $$ (2x-1)^2 = frac{2+sqrt{2}}{4} $$



                Note that $2x-1 = 2cos^2left(frac{pi}{16}right)-1 = cos left(frac{pi}{8}right) > 0$ so we take the positive root



                $$ 2x-1 = frac{sqrt{2+sqrt{2}}}{2} $$



                $$ x = frac{2+sqrt{2+sqrt{2}}}{4} $$



                Then we have



                $$ sinleft(frac{pi}{16}right) = sqrt{1-x} = frac{sqrt{2-sqrt{2+sqrt{2}}}}{2} $$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Use the second identity:



                  $$ 8cos^4left(frac{pi}{16}right) - 8cos^2left(frac{pi}{16}right) + 1 = cosleft(frac{pi}{4}right) = frac{sqrt2}{2} $$



                  Let $x = cos^2left(frac{pi}{16}right)$ then



                  $$ 8x^2-8x = frac{sqrt{2}-2}{2} $$



                  Completing the square:
                  $$ 4x^2 - 4x + 1 = frac{sqrt{2}-2}{4}+1 = frac{2+sqrt{2}}{4} $$



                  $$ (2x-1)^2 = frac{2+sqrt{2}}{4} $$



                  Note that $2x-1 = 2cos^2left(frac{pi}{16}right)-1 = cos left(frac{pi}{8}right) > 0$ so we take the positive root



                  $$ 2x-1 = frac{sqrt{2+sqrt{2}}}{2} $$



                  $$ x = frac{2+sqrt{2+sqrt{2}}}{4} $$



                  Then we have



                  $$ sinleft(frac{pi}{16}right) = sqrt{1-x} = frac{sqrt{2-sqrt{2+sqrt{2}}}}{2} $$






                  share|cite|improve this answer









                  $endgroup$



                  Use the second identity:



                  $$ 8cos^4left(frac{pi}{16}right) - 8cos^2left(frac{pi}{16}right) + 1 = cosleft(frac{pi}{4}right) = frac{sqrt2}{2} $$



                  Let $x = cos^2left(frac{pi}{16}right)$ then



                  $$ 8x^2-8x = frac{sqrt{2}-2}{2} $$



                  Completing the square:
                  $$ 4x^2 - 4x + 1 = frac{sqrt{2}-2}{4}+1 = frac{2+sqrt{2}}{4} $$



                  $$ (2x-1)^2 = frac{2+sqrt{2}}{4} $$



                  Note that $2x-1 = 2cos^2left(frac{pi}{16}right)-1 = cos left(frac{pi}{8}right) > 0$ so we take the positive root



                  $$ 2x-1 = frac{sqrt{2+sqrt{2}}}{2} $$



                  $$ x = frac{2+sqrt{2+sqrt{2}}}{4} $$



                  Then we have



                  $$ sinleft(frac{pi}{16}right) = sqrt{1-x} = frac{sqrt{2-sqrt{2+sqrt{2}}}}{2} $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 13 '18 at 9:35









                  DylanDylan

                  12.9k31027




                  12.9k31027






























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