Prove that the sum of squares of sin and cos equals 1












1












$begingroup$


I need to prove $sum_{i=1}^n{{(r_i)}^2}=1$ when:



$r_{i}(theta)= cos(theta_{1}) $if $i=1$



$ cos(theta_{i})prod_{k=1}^{i-1} sin(theta_{k}) $if $2 leq i leq n-1$



$ sin(theta_{n-1})prod_{k=1}^{n-2} sin(theta_{k}) $if $i=n$



So my question is, how do I prove that:



$(cos(theta_1))^2+(cos(theta_2)sin(theta_1))^2+(cos(theta_3)sin(theta_1)sin(theta_2))^2+...+(cos(theta_{n-1})sin(theta_{1})...sin(theta_{n-2}))^2+(sin(theta_{1})...sin(theta_{n-1}))^2 =1$



It seems obvious to do something with $sin^2 x + cos^2 x = 1$, but I cannot understand what exactly I should do with it.






trigonometry







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$endgroup$








  • 1




    $begingroup$
    I'll try something like grouping 2 last terms, using trig identity, then having one term less, do it again and so on
    $endgroup$
    – Dominik Kutek
    Jan 5 at 22:26
















1












$begingroup$


I need to prove $sum_{i=1}^n{{(r_i)}^2}=1$ when:



$r_{i}(theta)= cos(theta_{1}) $if $i=1$



$ cos(theta_{i})prod_{k=1}^{i-1} sin(theta_{k}) $if $2 leq i leq n-1$



$ sin(theta_{n-1})prod_{k=1}^{n-2} sin(theta_{k}) $if $i=n$



So my question is, how do I prove that:



$(cos(theta_1))^2+(cos(theta_2)sin(theta_1))^2+(cos(theta_3)sin(theta_1)sin(theta_2))^2+...+(cos(theta_{n-1})sin(theta_{1})...sin(theta_{n-2}))^2+(sin(theta_{1})...sin(theta_{n-1}))^2 =1$



It seems obvious to do something with $sin^2 x + cos^2 x = 1$, but I cannot understand what exactly I should do with it.






trigonometry







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    I'll try something like grouping 2 last terms, using trig identity, then having one term less, do it again and so on
    $endgroup$
    – Dominik Kutek
    Jan 5 at 22:26














1












1








1





$begingroup$


I need to prove $sum_{i=1}^n{{(r_i)}^2}=1$ when:



$r_{i}(theta)= cos(theta_{1}) $if $i=1$



$ cos(theta_{i})prod_{k=1}^{i-1} sin(theta_{k}) $if $2 leq i leq n-1$



$ sin(theta_{n-1})prod_{k=1}^{n-2} sin(theta_{k}) $if $i=n$



So my question is, how do I prove that:



$(cos(theta_1))^2+(cos(theta_2)sin(theta_1))^2+(cos(theta_3)sin(theta_1)sin(theta_2))^2+...+(cos(theta_{n-1})sin(theta_{1})...sin(theta_{n-2}))^2+(sin(theta_{1})...sin(theta_{n-1}))^2 =1$



It seems obvious to do something with $sin^2 x + cos^2 x = 1$, but I cannot understand what exactly I should do with it.






trigonometry







share|cite|improve this question









$endgroup$




I need to prove $sum_{i=1}^n{{(r_i)}^2}=1$ when:



$r_{i}(theta)= cos(theta_{1}) $if $i=1$



$ cos(theta_{i})prod_{k=1}^{i-1} sin(theta_{k}) $if $2 leq i leq n-1$



$ sin(theta_{n-1})prod_{k=1}^{n-2} sin(theta_{k}) $if $i=n$



So my question is, how do I prove that:



$(cos(theta_1))^2+(cos(theta_2)sin(theta_1))^2+(cos(theta_3)sin(theta_1)sin(theta_2))^2+...+(cos(theta_{n-1})sin(theta_{1})...sin(theta_{n-2}))^2+(sin(theta_{1})...sin(theta_{n-1}))^2 =1$



It seems obvious to do something with $sin^2 x + cos^2 x = 1$, but I cannot understand what exactly I should do with it.






trigonometry




trigonometry






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










asked Jan 5 at 22:22









HannahHannah

82




82








  • 1




    $begingroup$
    I'll try something like grouping 2 last terms, using trig identity, then having one term less, do it again and so on
    $endgroup$
    – Dominik Kutek
    Jan 5 at 22:26














  • 1




    $begingroup$
    I'll try something like grouping 2 last terms, using trig identity, then having one term less, do it again and so on
    $endgroup$
    – Dominik Kutek
    Jan 5 at 22:26








1




1




$begingroup$
I'll try something like grouping 2 last terms, using trig identity, then having one term less, do it again and so on
$endgroup$
– Dominik Kutek
Jan 5 at 22:26




$begingroup$
I'll try something like grouping 2 last terms, using trig identity, then having one term less, do it again and so on
$endgroup$
– Dominik Kutek
Jan 5 at 22:26










2 Answers
2






active

oldest

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1












$begingroup$

Firsly, note that by considering $2$ last terms ( that is for $i=n-1$ and $i=n$):



$(cos(theta_{n-1})*prod_{k=1}^{n-2}sin(theta_k))^2 + (sin(theta_{n-1})*prod_{k=1}^{n-2}sin(theta_k))^2$ =



$(cos^2(theta_{n-1})+sin^2(theta_{n-1}))*(prod_{k=1}^{n-2}sin(theta_k))^2 $ =



$(prod_{k=1}^{n-2}sin(theta_k))^2$



Now we're left with $n-1$ terms ( the one we get after grouping last $2$ terms and first $n-2$ terms).



Continue analogously, combine the new term with $n-2$'th one:



$(prod_{k=1}^{n-2}sin(theta_k))^2$ + $(cos(theta_{n-2})prod_{k=1}^{n-3}sin(theta_k))^2$ =



$(sin(theta_{n-2})(prod_{k=1}^{n-3}sin(theta_k))^2$ + $(cos(theta_{n-2})prod_{k=1}^{n-3}sin(theta_k))^2$ =
$(sin^2(theta_{n-2})+cos^2(theta_{n-2}))(prod_{k=1}^{n-3}sin(theta_k))^2$ =



=$(prod_{k=1}^{n-3}sin(theta_k))^2$



By continuing so (until the second term, cause it is of the same nature as every ${2,3,...,n-1}$, we get:
first term + created term ( which is $sin^2(theta_{n-1-(n-2)})$ ), so:



$cos^2(theta_1) + sin^2(theta_1) = 1 $



$1 = 1 $



Q.E.D






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    I think this is not too hard to prove inductively. The $n=2$ case is precisely $sin^2 + cos^2 = 1$, so we will use this as base case. Let's assume the statement holds for $n$ and prove it for $n+1$. We denote with $r^(n+1)_k$ and $r^(n)_k$ the $r_k$ from the cases $n$ and $n+1$ respectively.
    $$begin{align}
    (r^{(n+1)}_{n+1})^2 + (r^{(n+1)}_{n})^2 &= sin(theta_{n})^2 left( prod_{k=1}^{n+1-2} sin(theta_k) right)^2 + cos(theta_{n})^2 left( prod_{k=1}^{n+1-2} sin(theta_k) right)^2 \
    &= left( prod_{k=1}^{n-1} sin(theta_k) right)^2 = (r^{(n)}_n)^2
    end{align}$$



    Now note that for $k leq n$: $r^{(n+1)}_k = r^{(n)}_k$ and therefore
    $$(r^{(n+1)}_1)^2 + dots (r^{(n+1)}_n)^2 + (r^{(n+1)}_{n+1})^2 = (r^{(n)}_1)^2 + dots + (r^{(n)}_n)^2 = 1$$
    by induction hypothesis.






    share|cite|improve this answer









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      2 Answers
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      2 Answers
      2






      active

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      active

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      active

      oldest

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      1












      $begingroup$

      Firsly, note that by considering $2$ last terms ( that is for $i=n-1$ and $i=n$):



      $(cos(theta_{n-1})*prod_{k=1}^{n-2}sin(theta_k))^2 + (sin(theta_{n-1})*prod_{k=1}^{n-2}sin(theta_k))^2$ =



      $(cos^2(theta_{n-1})+sin^2(theta_{n-1}))*(prod_{k=1}^{n-2}sin(theta_k))^2 $ =



      $(prod_{k=1}^{n-2}sin(theta_k))^2$



      Now we're left with $n-1$ terms ( the one we get after grouping last $2$ terms and first $n-2$ terms).



      Continue analogously, combine the new term with $n-2$'th one:



      $(prod_{k=1}^{n-2}sin(theta_k))^2$ + $(cos(theta_{n-2})prod_{k=1}^{n-3}sin(theta_k))^2$ =



      $(sin(theta_{n-2})(prod_{k=1}^{n-3}sin(theta_k))^2$ + $(cos(theta_{n-2})prod_{k=1}^{n-3}sin(theta_k))^2$ =
      $(sin^2(theta_{n-2})+cos^2(theta_{n-2}))(prod_{k=1}^{n-3}sin(theta_k))^2$ =



      =$(prod_{k=1}^{n-3}sin(theta_k))^2$



      By continuing so (until the second term, cause it is of the same nature as every ${2,3,...,n-1}$, we get:
      first term + created term ( which is $sin^2(theta_{n-1-(n-2)})$ ), so:



      $cos^2(theta_1) + sin^2(theta_1) = 1 $



      $1 = 1 $



      Q.E.D






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Firsly, note that by considering $2$ last terms ( that is for $i=n-1$ and $i=n$):



        $(cos(theta_{n-1})*prod_{k=1}^{n-2}sin(theta_k))^2 + (sin(theta_{n-1})*prod_{k=1}^{n-2}sin(theta_k))^2$ =



        $(cos^2(theta_{n-1})+sin^2(theta_{n-1}))*(prod_{k=1}^{n-2}sin(theta_k))^2 $ =



        $(prod_{k=1}^{n-2}sin(theta_k))^2$



        Now we're left with $n-1$ terms ( the one we get after grouping last $2$ terms and first $n-2$ terms).



        Continue analogously, combine the new term with $n-2$'th one:



        $(prod_{k=1}^{n-2}sin(theta_k))^2$ + $(cos(theta_{n-2})prod_{k=1}^{n-3}sin(theta_k))^2$ =



        $(sin(theta_{n-2})(prod_{k=1}^{n-3}sin(theta_k))^2$ + $(cos(theta_{n-2})prod_{k=1}^{n-3}sin(theta_k))^2$ =
        $(sin^2(theta_{n-2})+cos^2(theta_{n-2}))(prod_{k=1}^{n-3}sin(theta_k))^2$ =



        =$(prod_{k=1}^{n-3}sin(theta_k))^2$



        By continuing so (until the second term, cause it is of the same nature as every ${2,3,...,n-1}$, we get:
        first term + created term ( which is $sin^2(theta_{n-1-(n-2)})$ ), so:



        $cos^2(theta_1) + sin^2(theta_1) = 1 $



        $1 = 1 $



        Q.E.D






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Firsly, note that by considering $2$ last terms ( that is for $i=n-1$ and $i=n$):



          $(cos(theta_{n-1})*prod_{k=1}^{n-2}sin(theta_k))^2 + (sin(theta_{n-1})*prod_{k=1}^{n-2}sin(theta_k))^2$ =



          $(cos^2(theta_{n-1})+sin^2(theta_{n-1}))*(prod_{k=1}^{n-2}sin(theta_k))^2 $ =



          $(prod_{k=1}^{n-2}sin(theta_k))^2$



          Now we're left with $n-1$ terms ( the one we get after grouping last $2$ terms and first $n-2$ terms).



          Continue analogously, combine the new term with $n-2$'th one:



          $(prod_{k=1}^{n-2}sin(theta_k))^2$ + $(cos(theta_{n-2})prod_{k=1}^{n-3}sin(theta_k))^2$ =



          $(sin(theta_{n-2})(prod_{k=1}^{n-3}sin(theta_k))^2$ + $(cos(theta_{n-2})prod_{k=1}^{n-3}sin(theta_k))^2$ =
          $(sin^2(theta_{n-2})+cos^2(theta_{n-2}))(prod_{k=1}^{n-3}sin(theta_k))^2$ =



          =$(prod_{k=1}^{n-3}sin(theta_k))^2$



          By continuing so (until the second term, cause it is of the same nature as every ${2,3,...,n-1}$, we get:
          first term + created term ( which is $sin^2(theta_{n-1-(n-2)})$ ), so:



          $cos^2(theta_1) + sin^2(theta_1) = 1 $



          $1 = 1 $



          Q.E.D






          share|cite|improve this answer









          $endgroup$



          Firsly, note that by considering $2$ last terms ( that is for $i=n-1$ and $i=n$):



          $(cos(theta_{n-1})*prod_{k=1}^{n-2}sin(theta_k))^2 + (sin(theta_{n-1})*prod_{k=1}^{n-2}sin(theta_k))^2$ =



          $(cos^2(theta_{n-1})+sin^2(theta_{n-1}))*(prod_{k=1}^{n-2}sin(theta_k))^2 $ =



          $(prod_{k=1}^{n-2}sin(theta_k))^2$



          Now we're left with $n-1$ terms ( the one we get after grouping last $2$ terms and first $n-2$ terms).



          Continue analogously, combine the new term with $n-2$'th one:



          $(prod_{k=1}^{n-2}sin(theta_k))^2$ + $(cos(theta_{n-2})prod_{k=1}^{n-3}sin(theta_k))^2$ =



          $(sin(theta_{n-2})(prod_{k=1}^{n-3}sin(theta_k))^2$ + $(cos(theta_{n-2})prod_{k=1}^{n-3}sin(theta_k))^2$ =
          $(sin^2(theta_{n-2})+cos^2(theta_{n-2}))(prod_{k=1}^{n-3}sin(theta_k))^2$ =



          =$(prod_{k=1}^{n-3}sin(theta_k))^2$



          By continuing so (until the second term, cause it is of the same nature as every ${2,3,...,n-1}$, we get:
          first term + created term ( which is $sin^2(theta_{n-1-(n-2)})$ ), so:



          $cos^2(theta_1) + sin^2(theta_1) = 1 $



          $1 = 1 $



          Q.E.D







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 5 at 22:44









          Dominik KutekDominik Kutek

          3345




          3345























              1












              $begingroup$

              I think this is not too hard to prove inductively. The $n=2$ case is precisely $sin^2 + cos^2 = 1$, so we will use this as base case. Let's assume the statement holds for $n$ and prove it for $n+1$. We denote with $r^(n+1)_k$ and $r^(n)_k$ the $r_k$ from the cases $n$ and $n+1$ respectively.
              $$begin{align}
              (r^{(n+1)}_{n+1})^2 + (r^{(n+1)}_{n})^2 &= sin(theta_{n})^2 left( prod_{k=1}^{n+1-2} sin(theta_k) right)^2 + cos(theta_{n})^2 left( prod_{k=1}^{n+1-2} sin(theta_k) right)^2 \
              &= left( prod_{k=1}^{n-1} sin(theta_k) right)^2 = (r^{(n)}_n)^2
              end{align}$$



              Now note that for $k leq n$: $r^{(n+1)}_k = r^{(n)}_k$ and therefore
              $$(r^{(n+1)}_1)^2 + dots (r^{(n+1)}_n)^2 + (r^{(n+1)}_{n+1})^2 = (r^{(n)}_1)^2 + dots + (r^{(n)}_n)^2 = 1$$
              by induction hypothesis.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                I think this is not too hard to prove inductively. The $n=2$ case is precisely $sin^2 + cos^2 = 1$, so we will use this as base case. Let's assume the statement holds for $n$ and prove it for $n+1$. We denote with $r^(n+1)_k$ and $r^(n)_k$ the $r_k$ from the cases $n$ and $n+1$ respectively.
                $$begin{align}
                (r^{(n+1)}_{n+1})^2 + (r^{(n+1)}_{n})^2 &= sin(theta_{n})^2 left( prod_{k=1}^{n+1-2} sin(theta_k) right)^2 + cos(theta_{n})^2 left( prod_{k=1}^{n+1-2} sin(theta_k) right)^2 \
                &= left( prod_{k=1}^{n-1} sin(theta_k) right)^2 = (r^{(n)}_n)^2
                end{align}$$



                Now note that for $k leq n$: $r^{(n+1)}_k = r^{(n)}_k$ and therefore
                $$(r^{(n+1)}_1)^2 + dots (r^{(n+1)}_n)^2 + (r^{(n+1)}_{n+1})^2 = (r^{(n)}_1)^2 + dots + (r^{(n)}_n)^2 = 1$$
                by induction hypothesis.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  I think this is not too hard to prove inductively. The $n=2$ case is precisely $sin^2 + cos^2 = 1$, so we will use this as base case. Let's assume the statement holds for $n$ and prove it for $n+1$. We denote with $r^(n+1)_k$ and $r^(n)_k$ the $r_k$ from the cases $n$ and $n+1$ respectively.
                  $$begin{align}
                  (r^{(n+1)}_{n+1})^2 + (r^{(n+1)}_{n})^2 &= sin(theta_{n})^2 left( prod_{k=1}^{n+1-2} sin(theta_k) right)^2 + cos(theta_{n})^2 left( prod_{k=1}^{n+1-2} sin(theta_k) right)^2 \
                  &= left( prod_{k=1}^{n-1} sin(theta_k) right)^2 = (r^{(n)}_n)^2
                  end{align}$$



                  Now note that for $k leq n$: $r^{(n+1)}_k = r^{(n)}_k$ and therefore
                  $$(r^{(n+1)}_1)^2 + dots (r^{(n+1)}_n)^2 + (r^{(n+1)}_{n+1})^2 = (r^{(n)}_1)^2 + dots + (r^{(n)}_n)^2 = 1$$
                  by induction hypothesis.






                  share|cite|improve this answer









                  $endgroup$



                  I think this is not too hard to prove inductively. The $n=2$ case is precisely $sin^2 + cos^2 = 1$, so we will use this as base case. Let's assume the statement holds for $n$ and prove it for $n+1$. We denote with $r^(n+1)_k$ and $r^(n)_k$ the $r_k$ from the cases $n$ and $n+1$ respectively.
                  $$begin{align}
                  (r^{(n+1)}_{n+1})^2 + (r^{(n+1)}_{n})^2 &= sin(theta_{n})^2 left( prod_{k=1}^{n+1-2} sin(theta_k) right)^2 + cos(theta_{n})^2 left( prod_{k=1}^{n+1-2} sin(theta_k) right)^2 \
                  &= left( prod_{k=1}^{n-1} sin(theta_k) right)^2 = (r^{(n)}_n)^2
                  end{align}$$



                  Now note that for $k leq n$: $r^{(n+1)}_k = r^{(n)}_k$ and therefore
                  $$(r^{(n+1)}_1)^2 + dots (r^{(n+1)}_n)^2 + (r^{(n+1)}_{n+1})^2 = (r^{(n)}_1)^2 + dots + (r^{(n)}_n)^2 = 1$$
                  by induction hypothesis.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 5 at 22:49









                  0x5390x539

                  1,445518




                  1,445518






























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                      .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
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