Rewrite $sec^2 (x) cdot sin^2 (x) $in terms of $cos (x)$












-2












$begingroup$


Write $sec^2 (x) cdot sin^2 (x)$ in terms of $cos (x)$.



I was not able to answer this question on a pre-calculus exam and my professor will not explain to me how to answer it; I wish to learn.



Here is what I did on the test:



$sec^2 (x) cdot sin^2 (x)$ = $frac{sqrt{1}}{sqrt{cos^2 (x)}} cdot sqrt{1-cos^2 (x)} = frac{1}{cos (x)} cdot frac{pmsqrt{1-cos^2}}{1} = frac{pmsqrt{1-cos^2}}{cos (x)}$



What did I do wrong here?



Thank you in advance.






algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    @WesleyStrik You seem to have lost a minus sign in the last step of your comment.
    $endgroup$
    – Andreas Blass
    Dec 17 '18 at 3:36










  • $begingroup$
    I got lost on the last step. Is there a name for this kind of problem? Maybe I can watch another example and try to apply the concept to this problem. Thank you.
    $endgroup$
    – anonstudent
    Dec 17 '18 at 4:41










  • $begingroup$
    The trick is literally to first express both of these in terms of cosines and then to simply take the product.
    $endgroup$
    – Wesley Strik
    Dec 17 '18 at 9:14










  • $begingroup$
    I cannot answer your question because some guy closed your question.
    $endgroup$
    – Wesley Strik
    Dec 17 '18 at 9:15






  • 1




    $begingroup$
    Because it says that I should put it in terms of cos(x), is it okay then that it is still squared? Do I have to square root?
    $endgroup$
    – anonstudent
    Dec 19 '18 at 6:09
















-2












$begingroup$


Write $sec^2 (x) cdot sin^2 (x)$ in terms of $cos (x)$.



I was not able to answer this question on a pre-calculus exam and my professor will not explain to me how to answer it; I wish to learn.



Here is what I did on the test:



$sec^2 (x) cdot sin^2 (x)$ = $frac{sqrt{1}}{sqrt{cos^2 (x)}} cdot sqrt{1-cos^2 (x)} = frac{1}{cos (x)} cdot frac{pmsqrt{1-cos^2}}{1} = frac{pmsqrt{1-cos^2}}{cos (x)}$



What did I do wrong here?



Thank you in advance.






algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    @WesleyStrik You seem to have lost a minus sign in the last step of your comment.
    $endgroup$
    – Andreas Blass
    Dec 17 '18 at 3:36










  • $begingroup$
    I got lost on the last step. Is there a name for this kind of problem? Maybe I can watch another example and try to apply the concept to this problem. Thank you.
    $endgroup$
    – anonstudent
    Dec 17 '18 at 4:41










  • $begingroup$
    The trick is literally to first express both of these in terms of cosines and then to simply take the product.
    $endgroup$
    – Wesley Strik
    Dec 17 '18 at 9:14










  • $begingroup$
    I cannot answer your question because some guy closed your question.
    $endgroup$
    – Wesley Strik
    Dec 17 '18 at 9:15






  • 1




    $begingroup$
    Because it says that I should put it in terms of cos(x), is it okay then that it is still squared? Do I have to square root?
    $endgroup$
    – anonstudent
    Dec 19 '18 at 6:09














-2












-2








-2


0



$begingroup$


Write $sec^2 (x) cdot sin^2 (x)$ in terms of $cos (x)$.



I was not able to answer this question on a pre-calculus exam and my professor will not explain to me how to answer it; I wish to learn.



Here is what I did on the test:



$sec^2 (x) cdot sin^2 (x)$ = $frac{sqrt{1}}{sqrt{cos^2 (x)}} cdot sqrt{1-cos^2 (x)} = frac{1}{cos (x)} cdot frac{pmsqrt{1-cos^2}}{1} = frac{pmsqrt{1-cos^2}}{cos (x)}$



What did I do wrong here?



Thank you in advance.






algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$




Write $sec^2 (x) cdot sin^2 (x)$ in terms of $cos (x)$.



I was not able to answer this question on a pre-calculus exam and my professor will not explain to me how to answer it; I wish to learn.



Here is what I did on the test:



$sec^2 (x) cdot sin^2 (x)$ = $frac{sqrt{1}}{sqrt{cos^2 (x)}} cdot sqrt{1-cos^2 (x)} = frac{1}{cos (x)} cdot frac{pmsqrt{1-cos^2}}{1} = frac{pmsqrt{1-cos^2}}{cos (x)}$



What did I do wrong here?



Thank you in advance.






algebra-precalculus trigonometry




algebra-precalculus trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 9:41









Wesley Strik

2,017423




2,017423










asked Dec 17 '18 at 0:04









anonstudentanonstudent

62




62








  • 1




    $begingroup$
    @WesleyStrik You seem to have lost a minus sign in the last step of your comment.
    $endgroup$
    – Andreas Blass
    Dec 17 '18 at 3:36










  • $begingroup$
    I got lost on the last step. Is there a name for this kind of problem? Maybe I can watch another example and try to apply the concept to this problem. Thank you.
    $endgroup$
    – anonstudent
    Dec 17 '18 at 4:41










  • $begingroup$
    The trick is literally to first express both of these in terms of cosines and then to simply take the product.
    $endgroup$
    – Wesley Strik
    Dec 17 '18 at 9:14










  • $begingroup$
    I cannot answer your question because some guy closed your question.
    $endgroup$
    – Wesley Strik
    Dec 17 '18 at 9:15






  • 1




    $begingroup$
    Because it says that I should put it in terms of cos(x), is it okay then that it is still squared? Do I have to square root?
    $endgroup$
    – anonstudent
    Dec 19 '18 at 6:09














  • 1




    $begingroup$
    @WesleyStrik You seem to have lost a minus sign in the last step of your comment.
    $endgroup$
    – Andreas Blass
    Dec 17 '18 at 3:36










  • $begingroup$
    I got lost on the last step. Is there a name for this kind of problem? Maybe I can watch another example and try to apply the concept to this problem. Thank you.
    $endgroup$
    – anonstudent
    Dec 17 '18 at 4:41










  • $begingroup$
    The trick is literally to first express both of these in terms of cosines and then to simply take the product.
    $endgroup$
    – Wesley Strik
    Dec 17 '18 at 9:14










  • $begingroup$
    I cannot answer your question because some guy closed your question.
    $endgroup$
    – Wesley Strik
    Dec 17 '18 at 9:15






  • 1




    $begingroup$
    Because it says that I should put it in terms of cos(x), is it okay then that it is still squared? Do I have to square root?
    $endgroup$
    – anonstudent
    Dec 19 '18 at 6:09








1




1




$begingroup$
@WesleyStrik You seem to have lost a minus sign in the last step of your comment.
$endgroup$
– Andreas Blass
Dec 17 '18 at 3:36




$begingroup$
@WesleyStrik You seem to have lost a minus sign in the last step of your comment.
$endgroup$
– Andreas Blass
Dec 17 '18 at 3:36












$begingroup$
I got lost on the last step. Is there a name for this kind of problem? Maybe I can watch another example and try to apply the concept to this problem. Thank you.
$endgroup$
– anonstudent
Dec 17 '18 at 4:41




$begingroup$
I got lost on the last step. Is there a name for this kind of problem? Maybe I can watch another example and try to apply the concept to this problem. Thank you.
$endgroup$
– anonstudent
Dec 17 '18 at 4:41












$begingroup$
The trick is literally to first express both of these in terms of cosines and then to simply take the product.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:14




$begingroup$
The trick is literally to first express both of these in terms of cosines and then to simply take the product.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:14












$begingroup$
I cannot answer your question because some guy closed your question.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:15




$begingroup$
I cannot answer your question because some guy closed your question.
$endgroup$
– Wesley Strik
Dec 17 '18 at 9:15




1




1




$begingroup$
Because it says that I should put it in terms of cos(x), is it okay then that it is still squared? Do I have to square root?
$endgroup$
– anonstudent
Dec 19 '18 at 6:09




$begingroup$
Because it says that I should put it in terms of cos(x), is it okay then that it is still squared? Do I have to square root?
$endgroup$
– anonstudent
Dec 19 '18 at 6:09










3 Answers
3






active

oldest

votes


















0












$begingroup$

$sec x = frac{1}{cos x} \ sin^2x = 1 - cos^2x$



Put it all together.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Notice that we have that:
    $$sec^2 x = frac{1}{cos^2 x} $$ $$ sin^2x = 1 - cos^2x$$



    So if we multiply the two and use distributivity:
    $$(sec^2 x) cdot(sin^2x )=(frac{1}{cos^2 x}) cdot(1 - cos^2x)=frac{1}{cos^2 x} -frac{cos^2 x}{cos^2 x}= frac{1}{cos^2 x} -1$$



    Notice now that we expressed this in terms of $cos(x)$, we now have a function in terms of $cos(x)$, namely:
    $$ f(cos x )=frac{1}{cos^2 x} -1$$
    Or if we let $z=cos(x)$
    $$ f(z )=frac{1}{z^2} -1$$
    would you not say that this is expressed in terms of $z$ $(=cos(x))$?






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      $z^2$ in the denominator in the last step.
      $endgroup$
      – Deepak
      Dec 20 '18 at 17:34



















    0












    $begingroup$

    $sec^2(x)bullet sin^2(x)={1over{cos^2(x)}}left(1-cos^2(x)right)={1over{cos^2(x)}}-1$






    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









      0












      $begingroup$

      $sec x = frac{1}{cos x} \ sin^2x = 1 - cos^2x$



      Put it all together.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        $sec x = frac{1}{cos x} \ sin^2x = 1 - cos^2x$



        Put it all together.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          $sec x = frac{1}{cos x} \ sin^2x = 1 - cos^2x$



          Put it all together.






          share|cite|improve this answer









          $endgroup$



          $sec x = frac{1}{cos x} \ sin^2x = 1 - cos^2x$



          Put it all together.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 17 '18 at 0:06









          Lucas HenriqueLucas Henrique

          1,059414




          1,059414























              0












              $begingroup$

              Notice that we have that:
              $$sec^2 x = frac{1}{cos^2 x} $$ $$ sin^2x = 1 - cos^2x$$



              So if we multiply the two and use distributivity:
              $$(sec^2 x) cdot(sin^2x )=(frac{1}{cos^2 x}) cdot(1 - cos^2x)=frac{1}{cos^2 x} -frac{cos^2 x}{cos^2 x}= frac{1}{cos^2 x} -1$$



              Notice now that we expressed this in terms of $cos(x)$, we now have a function in terms of $cos(x)$, namely:
              $$ f(cos x )=frac{1}{cos^2 x} -1$$
              Or if we let $z=cos(x)$
              $$ f(z )=frac{1}{z^2} -1$$
              would you not say that this is expressed in terms of $z$ $(=cos(x))$?






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                $z^2$ in the denominator in the last step.
                $endgroup$
                – Deepak
                Dec 20 '18 at 17:34
















              0












              $begingroup$

              Notice that we have that:
              $$sec^2 x = frac{1}{cos^2 x} $$ $$ sin^2x = 1 - cos^2x$$



              So if we multiply the two and use distributivity:
              $$(sec^2 x) cdot(sin^2x )=(frac{1}{cos^2 x}) cdot(1 - cos^2x)=frac{1}{cos^2 x} -frac{cos^2 x}{cos^2 x}= frac{1}{cos^2 x} -1$$



              Notice now that we expressed this in terms of $cos(x)$, we now have a function in terms of $cos(x)$, namely:
              $$ f(cos x )=frac{1}{cos^2 x} -1$$
              Or if we let $z=cos(x)$
              $$ f(z )=frac{1}{z^2} -1$$
              would you not say that this is expressed in terms of $z$ $(=cos(x))$?






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                $z^2$ in the denominator in the last step.
                $endgroup$
                – Deepak
                Dec 20 '18 at 17:34














              0












              0








              0





              $begingroup$

              Notice that we have that:
              $$sec^2 x = frac{1}{cos^2 x} $$ $$ sin^2x = 1 - cos^2x$$



              So if we multiply the two and use distributivity:
              $$(sec^2 x) cdot(sin^2x )=(frac{1}{cos^2 x}) cdot(1 - cos^2x)=frac{1}{cos^2 x} -frac{cos^2 x}{cos^2 x}= frac{1}{cos^2 x} -1$$



              Notice now that we expressed this in terms of $cos(x)$, we now have a function in terms of $cos(x)$, namely:
              $$ f(cos x )=frac{1}{cos^2 x} -1$$
              Or if we let $z=cos(x)$
              $$ f(z )=frac{1}{z^2} -1$$
              would you not say that this is expressed in terms of $z$ $(=cos(x))$?






              share|cite|improve this answer











              $endgroup$



              Notice that we have that:
              $$sec^2 x = frac{1}{cos^2 x} $$ $$ sin^2x = 1 - cos^2x$$



              So if we multiply the two and use distributivity:
              $$(sec^2 x) cdot(sin^2x )=(frac{1}{cos^2 x}) cdot(1 - cos^2x)=frac{1}{cos^2 x} -frac{cos^2 x}{cos^2 x}= frac{1}{cos^2 x} -1$$



              Notice now that we expressed this in terms of $cos(x)$, we now have a function in terms of $cos(x)$, namely:
              $$ f(cos x )=frac{1}{cos^2 x} -1$$
              Or if we let $z=cos(x)$
              $$ f(z )=frac{1}{z^2} -1$$
              would you not say that this is expressed in terms of $z$ $(=cos(x))$?







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 21 '18 at 6:42

























              answered Dec 20 '18 at 17:26









              Wesley StrikWesley Strik

              2,017423




              2,017423








              • 1




                $begingroup$
                $z^2$ in the denominator in the last step.
                $endgroup$
                – Deepak
                Dec 20 '18 at 17:34














              • 1




                $begingroup$
                $z^2$ in the denominator in the last step.
                $endgroup$
                – Deepak
                Dec 20 '18 at 17:34








              1




              1




              $begingroup$
              $z^2$ in the denominator in the last step.
              $endgroup$
              – Deepak
              Dec 20 '18 at 17:34




              $begingroup$
              $z^2$ in the denominator in the last step.
              $endgroup$
              – Deepak
              Dec 20 '18 at 17:34











              0












              $begingroup$

              $sec^2(x)bullet sin^2(x)={1over{cos^2(x)}}left(1-cos^2(x)right)={1over{cos^2(x)}}-1$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $sec^2(x)bullet sin^2(x)={1over{cos^2(x)}}left(1-cos^2(x)right)={1over{cos^2(x)}}-1$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $sec^2(x)bullet sin^2(x)={1over{cos^2(x)}}left(1-cos^2(x)right)={1over{cos^2(x)}}-1$






                  share|cite|improve this answer









                  $endgroup$



                  $sec^2(x)bullet sin^2(x)={1over{cos^2(x)}}left(1-cos^2(x)right)={1over{cos^2(x)}}-1$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 6 at 19:36









                  Math LoverMath Lover

                  15910




                  15910






























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