Trig Identity Verification












-1














$$cos^2(t)-sin^2(t)=0$$



Using trig identity, we can write it equal to



$cos(2t) = 0$
where I get $2t = frac{pi}{2}, frac{3pi}{2}$
which means $t = frac{pi}{4}, frac{3pi}{4}$



or (without trig identity)



$$cos^2(t)=sin^2(t)$$
where I get $t = frac{pi}{4}, frac{3pi}{4}, frac{5pi}{4}, frac{7pi}{4}$



What am I doing wrong?






trigonometry







share|cite|improve this question
























  • It is not true that $cos(2t)=0implies 2t=frac{pi}{2},frac{3pi}{2}$.
    – mathlove
    Nov 30 '17 at 6:11








  • 1




    In your first approach these are all of the solutions inside the interval $tin[0,pi]$.
    – Dave
    Nov 30 '17 at 13:26


















-1














$$cos^2(t)-sin^2(t)=0$$



Using trig identity, we can write it equal to



$cos(2t) = 0$
where I get $2t = frac{pi}{2}, frac{3pi}{2}$
which means $t = frac{pi}{4}, frac{3pi}{4}$



or (without trig identity)



$$cos^2(t)=sin^2(t)$$
where I get $t = frac{pi}{4}, frac{3pi}{4}, frac{5pi}{4}, frac{7pi}{4}$



What am I doing wrong?






trigonometry







share|cite|improve this question
























  • It is not true that $cos(2t)=0implies 2t=frac{pi}{2},frac{3pi}{2}$.
    – mathlove
    Nov 30 '17 at 6:11








  • 1




    In your first approach these are all of the solutions inside the interval $tin[0,pi]$.
    – Dave
    Nov 30 '17 at 13:26
















-1












-1








-1







$$cos^2(t)-sin^2(t)=0$$



Using trig identity, we can write it equal to



$cos(2t) = 0$
where I get $2t = frac{pi}{2}, frac{3pi}{2}$
which means $t = frac{pi}{4}, frac{3pi}{4}$



or (without trig identity)



$$cos^2(t)=sin^2(t)$$
where I get $t = frac{pi}{4}, frac{3pi}{4}, frac{5pi}{4}, frac{7pi}{4}$



What am I doing wrong?






trigonometry







share|cite|improve this question















$$cos^2(t)-sin^2(t)=0$$



Using trig identity, we can write it equal to



$cos(2t) = 0$
where I get $2t = frac{pi}{2}, frac{3pi}{2}$
which means $t = frac{pi}{4}, frac{3pi}{4}$



or (without trig identity)



$$cos^2(t)=sin^2(t)$$
where I get $t = frac{pi}{4}, frac{3pi}{4}, frac{5pi}{4}, frac{7pi}{4}$



What am I doing wrong?






trigonometry




trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '17 at 6:43









Dylan

12.4k31026




12.4k31026










asked Nov 30 '17 at 6:07









mathguymathguy

496418




496418












  • It is not true that $cos(2t)=0implies 2t=frac{pi}{2},frac{3pi}{2}$.
    – mathlove
    Nov 30 '17 at 6:11








  • 1




    In your first approach these are all of the solutions inside the interval $tin[0,pi]$.
    – Dave
    Nov 30 '17 at 13:26




















  • It is not true that $cos(2t)=0implies 2t=frac{pi}{2},frac{3pi}{2}$.
    – mathlove
    Nov 30 '17 at 6:11








  • 1




    In your first approach these are all of the solutions inside the interval $tin[0,pi]$.
    – Dave
    Nov 30 '17 at 13:26


















It is not true that $cos(2t)=0implies 2t=frac{pi}{2},frac{3pi}{2}$.
– mathlove
Nov 30 '17 at 6:11






It is not true that $cos(2t)=0implies 2t=frac{pi}{2},frac{3pi}{2}$.
– mathlove
Nov 30 '17 at 6:11






1




1




In your first approach these are all of the solutions inside the interval $tin[0,pi]$.
– Dave
Nov 30 '17 at 13:26






In your first approach these are all of the solutions inside the interval $tin[0,pi]$.
– Dave
Nov 30 '17 at 13:26












3 Answers
3






active

oldest

votes


















1














$cos(2t) = 0$ implies that $2t = pi(n+1/2) $ where $n= 0, pm 1, pm 2, ... $ since $cos(pi(n+1/2)) = 0$ for those values of $n$. You're simply missing a few of these solutions above. In the same way your neglecting other solutions when you solve without trig identity. Does this answer your question?






share|cite|improve this answer































    1














    Method$#1:$
    $$2t=(2m+1)dfracpi2$$ where $m$ is any integer



    Method$#2:$



    $$cos^2t=sin^2timpliessin^2t=1-sin^2t$$



    $$iffsin^2t=dfrac12=sin^2dfracpi4$$



    Now use Prove $ sin(A+B)sin(A-B)=sin^2A-sin^2B $,



    If $sinleft(t-dfracpi4right)=0, t-dfracpi4=npi$ where $n$ is any integer



    What if $sinleft(t+dfracpi4right)=0?$



    More generally, if $$sin^2x=sin^2Aiffcos^2x=cos^2Aifftan^2x=tan^2A$$



    Can you prove $x=rpipm A$ where $r$ is any integer






    share|cite|improve this answer





























      1














      Alternative solution:



      $$cos^2(t)-sin^2(t)=0$$



      $$sin^2(t)=cos^2(t)$$



      $$tan^2(t)=1$$



      $$tan(t)= pm 1$$



      thus $$t= frac{π}4,frac{3π}4,frac{5π}4,frac{7π}4$$



      NOTE:



      keep attention when from here
      $$sin^2(t)=cos^2(t)$$
      dividing by $cos^2(t)$ and obtain
      $$tan^2(t)=1$$
      you are assuming that $cos^2(t) neq0$ that's a correct assumption here because $t=pmfrac{π}2$ do not satisfy the original equation.






      share|cite|improve this answer





















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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1














        $cos(2t) = 0$ implies that $2t = pi(n+1/2) $ where $n= 0, pm 1, pm 2, ... $ since $cos(pi(n+1/2)) = 0$ for those values of $n$. You're simply missing a few of these solutions above. In the same way your neglecting other solutions when you solve without trig identity. Does this answer your question?






        share|cite|improve this answer




























          1














          $cos(2t) = 0$ implies that $2t = pi(n+1/2) $ where $n= 0, pm 1, pm 2, ... $ since $cos(pi(n+1/2)) = 0$ for those values of $n$. You're simply missing a few of these solutions above. In the same way your neglecting other solutions when you solve without trig identity. Does this answer your question?






          share|cite|improve this answer


























            1












            1








            1






            $cos(2t) = 0$ implies that $2t = pi(n+1/2) $ where $n= 0, pm 1, pm 2, ... $ since $cos(pi(n+1/2)) = 0$ for those values of $n$. You're simply missing a few of these solutions above. In the same way your neglecting other solutions when you solve without trig identity. Does this answer your question?






            share|cite|improve this answer














            $cos(2t) = 0$ implies that $2t = pi(n+1/2) $ where $n= 0, pm 1, pm 2, ... $ since $cos(pi(n+1/2)) = 0$ for those values of $n$. You're simply missing a few of these solutions above. In the same way your neglecting other solutions when you solve without trig identity. Does this answer your question?







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 30 '17 at 6:33

























            answered Nov 30 '17 at 6:12









            ThambiThambi

            284




            284























                1














                Method$#1:$
                $$2t=(2m+1)dfracpi2$$ where $m$ is any integer



                Method$#2:$



                $$cos^2t=sin^2timpliessin^2t=1-sin^2t$$



                $$iffsin^2t=dfrac12=sin^2dfracpi4$$



                Now use Prove $ sin(A+B)sin(A-B)=sin^2A-sin^2B $,



                If $sinleft(t-dfracpi4right)=0, t-dfracpi4=npi$ where $n$ is any integer



                What if $sinleft(t+dfracpi4right)=0?$



                More generally, if $$sin^2x=sin^2Aiffcos^2x=cos^2Aifftan^2x=tan^2A$$



                Can you prove $x=rpipm A$ where $r$ is any integer






                share|cite|improve this answer


























                  1














                  Method$#1:$
                  $$2t=(2m+1)dfracpi2$$ where $m$ is any integer



                  Method$#2:$



                  $$cos^2t=sin^2timpliessin^2t=1-sin^2t$$



                  $$iffsin^2t=dfrac12=sin^2dfracpi4$$



                  Now use Prove $ sin(A+B)sin(A-B)=sin^2A-sin^2B $,



                  If $sinleft(t-dfracpi4right)=0, t-dfracpi4=npi$ where $n$ is any integer



                  What if $sinleft(t+dfracpi4right)=0?$



                  More generally, if $$sin^2x=sin^2Aiffcos^2x=cos^2Aifftan^2x=tan^2A$$



                  Can you prove $x=rpipm A$ where $r$ is any integer






                  share|cite|improve this answer
























                    1












                    1








                    1






                    Method$#1:$
                    $$2t=(2m+1)dfracpi2$$ where $m$ is any integer



                    Method$#2:$



                    $$cos^2t=sin^2timpliessin^2t=1-sin^2t$$



                    $$iffsin^2t=dfrac12=sin^2dfracpi4$$



                    Now use Prove $ sin(A+B)sin(A-B)=sin^2A-sin^2B $,



                    If $sinleft(t-dfracpi4right)=0, t-dfracpi4=npi$ where $n$ is any integer



                    What if $sinleft(t+dfracpi4right)=0?$



                    More generally, if $$sin^2x=sin^2Aiffcos^2x=cos^2Aifftan^2x=tan^2A$$



                    Can you prove $x=rpipm A$ where $r$ is any integer






                    share|cite|improve this answer












                    Method$#1:$
                    $$2t=(2m+1)dfracpi2$$ where $m$ is any integer



                    Method$#2:$



                    $$cos^2t=sin^2timpliessin^2t=1-sin^2t$$



                    $$iffsin^2t=dfrac12=sin^2dfracpi4$$



                    Now use Prove $ sin(A+B)sin(A-B)=sin^2A-sin^2B $,



                    If $sinleft(t-dfracpi4right)=0, t-dfracpi4=npi$ where $n$ is any integer



                    What if $sinleft(t+dfracpi4right)=0?$



                    More generally, if $$sin^2x=sin^2Aiffcos^2x=cos^2Aifftan^2x=tan^2A$$



                    Can you prove $x=rpipm A$ where $r$ is any integer







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 30 '17 at 6:14









                    lab bhattacharjeelab bhattacharjee

                    224k15156274




                    224k15156274























                        1














                        Alternative solution:



                        $$cos^2(t)-sin^2(t)=0$$



                        $$sin^2(t)=cos^2(t)$$



                        $$tan^2(t)=1$$



                        $$tan(t)= pm 1$$



                        thus $$t= frac{π}4,frac{3π}4,frac{5π}4,frac{7π}4$$



                        NOTE:



                        keep attention when from here
                        $$sin^2(t)=cos^2(t)$$
                        dividing by $cos^2(t)$ and obtain
                        $$tan^2(t)=1$$
                        you are assuming that $cos^2(t) neq0$ that's a correct assumption here because $t=pmfrac{π}2$ do not satisfy the original equation.






                        share|cite|improve this answer


























                          1














                          Alternative solution:



                          $$cos^2(t)-sin^2(t)=0$$



                          $$sin^2(t)=cos^2(t)$$



                          $$tan^2(t)=1$$



                          $$tan(t)= pm 1$$



                          thus $$t= frac{π}4,frac{3π}4,frac{5π}4,frac{7π}4$$



                          NOTE:



                          keep attention when from here
                          $$sin^2(t)=cos^2(t)$$
                          dividing by $cos^2(t)$ and obtain
                          $$tan^2(t)=1$$
                          you are assuming that $cos^2(t) neq0$ that's a correct assumption here because $t=pmfrac{π}2$ do not satisfy the original equation.






                          share|cite|improve this answer
























                            1












                            1








                            1






                            Alternative solution:



                            $$cos^2(t)-sin^2(t)=0$$



                            $$sin^2(t)=cos^2(t)$$



                            $$tan^2(t)=1$$



                            $$tan(t)= pm 1$$



                            thus $$t= frac{π}4,frac{3π}4,frac{5π}4,frac{7π}4$$



                            NOTE:



                            keep attention when from here
                            $$sin^2(t)=cos^2(t)$$
                            dividing by $cos^2(t)$ and obtain
                            $$tan^2(t)=1$$
                            you are assuming that $cos^2(t) neq0$ that's a correct assumption here because $t=pmfrac{π}2$ do not satisfy the original equation.






                            share|cite|improve this answer












                            Alternative solution:



                            $$cos^2(t)-sin^2(t)=0$$



                            $$sin^2(t)=cos^2(t)$$



                            $$tan^2(t)=1$$



                            $$tan(t)= pm 1$$



                            thus $$t= frac{π}4,frac{3π}4,frac{5π}4,frac{7π}4$$



                            NOTE:



                            keep attention when from here
                            $$sin^2(t)=cos^2(t)$$
                            dividing by $cos^2(t)$ and obtain
                            $$tan^2(t)=1$$
                            you are assuming that $cos^2(t) neq0$ that's a correct assumption here because $t=pmfrac{π}2$ do not satisfy the original equation.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 30 '17 at 6:32









                            gimusigimusi

                            1




                            1






























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