Simplify tan$theta$ cos$theta$












0












$begingroup$


How do I simplify tan$theta$ cos$theta$ ?



Why is this so hard to do? What pieces of information should I know before doing these?



Can someone just tell me were am I going wrong? I have 5 days to master this before my SAT Practice.



I would brush up on 'basics' but i don't even know how this related to anything ive learnt in trig before...I literally google every identity, am i supposed to remember or somehow derive them??






algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    Do you remember the definition of tan?
    $endgroup$
    – Mathmo123
    Jul 22 '14 at 10:27






  • 1




    $begingroup$
    Do you know how tan relates to sin and cos? (4 days, 23h, 58min)
    $endgroup$
    – Lord_Gestalter
    Jul 22 '14 at 10:28










  • $begingroup$
    sosmath.com/trig/Trig5/trig5/trig5.html
    $endgroup$
    – Scientifica
    Jul 22 '14 at 10:28










  • $begingroup$
    To use the 5 days brushing up the basics will be worth more than collecting and memorising possible answers
    $endgroup$
    – Lord_Gestalter
    Jul 22 '14 at 10:34










  • $begingroup$
    I saw your earlier posts. You are going wrong, by not concentrating on the basic definition of sine, cos and tan. If I was you, I would take a pen and paper and start by drawing triangle and learn the ratios of sin,cos and tan. It would not be called maths, if it was not tricky! :)
    $endgroup$
    – MonK
    Jul 22 '14 at 10:35
















0












$begingroup$


How do I simplify tan$theta$ cos$theta$ ?



Why is this so hard to do? What pieces of information should I know before doing these?



Can someone just tell me were am I going wrong? I have 5 days to master this before my SAT Practice.



I would brush up on 'basics' but i don't even know how this related to anything ive learnt in trig before...I literally google every identity, am i supposed to remember or somehow derive them??






algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    Do you remember the definition of tan?
    $endgroup$
    – Mathmo123
    Jul 22 '14 at 10:27






  • 1




    $begingroup$
    Do you know how tan relates to sin and cos? (4 days, 23h, 58min)
    $endgroup$
    – Lord_Gestalter
    Jul 22 '14 at 10:28










  • $begingroup$
    sosmath.com/trig/Trig5/trig5/trig5.html
    $endgroup$
    – Scientifica
    Jul 22 '14 at 10:28










  • $begingroup$
    To use the 5 days brushing up the basics will be worth more than collecting and memorising possible answers
    $endgroup$
    – Lord_Gestalter
    Jul 22 '14 at 10:34










  • $begingroup$
    I saw your earlier posts. You are going wrong, by not concentrating on the basic definition of sine, cos and tan. If I was you, I would take a pen and paper and start by drawing triangle and learn the ratios of sin,cos and tan. It would not be called maths, if it was not tricky! :)
    $endgroup$
    – MonK
    Jul 22 '14 at 10:35














0












0








0





$begingroup$


How do I simplify tan$theta$ cos$theta$ ?



Why is this so hard to do? What pieces of information should I know before doing these?



Can someone just tell me were am I going wrong? I have 5 days to master this before my SAT Practice.



I would brush up on 'basics' but i don't even know how this related to anything ive learnt in trig before...I literally google every identity, am i supposed to remember or somehow derive them??






algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$




How do I simplify tan$theta$ cos$theta$ ?



Why is this so hard to do? What pieces of information should I know before doing these?



Can someone just tell me were am I going wrong? I have 5 days to master this before my SAT Practice.



I would brush up on 'basics' but i don't even know how this related to anything ive learnt in trig before...I literally google every identity, am i supposed to remember or somehow derive them??






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 Jul 22 '14 at 10:59









Brad

4,27021448




4,27021448










asked Jul 22 '14 at 10:26









sashasasha

214




214








  • 5




    $begingroup$
    Do you remember the definition of tan?
    $endgroup$
    – Mathmo123
    Jul 22 '14 at 10:27






  • 1




    $begingroup$
    Do you know how tan relates to sin and cos? (4 days, 23h, 58min)
    $endgroup$
    – Lord_Gestalter
    Jul 22 '14 at 10:28










  • $begingroup$
    sosmath.com/trig/Trig5/trig5/trig5.html
    $endgroup$
    – Scientifica
    Jul 22 '14 at 10:28










  • $begingroup$
    To use the 5 days brushing up the basics will be worth more than collecting and memorising possible answers
    $endgroup$
    – Lord_Gestalter
    Jul 22 '14 at 10:34










  • $begingroup$
    I saw your earlier posts. You are going wrong, by not concentrating on the basic definition of sine, cos and tan. If I was you, I would take a pen and paper and start by drawing triangle and learn the ratios of sin,cos and tan. It would not be called maths, if it was not tricky! :)
    $endgroup$
    – MonK
    Jul 22 '14 at 10:35














  • 5




    $begingroup$
    Do you remember the definition of tan?
    $endgroup$
    – Mathmo123
    Jul 22 '14 at 10:27






  • 1




    $begingroup$
    Do you know how tan relates to sin and cos? (4 days, 23h, 58min)
    $endgroup$
    – Lord_Gestalter
    Jul 22 '14 at 10:28










  • $begingroup$
    sosmath.com/trig/Trig5/trig5/trig5.html
    $endgroup$
    – Scientifica
    Jul 22 '14 at 10:28










  • $begingroup$
    To use the 5 days brushing up the basics will be worth more than collecting and memorising possible answers
    $endgroup$
    – Lord_Gestalter
    Jul 22 '14 at 10:34










  • $begingroup$
    I saw your earlier posts. You are going wrong, by not concentrating on the basic definition of sine, cos and tan. If I was you, I would take a pen and paper and start by drawing triangle and learn the ratios of sin,cos and tan. It would not be called maths, if it was not tricky! :)
    $endgroup$
    – MonK
    Jul 22 '14 at 10:35








5




5




$begingroup$
Do you remember the definition of tan?
$endgroup$
– Mathmo123
Jul 22 '14 at 10:27




$begingroup$
Do you remember the definition of tan?
$endgroup$
– Mathmo123
Jul 22 '14 at 10:27




1




1




$begingroup$
Do you know how tan relates to sin and cos? (4 days, 23h, 58min)
$endgroup$
– Lord_Gestalter
Jul 22 '14 at 10:28




$begingroup$
Do you know how tan relates to sin and cos? (4 days, 23h, 58min)
$endgroup$
– Lord_Gestalter
Jul 22 '14 at 10:28












$begingroup$
sosmath.com/trig/Trig5/trig5/trig5.html
$endgroup$
– Scientifica
Jul 22 '14 at 10:28




$begingroup$
sosmath.com/trig/Trig5/trig5/trig5.html
$endgroup$
– Scientifica
Jul 22 '14 at 10:28












$begingroup$
To use the 5 days brushing up the basics will be worth more than collecting and memorising possible answers
$endgroup$
– Lord_Gestalter
Jul 22 '14 at 10:34




$begingroup$
To use the 5 days brushing up the basics will be worth more than collecting and memorising possible answers
$endgroup$
– Lord_Gestalter
Jul 22 '14 at 10:34












$begingroup$
I saw your earlier posts. You are going wrong, by not concentrating on the basic definition of sine, cos and tan. If I was you, I would take a pen and paper and start by drawing triangle and learn the ratios of sin,cos and tan. It would not be called maths, if it was not tricky! :)
$endgroup$
– MonK
Jul 22 '14 at 10:35




$begingroup$
I saw your earlier posts. You are going wrong, by not concentrating on the basic definition of sine, cos and tan. If I was you, I would take a pen and paper and start by drawing triangle and learn the ratios of sin,cos and tan. It would not be called maths, if it was not tricky! :)
$endgroup$
– MonK
Jul 22 '14 at 10:35










4 Answers
4






active

oldest

votes


















2












$begingroup$

Hint: What is the definition of the $tan$ function?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Tan = Opp / Adj..
    $endgroup$
    – sasha
    Jul 22 '14 at 10:38






  • 1




    $begingroup$
    @sasha That is a rough definition only true for angles from $0$ to $fracpi2$. For other corners, the $tan$ is defined via the $cos$ and $sin$ function. If you are practising for SATs, I advise you to look that definition up.
    $endgroup$
    – 5xum
    Jul 22 '14 at 10:40










  • $begingroup$
    @sasha $tantheta = opp/adj = displaystylefrac{opp}{adj}cdotfrac{frac{1}{hyp}}{frac{1}{hyp}} = frac{frac{opp}{hyp}}{frac{opp}{hyp}}=frac{sintheta}{costheta}$
    $endgroup$
    – John Joy
    Jul 22 '14 at 13:30



















5












$begingroup$

Notice first that the $tan$ function is defined on $D:=Bbb Rsetminus{fracpi2+kpi,; kinBbb Z}$ and that $tantheta=frac{sintheta}{costheta}$ so



$$tanthetacostheta=sintheta,quadforallthetain D$$
and be careful the trap is to give a wrong domain.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am used to write - rather than $$ for excluding. Anyway +1
    $endgroup$
    – mrs
    Jul 22 '14 at 12:21





















1












$begingroup$

Elementary school proof for $theta in ]0, pi/2[$:



Consider a right triangle with angles $pi/2, theta, pi/2-theta$ and edges lengths $a,b,c>0$ as in the figure below



Right triangle



Then
$$tan(theta)= frac{a}{c}, quad cos(theta) = frac{c}{b}, quad sin(theta) = frac{a}{b},$$
thus
$$tan(theta) cos(theta) = frac{a}{c}cdot frac{c}{b} = frac{a}{b} = sin(theta).$$






share|cite|improve this answer









$endgroup$





















    -1












    $begingroup$

    There are two ways to do this.
    Here is one way:



    Saying that



    $tan(x)=o/a $ and $ cos(x)=a/h$, we get $oa/ha = o/h=sin x$ so the answer is $sin x$!






    share|cite|improve this answer











    $endgroup$













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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Hint: What is the definition of the $tan$ function?






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Tan = Opp / Adj..
        $endgroup$
        – sasha
        Jul 22 '14 at 10:38






      • 1




        $begingroup$
        @sasha That is a rough definition only true for angles from $0$ to $fracpi2$. For other corners, the $tan$ is defined via the $cos$ and $sin$ function. If you are practising for SATs, I advise you to look that definition up.
        $endgroup$
        – 5xum
        Jul 22 '14 at 10:40










      • $begingroup$
        @sasha $tantheta = opp/adj = displaystylefrac{opp}{adj}cdotfrac{frac{1}{hyp}}{frac{1}{hyp}} = frac{frac{opp}{hyp}}{frac{opp}{hyp}}=frac{sintheta}{costheta}$
        $endgroup$
        – John Joy
        Jul 22 '14 at 13:30
















      2












      $begingroup$

      Hint: What is the definition of the $tan$ function?






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Tan = Opp / Adj..
        $endgroup$
        – sasha
        Jul 22 '14 at 10:38






      • 1




        $begingroup$
        @sasha That is a rough definition only true for angles from $0$ to $fracpi2$. For other corners, the $tan$ is defined via the $cos$ and $sin$ function. If you are practising for SATs, I advise you to look that definition up.
        $endgroup$
        – 5xum
        Jul 22 '14 at 10:40










      • $begingroup$
        @sasha $tantheta = opp/adj = displaystylefrac{opp}{adj}cdotfrac{frac{1}{hyp}}{frac{1}{hyp}} = frac{frac{opp}{hyp}}{frac{opp}{hyp}}=frac{sintheta}{costheta}$
        $endgroup$
        – John Joy
        Jul 22 '14 at 13:30














      2












      2








      2





      $begingroup$

      Hint: What is the definition of the $tan$ function?






      share|cite|improve this answer









      $endgroup$



      Hint: What is the definition of the $tan$ function?







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jul 22 '14 at 10:28









      5xum5xum

      90.6k394161




      90.6k394161












      • $begingroup$
        Tan = Opp / Adj..
        $endgroup$
        – sasha
        Jul 22 '14 at 10:38






      • 1




        $begingroup$
        @sasha That is a rough definition only true for angles from $0$ to $fracpi2$. For other corners, the $tan$ is defined via the $cos$ and $sin$ function. If you are practising for SATs, I advise you to look that definition up.
        $endgroup$
        – 5xum
        Jul 22 '14 at 10:40










      • $begingroup$
        @sasha $tantheta = opp/adj = displaystylefrac{opp}{adj}cdotfrac{frac{1}{hyp}}{frac{1}{hyp}} = frac{frac{opp}{hyp}}{frac{opp}{hyp}}=frac{sintheta}{costheta}$
        $endgroup$
        – John Joy
        Jul 22 '14 at 13:30


















      • $begingroup$
        Tan = Opp / Adj..
        $endgroup$
        – sasha
        Jul 22 '14 at 10:38






      • 1




        $begingroup$
        @sasha That is a rough definition only true for angles from $0$ to $fracpi2$. For other corners, the $tan$ is defined via the $cos$ and $sin$ function. If you are practising for SATs, I advise you to look that definition up.
        $endgroup$
        – 5xum
        Jul 22 '14 at 10:40










      • $begingroup$
        @sasha $tantheta = opp/adj = displaystylefrac{opp}{adj}cdotfrac{frac{1}{hyp}}{frac{1}{hyp}} = frac{frac{opp}{hyp}}{frac{opp}{hyp}}=frac{sintheta}{costheta}$
        $endgroup$
        – John Joy
        Jul 22 '14 at 13:30
















      $begingroup$
      Tan = Opp / Adj..
      $endgroup$
      – sasha
      Jul 22 '14 at 10:38




      $begingroup$
      Tan = Opp / Adj..
      $endgroup$
      – sasha
      Jul 22 '14 at 10:38




      1




      1




      $begingroup$
      @sasha That is a rough definition only true for angles from $0$ to $fracpi2$. For other corners, the $tan$ is defined via the $cos$ and $sin$ function. If you are practising for SATs, I advise you to look that definition up.
      $endgroup$
      – 5xum
      Jul 22 '14 at 10:40




      $begingroup$
      @sasha That is a rough definition only true for angles from $0$ to $fracpi2$. For other corners, the $tan$ is defined via the $cos$ and $sin$ function. If you are practising for SATs, I advise you to look that definition up.
      $endgroup$
      – 5xum
      Jul 22 '14 at 10:40












      $begingroup$
      @sasha $tantheta = opp/adj = displaystylefrac{opp}{adj}cdotfrac{frac{1}{hyp}}{frac{1}{hyp}} = frac{frac{opp}{hyp}}{frac{opp}{hyp}}=frac{sintheta}{costheta}$
      $endgroup$
      – John Joy
      Jul 22 '14 at 13:30




      $begingroup$
      @sasha $tantheta = opp/adj = displaystylefrac{opp}{adj}cdotfrac{frac{1}{hyp}}{frac{1}{hyp}} = frac{frac{opp}{hyp}}{frac{opp}{hyp}}=frac{sintheta}{costheta}$
      $endgroup$
      – John Joy
      Jul 22 '14 at 13:30











      5












      $begingroup$

      Notice first that the $tan$ function is defined on $D:=Bbb Rsetminus{fracpi2+kpi,; kinBbb Z}$ and that $tantheta=frac{sintheta}{costheta}$ so



      $$tanthetacostheta=sintheta,quadforallthetain D$$
      and be careful the trap is to give a wrong domain.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I am used to write - rather than $$ for excluding. Anyway +1
        $endgroup$
        – mrs
        Jul 22 '14 at 12:21


















      5












      $begingroup$

      Notice first that the $tan$ function is defined on $D:=Bbb Rsetminus{fracpi2+kpi,; kinBbb Z}$ and that $tantheta=frac{sintheta}{costheta}$ so



      $$tanthetacostheta=sintheta,quadforallthetain D$$
      and be careful the trap is to give a wrong domain.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I am used to write - rather than $$ for excluding. Anyway +1
        $endgroup$
        – mrs
        Jul 22 '14 at 12:21
















      5












      5








      5





      $begingroup$

      Notice first that the $tan$ function is defined on $D:=Bbb Rsetminus{fracpi2+kpi,; kinBbb Z}$ and that $tantheta=frac{sintheta}{costheta}$ so



      $$tanthetacostheta=sintheta,quadforallthetain D$$
      and be careful the trap is to give a wrong domain.






      share|cite|improve this answer









      $endgroup$



      Notice first that the $tan$ function is defined on $D:=Bbb Rsetminus{fracpi2+kpi,; kinBbb Z}$ and that $tantheta=frac{sintheta}{costheta}$ so



      $$tanthetacostheta=sintheta,quadforallthetain D$$
      and be careful the trap is to give a wrong domain.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jul 22 '14 at 10:36







      user63181



















      • $begingroup$
        I am used to write - rather than $$ for excluding. Anyway +1
        $endgroup$
        – mrs
        Jul 22 '14 at 12:21




















      • $begingroup$
        I am used to write - rather than $$ for excluding. Anyway +1
        $endgroup$
        – mrs
        Jul 22 '14 at 12:21


















      $begingroup$
      I am used to write - rather than $$ for excluding. Anyway +1
      $endgroup$
      – mrs
      Jul 22 '14 at 12:21






      $begingroup$
      I am used to write - rather than $$ for excluding. Anyway +1
      $endgroup$
      – mrs
      Jul 22 '14 at 12:21













      1












      $begingroup$

      Elementary school proof for $theta in ]0, pi/2[$:



      Consider a right triangle with angles $pi/2, theta, pi/2-theta$ and edges lengths $a,b,c>0$ as in the figure below



      Right triangle



      Then
      $$tan(theta)= frac{a}{c}, quad cos(theta) = frac{c}{b}, quad sin(theta) = frac{a}{b},$$
      thus
      $$tan(theta) cos(theta) = frac{a}{c}cdot frac{c}{b} = frac{a}{b} = sin(theta).$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Elementary school proof for $theta in ]0, pi/2[$:



        Consider a right triangle with angles $pi/2, theta, pi/2-theta$ and edges lengths $a,b,c>0$ as in the figure below



        Right triangle



        Then
        $$tan(theta)= frac{a}{c}, quad cos(theta) = frac{c}{b}, quad sin(theta) = frac{a}{b},$$
        thus
        $$tan(theta) cos(theta) = frac{a}{c}cdot frac{c}{b} = frac{a}{b} = sin(theta).$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Elementary school proof for $theta in ]0, pi/2[$:



          Consider a right triangle with angles $pi/2, theta, pi/2-theta$ and edges lengths $a,b,c>0$ as in the figure below



          Right triangle



          Then
          $$tan(theta)= frac{a}{c}, quad cos(theta) = frac{c}{b}, quad sin(theta) = frac{a}{b},$$
          thus
          $$tan(theta) cos(theta) = frac{a}{c}cdot frac{c}{b} = frac{a}{b} = sin(theta).$$






          share|cite|improve this answer









          $endgroup$



          Elementary school proof for $theta in ]0, pi/2[$:



          Consider a right triangle with angles $pi/2, theta, pi/2-theta$ and edges lengths $a,b,c>0$ as in the figure below



          Right triangle



          Then
          $$tan(theta)= frac{a}{c}, quad cos(theta) = frac{c}{b}, quad sin(theta) = frac{a}{b},$$
          thus
          $$tan(theta) cos(theta) = frac{a}{c}cdot frac{c}{b} = frac{a}{b} = sin(theta).$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jul 22 '14 at 12:02









          SurbSurb

          37.7k94375




          37.7k94375























              -1












              $begingroup$

              There are two ways to do this.
              Here is one way:



              Saying that



              $tan(x)=o/a $ and $ cos(x)=a/h$, we get $oa/ha = o/h=sin x$ so the answer is $sin x$!






              share|cite|improve this answer











              $endgroup$


















                -1












                $begingroup$

                There are two ways to do this.
                Here is one way:



                Saying that



                $tan(x)=o/a $ and $ cos(x)=a/h$, we get $oa/ha = o/h=sin x$ so the answer is $sin x$!






                share|cite|improve this answer











                $endgroup$
















                  -1












                  -1








                  -1





                  $begingroup$

                  There are two ways to do this.
                  Here is one way:



                  Saying that



                  $tan(x)=o/a $ and $ cos(x)=a/h$, we get $oa/ha = o/h=sin x$ so the answer is $sin x$!






                  share|cite|improve this answer











                  $endgroup$



                  There are two ways to do this.
                  Here is one way:



                  Saying that



                  $tan(x)=o/a $ and $ cos(x)=a/h$, we get $oa/ha = o/h=sin x$ so the answer is $sin x$!







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 12 '18 at 1:15









                  amWhy

                  1




                  1










                  answered Dec 12 '18 at 1:09









                  A.usernemaA.usernema

                  113




                  113






























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