Find $sin(theta)$ when $sin(theta) > 0$, $tan(theta)=7/24$.












0












$begingroup$


Find $sin(theta)$ when $sin(theta) > 0$,
$tan(theta)=7/24$.



I have no idea what to do.






trigonometry







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    You're given the ratio between sine and cosine. Also the sum of their squares is one...
    $endgroup$
    – Matteo
    Jan 18 at 12:21










  • $begingroup$
    Hint: $sec^2(x)=1+tan^2(x)$ and $sin(x)=frac{tan(x)}{sec(x)}$
    $endgroup$
    – robjohn♦
    Jan 18 at 12:24










  • $begingroup$
    Please, 2 months later, select among the good answers by checking, the one that agrees you the most. It is by this little discipline that such a site can live and bring information to others.
    $endgroup$
    – Jean Marie
    Mar 10 at 22:30
















0












$begingroup$


Find $sin(theta)$ when $sin(theta) > 0$,
$tan(theta)=7/24$.



I have no idea what to do.






trigonometry







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    You're given the ratio between sine and cosine. Also the sum of their squares is one...
    $endgroup$
    – Matteo
    Jan 18 at 12:21










  • $begingroup$
    Hint: $sec^2(x)=1+tan^2(x)$ and $sin(x)=frac{tan(x)}{sec(x)}$
    $endgroup$
    – robjohn♦
    Jan 18 at 12:24










  • $begingroup$
    Please, 2 months later, select among the good answers by checking, the one that agrees you the most. It is by this little discipline that such a site can live and bring information to others.
    $endgroup$
    – Jean Marie
    Mar 10 at 22:30














0












0








0





$begingroup$


Find $sin(theta)$ when $sin(theta) > 0$,
$tan(theta)=7/24$.



I have no idea what to do.






trigonometry







share|cite|improve this question











$endgroup$




Find $sin(theta)$ when $sin(theta) > 0$,
$tan(theta)=7/24$.



I have no idea what to do.






trigonometry




trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 12:21









Yanko

8,5742830




8,5742830










asked Jan 18 at 12:18









Renee HaasRenee Haas

11




11








  • 3




    $begingroup$
    You're given the ratio between sine and cosine. Also the sum of their squares is one...
    $endgroup$
    – Matteo
    Jan 18 at 12:21










  • $begingroup$
    Hint: $sec^2(x)=1+tan^2(x)$ and $sin(x)=frac{tan(x)}{sec(x)}$
    $endgroup$
    – robjohn♦
    Jan 18 at 12:24










  • $begingroup$
    Please, 2 months later, select among the good answers by checking, the one that agrees you the most. It is by this little discipline that such a site can live and bring information to others.
    $endgroup$
    – Jean Marie
    Mar 10 at 22:30














  • 3




    $begingroup$
    You're given the ratio between sine and cosine. Also the sum of their squares is one...
    $endgroup$
    – Matteo
    Jan 18 at 12:21










  • $begingroup$
    Hint: $sec^2(x)=1+tan^2(x)$ and $sin(x)=frac{tan(x)}{sec(x)}$
    $endgroup$
    – robjohn♦
    Jan 18 at 12:24










  • $begingroup$
    Please, 2 months later, select among the good answers by checking, the one that agrees you the most. It is by this little discipline that such a site can live and bring information to others.
    $endgroup$
    – Jean Marie
    Mar 10 at 22:30








3




3




$begingroup$
You're given the ratio between sine and cosine. Also the sum of their squares is one...
$endgroup$
– Matteo
Jan 18 at 12:21




$begingroup$
You're given the ratio between sine and cosine. Also the sum of their squares is one...
$endgroup$
– Matteo
Jan 18 at 12:21












$begingroup$
Hint: $sec^2(x)=1+tan^2(x)$ and $sin(x)=frac{tan(x)}{sec(x)}$
$endgroup$
– robjohn♦
Jan 18 at 12:24




$begingroup$
Hint: $sec^2(x)=1+tan^2(x)$ and $sin(x)=frac{tan(x)}{sec(x)}$
$endgroup$
– robjohn♦
Jan 18 at 12:24












$begingroup$
Please, 2 months later, select among the good answers by checking, the one that agrees you the most. It is by this little discipline that such a site can live and bring information to others.
$endgroup$
– Jean Marie
Mar 10 at 22:30




$begingroup$
Please, 2 months later, select among the good answers by checking, the one that agrees you the most. It is by this little discipline that such a site can live and bring information to others.
$endgroup$
– Jean Marie
Mar 10 at 22:30










3 Answers
3






active

oldest

votes


















5












$begingroup$

Hint



use $$sin theta ={tanthetaover sqrt{1+tan^2theta}}$$whenever $sin theta , tan theta ge 0$.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Draw a right triangle that shows $tan theta = 7/24.$ There are infinitely many, but choosing the one with legs $7$ and $24$ is a swell choice. Now use Pythagorean Theorem to find the length of the hypotenuse. Finding the values of any of the other trig functions should be easy now.



    The only snag is that you have to think about which quadrant you're in. But you have $sin theta$ and $tan theta$ both positive, so you're in the first quadrant.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Use $cos^2theta = 1 - sin^2theta$ and $tantheta = frac{sintheta}{costheta}=c$. Then $sintheta = csqrt{1-sin^2theta}$, then $sin^2theta = c^2-c^2sin^2theta$, and finally $sintheta = frac{c}{sqrt{1+c^2}}$. Note that when taking the roots I have taken the positive such since $sintheta>0$ and also $tantheta >0$, which implies $costheta > 0$.






      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









        5












        $begingroup$

        Hint



        use $$sin theta ={tanthetaover sqrt{1+tan^2theta}}$$whenever $sin theta , tan theta ge 0$.






        share|cite|improve this answer









        $endgroup$


















          5












          $begingroup$

          Hint



          use $$sin theta ={tanthetaover sqrt{1+tan^2theta}}$$whenever $sin theta , tan theta ge 0$.






          share|cite|improve this answer









          $endgroup$
















            5












            5








            5





            $begingroup$

            Hint



            use $$sin theta ={tanthetaover sqrt{1+tan^2theta}}$$whenever $sin theta , tan theta ge 0$.






            share|cite|improve this answer









            $endgroup$



            Hint



            use $$sin theta ={tanthetaover sqrt{1+tan^2theta}}$$whenever $sin theta , tan theta ge 0$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 18 at 12:23









            Mostafa AyazMostafa Ayaz

            18.1k31040




            18.1k31040























                3












                $begingroup$

                Draw a right triangle that shows $tan theta = 7/24.$ There are infinitely many, but choosing the one with legs $7$ and $24$ is a swell choice. Now use Pythagorean Theorem to find the length of the hypotenuse. Finding the values of any of the other trig functions should be easy now.



                The only snag is that you have to think about which quadrant you're in. But you have $sin theta$ and $tan theta$ both positive, so you're in the first quadrant.






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  Draw a right triangle that shows $tan theta = 7/24.$ There are infinitely many, but choosing the one with legs $7$ and $24$ is a swell choice. Now use Pythagorean Theorem to find the length of the hypotenuse. Finding the values of any of the other trig functions should be easy now.



                  The only snag is that you have to think about which quadrant you're in. But you have $sin theta$ and $tan theta$ both positive, so you're in the first quadrant.






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    Draw a right triangle that shows $tan theta = 7/24.$ There are infinitely many, but choosing the one with legs $7$ and $24$ is a swell choice. Now use Pythagorean Theorem to find the length of the hypotenuse. Finding the values of any of the other trig functions should be easy now.



                    The only snag is that you have to think about which quadrant you're in. But you have $sin theta$ and $tan theta$ both positive, so you're in the first quadrant.






                    share|cite|improve this answer









                    $endgroup$



                    Draw a right triangle that shows $tan theta = 7/24.$ There are infinitely many, but choosing the one with legs $7$ and $24$ is a swell choice. Now use Pythagorean Theorem to find the length of the hypotenuse. Finding the values of any of the other trig functions should be easy now.



                    The only snag is that you have to think about which quadrant you're in. But you have $sin theta$ and $tan theta$ both positive, so you're in the first quadrant.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 18 at 12:28









                    B. GoddardB. Goddard

                    20.2k21543




                    20.2k21543























                        1












                        $begingroup$

                        Use $cos^2theta = 1 - sin^2theta$ and $tantheta = frac{sintheta}{costheta}=c$. Then $sintheta = csqrt{1-sin^2theta}$, then $sin^2theta = c^2-c^2sin^2theta$, and finally $sintheta = frac{c}{sqrt{1+c^2}}$. Note that when taking the roots I have taken the positive such since $sintheta>0$ and also $tantheta >0$, which implies $costheta > 0$.






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          Use $cos^2theta = 1 - sin^2theta$ and $tantheta = frac{sintheta}{costheta}=c$. Then $sintheta = csqrt{1-sin^2theta}$, then $sin^2theta = c^2-c^2sin^2theta$, and finally $sintheta = frac{c}{sqrt{1+c^2}}$. Note that when taking the roots I have taken the positive such since $sintheta>0$ and also $tantheta >0$, which implies $costheta > 0$.






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Use $cos^2theta = 1 - sin^2theta$ and $tantheta = frac{sintheta}{costheta}=c$. Then $sintheta = csqrt{1-sin^2theta}$, then $sin^2theta = c^2-c^2sin^2theta$, and finally $sintheta = frac{c}{sqrt{1+c^2}}$. Note that when taking the roots I have taken the positive such since $sintheta>0$ and also $tantheta >0$, which implies $costheta > 0$.






                            share|cite|improve this answer











                            $endgroup$



                            Use $cos^2theta = 1 - sin^2theta$ and $tantheta = frac{sintheta}{costheta}=c$. Then $sintheta = csqrt{1-sin^2theta}$, then $sin^2theta = c^2-c^2sin^2theta$, and finally $sintheta = frac{c}{sqrt{1+c^2}}$. Note that when taking the roots I have taken the positive such since $sintheta>0$ and also $tantheta >0$, which implies $costheta > 0$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jan 18 at 12:34

























                            answered Jan 18 at 12:27









                            lightxbulblightxbulb

                            1,169311




                            1,169311






























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