Given that $cot(m)=0.75$ and $cos(m)<0$, what is the value of $sin(m)$?

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One of my homework problems is "Given that $cot(m)=0.75$ and $cos(m)<0$, what is the value of $sin(m)$?" I keep getting $sin m=frac-45=-.8$ which isn't an option



My options are:

A- $ -.5625$

B-$ -1.25$

C-$ -.25$

D-$ -.8$

E- None of the above



Edit: Because cotangent is positive and cosine is negative I know the angle is in the 3rd quadrant, I then used cotangent to get two of the dimensions, $3$ as the adjacent, and $4$ as the opposite. This gave me $5$ as the hypotenuse. Since sine is opposite divided by hypotenuse, I got $sin(m)=sin(4/5)=-.8$




algebra-precalculus trigonometry




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  • Try drawing a unit circle, and explicitly draw the angle out, double check the definitions of trig functions. Please explain why you get $sin (-4/5)$, as I cannot imagine how.
    – Trebor
    Aug 19 at 7:20










  • I think OP means $sin m= -0.8$
    – Mohammad Zuhair Khan
    Aug 19 at 7:26






  • 1




    isn't $-frac45 = -frac810 = -0.8$?
    – Ronald
    Aug 19 at 7:32






  • 2




    You should have written $sin m = -frac45 = -0.8$.
    – N. F. Taussig
    Aug 19 at 7:35






  • 1




    You should also change the final line. However, the key point is that your revised answer is option D.
    – N. F. Taussig
    Aug 19 at 7:40














up vote
1
down vote

favorite












One of my homework problems is "Given that $cot(m)=0.75$ and $cos(m)<0$, what is the value of $sin(m)$?" I keep getting $sin m=frac-45=-.8$ which isn't an option



My options are:

A- $ -.5625$

B-$ -1.25$

C-$ -.25$

D-$ -.8$

E- None of the above



Edit: Because cotangent is positive and cosine is negative I know the angle is in the 3rd quadrant, I then used cotangent to get two of the dimensions, $3$ as the adjacent, and $4$ as the opposite. This gave me $5$ as the hypotenuse. Since sine is opposite divided by hypotenuse, I got $sin(m)=sin(4/5)=-.8$




algebra-precalculus trigonometry




share|cite|improve this question






















  • Try drawing a unit circle, and explicitly draw the angle out, double check the definitions of trig functions. Please explain why you get $sin (-4/5)$, as I cannot imagine how.
    – Trebor
    Aug 19 at 7:20










  • I think OP means $sin m= -0.8$
    – Mohammad Zuhair Khan
    Aug 19 at 7:26






  • 1




    isn't $-frac45 = -frac810 = -0.8$?
    – Ronald
    Aug 19 at 7:32






  • 2




    You should have written $sin m = -frac45 = -0.8$.
    – N. F. Taussig
    Aug 19 at 7:35






  • 1




    You should also change the final line. However, the key point is that your revised answer is option D.
    – N. F. Taussig
    Aug 19 at 7:40












up vote
1
down vote

favorite









up vote
1
down vote

favorite











One of my homework problems is "Given that $cot(m)=0.75$ and $cos(m)<0$, what is the value of $sin(m)$?" I keep getting $sin m=frac-45=-.8$ which isn't an option



My options are:

A- $ -.5625$

B-$ -1.25$

C-$ -.25$

D-$ -.8$

E- None of the above



Edit: Because cotangent is positive and cosine is negative I know the angle is in the 3rd quadrant, I then used cotangent to get two of the dimensions, $3$ as the adjacent, and $4$ as the opposite. This gave me $5$ as the hypotenuse. Since sine is opposite divided by hypotenuse, I got $sin(m)=sin(4/5)=-.8$




algebra-precalculus trigonometry




share|cite|improve this question














One of my homework problems is "Given that $cot(m)=0.75$ and $cos(m)<0$, what is the value of $sin(m)$?" I keep getting $sin m=frac-45=-.8$ which isn't an option



My options are:

A- $ -.5625$

B-$ -1.25$

C-$ -.25$

D-$ -.8$

E- None of the above



Edit: Because cotangent is positive and cosine is negative I know the angle is in the 3rd quadrant, I then used cotangent to get two of the dimensions, $3$ as the adjacent, and $4$ as the opposite. This gave me $5$ as the hypotenuse. Since sine is opposite divided by hypotenuse, I got $sin(m)=sin(4/5)=-.8$





algebra-precalculus trigonometry





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share|cite|improve this question




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edited Aug 20 at 18:43









thesmallprint

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2,2341617










asked Aug 19 at 7:14









Pagaley 12

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  • Try drawing a unit circle, and explicitly draw the angle out, double check the definitions of trig functions. Please explain why you get $sin (-4/5)$, as I cannot imagine how.
    – Trebor
    Aug 19 at 7:20










  • I think OP means $sin m= -0.8$
    – Mohammad Zuhair Khan
    Aug 19 at 7:26






  • 1




    isn't $-frac45 = -frac810 = -0.8$?
    – Ronald
    Aug 19 at 7:32






  • 2




    You should have written $sin m = -frac45 = -0.8$.
    – N. F. Taussig
    Aug 19 at 7:35






  • 1




    You should also change the final line. However, the key point is that your revised answer is option D.
    – N. F. Taussig
    Aug 19 at 7:40
















  • Try drawing a unit circle, and explicitly draw the angle out, double check the definitions of trig functions. Please explain why you get $sin (-4/5)$, as I cannot imagine how.
    – Trebor
    Aug 19 at 7:20










  • I think OP means $sin m= -0.8$
    – Mohammad Zuhair Khan
    Aug 19 at 7:26






  • 1




    isn't $-frac45 = -frac810 = -0.8$?
    – Ronald
    Aug 19 at 7:32






  • 2




    You should have written $sin m = -frac45 = -0.8$.
    – N. F. Taussig
    Aug 19 at 7:35






  • 1




    You should also change the final line. However, the key point is that your revised answer is option D.
    – N. F. Taussig
    Aug 19 at 7:40















Try drawing a unit circle, and explicitly draw the angle out, double check the definitions of trig functions. Please explain why you get $sin (-4/5)$, as I cannot imagine how.
– Trebor
Aug 19 at 7:20




Try drawing a unit circle, and explicitly draw the angle out, double check the definitions of trig functions. Please explain why you get $sin (-4/5)$, as I cannot imagine how.
– Trebor
Aug 19 at 7:20












I think OP means $sin m= -0.8$
– Mohammad Zuhair Khan
Aug 19 at 7:26




I think OP means $sin m= -0.8$
– Mohammad Zuhair Khan
Aug 19 at 7:26




1




1




isn't $-frac45 = -frac810 = -0.8$?
– Ronald
Aug 19 at 7:32




isn't $-frac45 = -frac810 = -0.8$?
– Ronald
Aug 19 at 7:32




2




2




You should have written $sin m = -frac45 = -0.8$.
– N. F. Taussig
Aug 19 at 7:35




You should have written $sin m = -frac45 = -0.8$.
– N. F. Taussig
Aug 19 at 7:35




1




1




You should also change the final line. However, the key point is that your revised answer is option D.
– N. F. Taussig
Aug 19 at 7:40




You should also change the final line. However, the key point is that your revised answer is option D.
– N. F. Taussig
Aug 19 at 7:40










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By looking at the signs of $cot(m)$ and $cos(m)$ you can see from this link that $sin(m)$ should be negative. It's great that you used 3 and 4 to get a 5 for the hypotenuse and get the 0.8 but I suggest you always beware of the differences between angle measures (or arc length measures) and line segment length measures. The trigonometric functions always use angles as arguments, so in $sin(theta)$, $theta$ is an angle. When we are using formulas for calculating the value of a trigonometric function in an angle then we are using lengths of line segments. So, if we say $sin(theta) = opposite over hypotenuse$ then $theta$ is an angle but opposite and hypotenuse are length of line segments. Bottom line: lengths of line segments or rations of line segments (like $opposite over hypotenuse$) should never be evaluated in a trigonometric function.



I suggest you check the graphs of trigonometric functions to get a feeling of how these functions behave with different values. I find this much more insightful than checking the tables of quadrants.






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    By looking at the signs of $cot(m)$ and $cos(m)$ you can see from this link that $sin(m)$ should be negative. It's great that you used 3 and 4 to get a 5 for the hypotenuse and get the 0.8 but I suggest you always beware of the differences between angle measures (or arc length measures) and line segment length measures. The trigonometric functions always use angles as arguments, so in $sin(theta)$, $theta$ is an angle. When we are using formulas for calculating the value of a trigonometric function in an angle then we are using lengths of line segments. So, if we say $sin(theta) = opposite over hypotenuse$ then $theta$ is an angle but opposite and hypotenuse are length of line segments. Bottom line: lengths of line segments or rations of line segments (like $opposite over hypotenuse$) should never be evaluated in a trigonometric function.



    I suggest you check the graphs of trigonometric functions to get a feeling of how these functions behave with different values. I find this much more insightful than checking the tables of quadrants.






    share|cite|improve this answer


























      up vote
      0
      down vote













      By looking at the signs of $cot(m)$ and $cos(m)$ you can see from this link that $sin(m)$ should be negative. It's great that you used 3 and 4 to get a 5 for the hypotenuse and get the 0.8 but I suggest you always beware of the differences between angle measures (or arc length measures) and line segment length measures. The trigonometric functions always use angles as arguments, so in $sin(theta)$, $theta$ is an angle. When we are using formulas for calculating the value of a trigonometric function in an angle then we are using lengths of line segments. So, if we say $sin(theta) = opposite over hypotenuse$ then $theta$ is an angle but opposite and hypotenuse are length of line segments. Bottom line: lengths of line segments or rations of line segments (like $opposite over hypotenuse$) should never be evaluated in a trigonometric function.



      I suggest you check the graphs of trigonometric functions to get a feeling of how these functions behave with different values. I find this much more insightful than checking the tables of quadrants.






      share|cite|improve this answer
























        up vote
        0
        down vote










        up vote
        0
        down vote









        By looking at the signs of $cot(m)$ and $cos(m)$ you can see from this link that $sin(m)$ should be negative. It's great that you used 3 and 4 to get a 5 for the hypotenuse and get the 0.8 but I suggest you always beware of the differences between angle measures (or arc length measures) and line segment length measures. The trigonometric functions always use angles as arguments, so in $sin(theta)$, $theta$ is an angle. When we are using formulas for calculating the value of a trigonometric function in an angle then we are using lengths of line segments. So, if we say $sin(theta) = opposite over hypotenuse$ then $theta$ is an angle but opposite and hypotenuse are length of line segments. Bottom line: lengths of line segments or rations of line segments (like $opposite over hypotenuse$) should never be evaluated in a trigonometric function.



        I suggest you check the graphs of trigonometric functions to get a feeling of how these functions behave with different values. I find this much more insightful than checking the tables of quadrants.






        share|cite|improve this answer














        By looking at the signs of $cot(m)$ and $cos(m)$ you can see from this link that $sin(m)$ should be negative. It's great that you used 3 and 4 to get a 5 for the hypotenuse and get the 0.8 but I suggest you always beware of the differences between angle measures (or arc length measures) and line segment length measures. The trigonometric functions always use angles as arguments, so in $sin(theta)$, $theta$ is an angle. When we are using formulas for calculating the value of a trigonometric function in an angle then we are using lengths of line segments. So, if we say $sin(theta) = opposite over hypotenuse$ then $theta$ is an angle but opposite and hypotenuse are length of line segments. Bottom line: lengths of line segments or rations of line segments (like $opposite over hypotenuse$) should never be evaluated in a trigonometric function.



        I suggest you check the graphs of trigonometric functions to get a feeling of how these functions behave with different values. I find this much more insightful than checking the tables of quadrants.







        share|cite|improve this answer














        share|cite|improve this answer



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        edited Aug 20 at 18:21

























        answered Aug 19 at 8:00









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