I am working on a trig problem:
Determine the values of cos Θ and tan Θ if sin Θ = m/n, a negative fraction. Since sin Θ is negative, Θ is in quadrant III or IV. Since sin = m/n, y = m and r = n. n = - √n^2 - m^2
My problem is with quadrant III, both x and y should be negative, thus
tan ɵ = - m / -√n^2 - m^2 = m √n^2 - m^2/ n^2 - m^2
Schaum's gives the following as tan Θ:
- m √n^2 m^2 / n^2 - m^2
I don't understand where they get the -m in the numerator. Any help is appreciated.
Determine the values of cos Θ and tan Θ if sin Θ = m/n, a negative fraction. Since sin Θ is negative, Θ is in quadrant III or IV. Since sin = m/n, y = m and r = n. n = - √n^2 - m^2
My problem is with quadrant III, both x and y should be negative, thus
tan ɵ = - m / -√n^2 - m^2 = m √n^2 - m^2/ n^2 - m^2
Schaum's gives the following as tan Θ:
- m √n^2 m^2 / n^2 - m^2
I don't understand where they get the -m in the numerator. Any help is appreciated.
-
Since sinΘ= m/n in quadrants III and IV, then m<0 and n>0
Y= m, r= n , and x= -sqr(n^2-m^2) in quadrant III, and x= sqr(n^2-m^2) in quadrant IV.
Then tanΘ= m/[-sqr(n^2-m^2)] in Q. III
Now you can rationalize and pull up the negative sign.
tan Θ= m/[ sqr(n^2-m^2)] in Q. IV
Hoping this helps!
Y= m, r= n , and x= -sqr(n^2-m^2) in quadrant III, and x= sqr(n^2-m^2) in quadrant IV.
Then tanΘ= m/[-sqr(n^2-m^2)] in Q. III
Now you can rationalize and pull up the negative sign.
tan Θ= m/[ sqr(n^2-m^2)] in Q. IV
Hoping this helps!