Question 1169169: Find tan, if sec=4, sin>0
Another question is if tan=7/4 and sin<0; what is csc and sin
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! problem number 1:
find tan(x) if sec(x) = 4 and sine is greater than 0.
tan(x) = sin(x) / cos(x)
sec(x) = 4
cos(x) = 1/sec(x) = 1/4
sin^2(x) + cos^2(x) = 1
sin^2(x) = 1 - cos^2(x) = 1 - (1/4)^2 = 1 - 1/16 = 15/16
sin(x) = sqrt(15/16) = sqrt(15)/4
tan(x) = (sqrt(15)/4)/(1/4) = (sqrt(15)/4)*4 = sqrt(15).
sin(x) is positive in the first and second quadrant.
in the first quadrant, sin(x) is positive and cos(x) is positive, therefore:
tan(x) = sqrt(15) in the first quadrant.
in the second quadrant, sin(x) is positive and cos(x) is negative, therefore:
tan(x) = -sqrt(15) in the second quadrant.
your solution appears to be that:
tan(x) = sqrt(15) when x is in the first quadrant and tan(x) = -sqrt(15) when x is in the second quadrant.
since sin(x) is only positive in the first and second quadrant, then those quadrants are not considered.
problem number 2:
find sin(x) and csc(x) if tan(x) = 7/4) and sine is less than 0.
tan(x) = (7/4)
tangent function = opposite side / adjacent side.
when tan(x) = 7/4, opposite side = 7 and adjacent side = 4.
hypotenuse = sqrt(7^2 + 4^2) = sqrt(65)
sin(x) = opposite side / hypotenuse = 7/sqrt(65)
tan(x) is positive in the first and third quadrants only.
in the first quadrant, sin(x) is positive, so first quadrant is not good.
in the third quadrant, sin(x) is negative, so third quadrant is good.
therefore, tan(x) = 7/4 in the third quadrant only if sin(x) has to be less than 0.
in the third quadrant, when tan(x) = 7/4, .....
sin(x) = -7/sqrt(65)
csc(x) = 1/sin(x) = 1/(-7/sqrt(65) = -sqrt(65)/7.
your solution appears to be that:
when tan(x) = 7/4) and sine is less than 0, then sin(x) = -7/sqrt(65) and csc(x) = -sqrt(65)/7
let me know if you have any questions or concerns about these answers.
theo
|
|
|
