Answer the Trig problems

Question

1. 2tan(theta) = x-2 ; Find cos(theta)
2. 2tan(theta) = 5 ; Find cos(theta)

Answer 1

2) 2tanT = 5

tanT = 5/2

tanT = opposite side over adjacent side

hypotenuse = +/-square root of sum of squares of the other two sides

= +/-sqr(5^2+2^2) =+/-sqr(25+4) =

+/-sqr29

tangents >0 in quadrants I and III

the angle Theta is in quadrant I or III

cosT = adjacent side over hypotenuse = 2/+/-sqr29 = +/-(sqr29)/29

cosine of Theta = +/-(sqr29)/29

cosT = -(sqr29)/29 or (sqr29)/29

1) 2tanT = x-2

tanT = (x-2)/2 = opposite side over adjacent side

hypotenuse = square root of sum of other sides squared

= sqr((x-2)^2 +2^2) = sqr(x^2-4x-4+4)

= +/-sqr(x^2 -4x)

cosT = adjacent side over hypotenuse

= 2/(+/-sqr(x^2-4x))

= +/-2sqr(x^2-4x)/(x^2-4)

or

T = tan^-1((x-2)/2)

cosT = cos(tan^-1((x-2)/2))