tangent of theta = u/sqrt 2 = opposite/adjacent
let u = 1
then v = sqrt 2 = adjacent leg
and hypotenuse = sqrt (1 + 2) = sqrt 3
then cos theta = sqrt(2/3)
Write the following as an algebraic expression in u, u>0.
cos(arctan(u/sqrt2))
2 answers
But Damon, that would only be the case when u = 1
e.g. when u = 4, we have tanØ = 4/√2
r^2 = 16 + 2 = 18
r = √18
then cosØ = √2/√18 = 1/3
in general case
r^2 = 2 + u^2
r = √(2 + u^2)
cos(arctan(u/√2)) = √2/√(2+u^2)
e.g. when u = 4, we have tanØ = 4/√2
r^2 = 16 + 2 = 18
r = √18
then cosØ = √2/√18 = 1/3
in general case
r^2 = 2 + u^2
r = √(2 + u^2)
cos(arctan(u/√2)) = √2/√(2+u^2)
1 answer