Asked by jill
Given that α is in Quadrant 1 and tan(α)=67, give an exact answer for the following:
sin(2α)=
cos(2α)=
tan(2α)=
sin(2α)=
cos(2α)=
tan(2α)=
Answers
Answered by
Damon
is 67 a typo or is it really that close to the y axis?
anyway find sin α and cos α from the tangent then
given sin α / cos α = k (here k = 67)
so sin α = k cos α
sin 2α = 2 sin α cos α = 2 k cos^2 α
cos 2α = cos^2 α - sin^2 α
tan 2α = sin 2α / cos2α
anyway find sin α and cos α from the tangent then
given sin α / cos α = k (here k = 67)
so sin α = k cos α
sin 2α = 2 sin α cos α = 2 k cos^2 α
cos 2α = cos^2 α - sin^2 α
tan 2α = sin 2α / cos2α
Answered by
jill
Its a typo. It's suppose to be 6/7 .
Answered by
Damon
well that seriously helps
opposite/adjacent = 6/7
then hypotenuse = sqrt (36 + 49) = sqrt 85
so sin α = 6/sqrt 85
and cos α = 7/sqrt85
use those in the equations I gave you
opposite/adjacent = 6/7
then hypotenuse = sqrt (36 + 49) = sqrt 85
so sin α = 6/sqrt 85
and cos α = 7/sqrt85
use those in the equations I gave you
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