Find the exact value of cos(2θ)and sin(2θ) if tan(θ) = (-5/12) and 270° < θ < 360

Answer 1

Use double angle identities:

sin 2θ = 2 tan θ / ( 1 + tan² θ )

cos 2θ = ( 1 - tan² θ ) / ( 1 + tan² θ )

In this case:

sin 2θ = 2 ∙ ( - 5 / 12 ) / [ 1 + ( - 5 / 12 )² ] =

- 2 ∙ 5 / 12 / ( 1 + 25 / 144 ) = - 10 / 12 / ( 144 / 144 + 25 / 144 ) =

- 10 / 12 / 169 / 144 = - 10 ∙ 144 / 12 ∙ 169 =

- 10 ∙ 12 ∙ 12 / 12 ∙ 169 = - 10 ∙ 12 / 169 = - 120 / 169

cos 2θ = ( 1 - tan² θ ) / ( 1 + tan² θ )

cos 2θ = [ 1 - ( - 5 / 12 )² ] / [ 1 + ( - 5 / 12 )² ] =

( 1 - 25 / 144 ) / ( 1 + 25 / 144 ) =

( 144 / 144 - 25 / 144 ) / ( 144 / 144 + 25 / 144 ) =

119 / 144 / 169 / 144 = 144 ∙ 119 / 144 ∙ 169 = 119 / 169

Answer 2

tanθ = -5/12 and θ is in the fourth quadrant
tanθ = y/x and in IV, x = 12 and y = -5
r^2 = x^2 + y^2 = 144+25 = 169
So r = 13
sinθ = -5/13 and cosθ = 12/13

sin 2θ = 2sinθcosθ = 2(-5/13)(12/13) = -120/169
cos 2θ = cos^2 θ - sin2 θ = 144/169 - 25/169 = 119/169