Tan[2arccos(5/13)] help

can somebody please help me understand how to answer the problem:
tan[2arccos(5/13)]
do I need to use the double angle formula?

thanks for your time!
-Mike

Hi.

Know your 5:12:13 triangles.
Since the cosine value of some angle is 5/13,
let arccos(5/13) = y

Rewriting we get:
cosy = 5/13

Thus, we obtain:
tan[2arccos(5/13)]
= tan[2( y )]<---- substituted arccos(5/13) with y.
= tan(2y)
= Yes, use the double angle formula for tangent
= 2tan(y) / (1 - tan²y)

Look at the triangle you drew where the cosine value at angle "y" is 5/13 (5 adjacent, 13 hypotenuse, 12 opposite, with reference angle "y")
Now find the tangent of y of the same triangle.
tany = 12/5

THus, we use tany = 12/5 and plug it into the double angle formula
= 2tan(y) / (1 - tan²y)
= 2(12/5) / (1 - (12/5)²)
= (24/5) / (1 - (144/25))
= (24/5) / ((25/25) - (144/25))
= (24/5) / (-119/25)
= (24/5)(-25/119)
= -120/119

Thus your tan[2arccos(5/13)] = -120/119

Hope this helps :D

Let me know if it's a bit confusing

Cosine = adjacent/hypotenuse = a/h

Write o for opposite side, and use Pythagoras

h^2 = o^2 + a^2 so o =12

So we want the tangent of the twice the angle whose cosine is 5/13

tan = sin/cos

cos 2x = cos^2 x - sin^2x = cos^2 x - 1 + cos^2 x

= 2*cos^2 x - 1 = 2*(5/13)^2 - 1 = (50 - 169)/169 = -119/169

sin 2x = 2*sin x * cos x

sin = o/h = 12/13

sin 2x = 2*(12/13)*(5/13) = 120/169

Thus tan = (120/169)/(-119/169) = -120/119 <<<<

tan[2arccos(5/13)]

ArcCos(5/13) results in an angle θ in quadrant I where
cosθ = (5/13) and
sinθ = (12/13)

tan(2θ) =
= sin(2θ) /cos(2θ)
= (2sinθcosθ) / (cos²θ - sin²θ)

= (2(12/13)(5/13)) / ((5/13)² - (12/13)²)

= (120/169) / (25/169 - 144/169)

= (120/169) / ( -119/169)

= 120 / -119

= - 120/119

Hope this helps you!

tan[2arccos(5/13)] = -120/119