Need help with a trig identity problem?

Question

The problem is: Use a trig identity to determine the value of cos2x when.... tan x= (4x) / ((x^2) - 4) . Any help would be great. Thanks

Answer 1

From tan x definition this means in aright angled triangle

opposite = 4x
adjacent = x^2 - 4

Hypotenuse^2 = 16x^2 + (x^2 - 4)^2 = x^4 + 8x^2 + 16 =(x^2 + 4)^2

Hypotenuse = x^2 + 4

Therefore

Sin x = 4x/(x^2 + 4) and Cos x = (x^2 - 4)/(x^2 + 4)

Now Cos 2x = Cos^2 x - Sin^2 x

= ((x^2 - 4)^2 - 16x^2)/(x^2 + 4)^2

I can't see any easy simplification this. Check the maths though - I did this on the run.

Answer 2

tan x = (4x) / (x² - 4)

sin²x + cos²x = (4x)²/h² + (x² - 4)²/h² = 1
(4x)² + (x² - 4)² = h²
16x² + x^4 - 8x² + 16 = h²
x^4 + 8x² + 16 = h²
(x² + 4)² = h²
±(x² + 4) = h

cos(2x) = 1 - 2sin²x = 1 - 2[(4x)/(x² + 4)]² = 1 - 32x²/(x² + 4)²

Answer 3

tan x = sin x/cos x
So, sin x = 4x and cos x = x^2 - 4
cos 2x = 1 - (sin x)^2 the identity
and cos 2x = 1 - (4x)^2 = 1 - 16x^2

Answer 4

Where's the other side of the equation ?

Answer 5

This is my answer:

http://img402.imageshack.us/img402/8092/respuesta8ug4.gif