Question

If cos θ=4/5 on the interval (3pi/2, 2pi) find the exact value of tan 2θ?

If you could show work that'd be great!

Answer 1

`Tan2theta= -24/7`

Explanation:

Considering this is in the fourth quadrant:
Cos is positive and sin is negative

The measure of the other leg: `3`
Why? Pythagorean triple of 3,4,5
But the 3 will be negative because it is supposed to represent the sin value

So: `Tantheta= -3/4`

Tangent double angle identity:
`Tan2theta= (2tantheta)/(1-tan^2theta)`

Therefore:
`Tan2theta= (2(-3/4))/(1-(-3/4)^2)`

`Tan2theta= (-6/4)/(16/16-9/16)`

`Tan2theta= (-6/4)/(7/16)`

`Tan2theta= (-6/4)*(16/7)= -96/28= -24/7`

Answer 2

Given `costheta=4/5 and theta in ((3pi)/2,2pi)i.e.Q_4`

So `costheta and Sec theta->+ve`

But `tantheta->-ve`

Now `tantheta=-sqrt(sec^2theta-1)=-sqrt((5/4)^2-1)=-3/4`

Now

`tan2theta=(2tantheta)/(1-tan^2theta)=(2xx(-3/4))/(1-(-3/4)^2)`

`=-2xx3/4xx16/7=-24/7`