Question

If `cos x=11/12` then find the value of `tan2x`?

Answer 1

`tan2x= = (11sqrt(23))/49`

Explanation:

Consider the following right angle triangle:

By elementary trigonometry we have:

`cos theta = "adj"/"hyp" = 11/12 => theta = x`

By Pythagoras:

`\ \ \ 12^2 = 11^2+h^2`
`:. h^2 = 144-121`
`:. h^2 = 23`
`:. \ \h = sqrt(23)`

And so;

`tan x = "opp"/"adj" = h/11 = sqrt(23)/11`

Using the identity `tan(2theta) -= (2tan theta)/(1-tan^2theta)` we have:
`tan(2x) = (2tan x)/(1-tan^2 x)`

`" " = (2sqrt(23)/11)/(1-(sqrt(23)/11)^2)`

`" " = (2sqrt(23)/11)/(1-23/121)`

`" " = (2sqrt(23)/11)/(98/121)`

`" " = (2sqrt(23))/11 * 121/98`

`" " = (11sqrt(23))/49`

For comparison, to verify the solution, if we use a calculator:

`\ \ \ \ \ \ \ cos x = 11/12 => x = 23.556^o`
`:. tan 2x= 1.0766`; and `(11sqrt(23))/49 = 1.0766`

Answer 2

`tan(2x) = +-(11sqrt(23))/49`

Explanation:

`cosx = 11/12`

Use the identity `cos2x = 2cos^2x-1`:

`cos2x = 2*121/144 -1 = (242-144)/144 = 98/144`

Use now the identity:

`tan^2(2x) = sin^2(2x)/cos^2(2x) = (1-cos^2(2x))/cos^2(2x) =1/cos^2(2x)-1`

so:

`tan^2(2x) = (144)^2/(98)^2 -1 = (20736-9604)/9604 =11132/9604`

Then:

`tan(2x) = +-sqrt(11132/9604) = +- sqrt(2*2*11*11*23)/98 = +-22sqrt(23)/98 =+-11/49sqrt(23)`