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

2 Answers
Mar 23, 2017

Answer:

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

Explanation:

Consider the following right angle triangle:

enter image source here

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 #

Mar 23, 2017

Answer:

#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)#