How do you find the values of #sin 2theta# and #cos 2theta# when #cos theta = 12/13#?

2 Answers
Aug 4, 2018

Answer:

Below

Explanation:

#theta# can be in the first quadrant #0<=theta<=90# or the fourth quadrant #270<=theta<=360#

If #theta# is in the first quadrant,
then
#sintheta=5/13#
#costheta=12/13#
#tantheta=5/12#

Therefore,
#sin2theta=2sinthetacostheta=2times5/13times12/13=120/169#

#cos2theta=cos^2theta-sin^2theta=(12/13)^2-(5/13)^2=144/169-25/169=119/169#

If #theta# is in the fourth quadrant,
then
#sintheta=-5/13#
#costheta=12/13#
#tantheta=-5/12#

Therefore,
#sin2theta=2sinthetacostheta=2times-5/13times12/13=-120/169#

#cos2theta=cos^2theta-sin^2theta=(12/13)^2-(-5/13)^2=144/169-25/169=119/169#

Aug 4, 2018

Answer:

#sin 2theta = 120/169, theta in Q_1 and - 120/169, theta in Q_4#.

#cos 2theta = 119/169#

Explanation:

See my answer in

https://socratic.org/questions/if-cos-a-5-13-how-do-you-find-sina-and-tana

As a continuation,

#sin theta = 5/13, theta in Q_1# and it is #- 5/13, theta in Q_4#.

#sin 2theta = 2 sin theta cos theta = 2 ( 5/13)(12/13)#

#= 120/169, theta in Q_1 and#

#= 2 ( -5/13)(12/13) = - 120/169, theta in Q_4#.

#cos 2theta = cos^2theta - sin^2theta = (12/13)^2 - (+-5/13)^2#

#= 119/169#