Question #3d0cd

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Answer 1 · 2017-04-06T15:39:45

#cos theta = - 3/( sqrt 10)#

#sin theta = +1 /sqrt 10#

#tan theta = -1/3#

#cos 2theta = 4/5#

a. #cos^2 theta - sin^2 theta = 4/5#

#cos^2 theta - (1 -cos^2 theta) = 4/5#

#2cos^2 theta - 1 = 4/5#

#2cos^2 theta = 4/5 + 1 = 9/5#

#cos^2 theta = 9/10#

#cos theta = -sqrt (9/10) = - 3/( sqrt 10)#-> quadrant II

b. #cos^2 theta - sin^2 theta = 4/5#

#(1- sin^2 theta) - sin^2 theta = 4/5#

#(1 - 2sin^2 theta) = 4/5#

#(1 - 4/5) = 2sin^2 theta#

#2sin^2 theta = 1/5#

#sin^2 theta = 1/10#

#sin theta = +sqrt(1/10) = +1 /sqrt 10#-> quadrant II

c. #tan theta = sin theta/cos theta = +1 /sqrt 10/- 3/( sqrt 10)#

#tan theta = -1/3#-> quadrant II