#sin x = 1/4# . Find cos x
#cos^2 x = 1 - sin^2 x = 1 - 1/16 = 15/16#
#cos x = +- sqrt15/4#
There are 2 values of cos x because if #sin x = 1/4#, x could either be in Quadrant 1 or in Quadrant 2
Use trig identity:
#2cos^2 (x/2) = 1 - cos 2a#
In this case:
#cos^2 (x/2) = 1/2 +- sqrt15/8 = 1/2 +- 0.484#
a. #cos^2 (x/2) = 0.984#
b. #cos^2 (x/2) = 0.016#
a. #cos (x/2) = sqrt(0.984) = 0.99#
b. #cos (x/2) = sqrt(0.016) = 0.127#
Check with calculator.
a. #cos (x/2) = 0.99# --> #x/2 = 7^@27# --> #x = 14^@53#
#sin (14^@52) = 0.25#. Proved
b. #cos (x/2) = 0.127# --> #x/2 = 82^@70# --> #x = 165^@41#
#sin (165^@41) = 0.25#. Proved
