How do you find the exact value for #sin (cos^-1 (-1/4))#?

1 Answer
Apr 15, 2015

#cos^-1(-1/4)# is a #t# in #[0, pi]# with #cos t = -1/4#

Since #t# is in quadrant 1 or 2, #sint# is positive.

The problem becomes: #cost=-1/4# and #sint >0#. Find #sint#.

Either draw a picture of #t# on the coordinate system,
or draw a right triangle with acute angle #t#, adjacent side 1 and hypotenuse 4
or use #sin^2t+cos^2 = 1# to get

#sint = sqrt15/4#