How do you find the value for #sin(arccos(-1/3))#?

2 Answers
Jun 11, 2015

Use calculator to evaluate #theta = arccos(-1/3)#, allow for #theta+pi# (since #cos(theta) = cos(theta+pi)#); use calculator to evaluate the two values of theta (within the #0# to #2pi# range.

Explanation:

Using caluclator
#color(white)("XXXX")##arccos(-1/3) = 1.910633#
#color(white)("XXXX")##color(white)("XXXX")#i.e #cos(1.910633) = -1/3#
#color(white)("XXXX")#note that
#color(white)("XXXX")##color(white)("XXXX")##cos(1.910633+pi)# also #= -1/3#

Using calculator evaluate:
#color(white)("XXXX")##sin(1.910633) = 0.942809#
and
#color(white)("XXXX")##sin(1.910633+pi) = sin(4.372552) = -0.942809#

Jun 11, 2015

Find# sin (arccos (-1/3))#

Explanation:

#cos x = -1/3 #--> arc x?

Calculator gives #x = +- 109.47#.
Now, find #sin (+- 109.47)#

sin (109.47) = 0.94
sin (-109.47) = -0.94