How do you simplify #sin(sin^-1(1/2)+cos^-1(3/5))#?

1 Answer
Alan P.
Jun 18, 2016

#sin(sin^(-1)(1/2)+cos^(-1)(3/5))=color(green)(1.3)#

Explanation:

#sin^(-1)# is by definition an angle in #[0,pi)#
within this interval #sin(sin^(-1)(theta))=theta#
#color(white)("XXX")rarr sin(sin^(-1)(1/2))=1/2#

#cos^(-1)# is by definition an angle in #(-pi/2,pi/2]#
within this range a #cos=y/r# of #3/5# implies a standard triangle with
#color(white)("XXX")x/r=sin=4/5#
So #cos^(-1)(3/5)=sin^(-1)(4/5)#
and, again, since this is in QI,
#color(white)("XXX")sin(cos^(-1)(3/5))=sin(sin^(-1)(4/5))=4/5#

Therefore
#sin(sin^(-1)(1/2)+cos^(-1)(3/5))#
#color(white)("XXX")=1/2+4/5#

#color(white)("XXX")=13/10 (=1 3/10 = 1.3)#

Related questions
See all questions in Basic Inverse Trigonometric Functions
Impact of this question
4109 views around the world
You can reuse this answer
Creative Commons License