Find the value of?

(1) #arccos(-sqrt3/2)-arcsin(-sqrt3/2)-arccos(1/2)+arcsin(sqrt3/2)#
(2) #arcsin(-1/2)+arcsin(-sqrt3/2)#
(3) #sin(arccos(-sqrt3/2))=sin150^o#

2 Answers
Dec 15, 2016

Answer:

(1) #210^o=(7pi)/12#

(2) #-90^o=-pi/2#

(3) #sin(arccos(-sqrt3/2))=1/2#

Explanation:

The trigonometric ratios of standard angles are given in

https://www.youtube.com/watch?v=RKETb3BzI6A
However, before we use this let us remember that range for inverse trigonometric functions are - #[-pi/2.pi/2]# for arcsin, arccsc, arctan and arccot, while for arccos and arcsec tange is #[0,p]#.

Considering this we solve above as follows:

(1) #arccos(-sqrt3/2)-arcsin(-sqrt3/2)-arccos(1/2)+arcsin(sqrt3/2)#

= #150^o-(-60^o)-60^o +60^o#

= #150^o +60^o-60^o +60^o=210^o# or #(7pi)/12#

(2) #arcsin(-1/2)+arcsin(-sqrt3/2)#

= #-30^o-60^o=-90^o=-pi/2#

(3) #sin(arccos(-sqrt3/2))=sin150^o=1/2#

Dec 15, 2016

google image

To solve this type of problem we are to remember the range of inverse trigonometric function as shown in above figure

1)
#arccos(-sqrt3/2)-arcsin(-sqrt2/2)-arccos(1/2)+arcsin(sqrt3/2)#

#=arccos(-cos(pi/6))-arcsin(-sin(pi/4))-arccos(cos(pi/3))+arcsin(sin(pi/3))#

#=arccos(cos(pi-pi/6))-arcsinsin(-pi/4)-arccos(cos(pi/3))+arcsin(sin(pi/3))#

#=(5pi)/6+pi/4-pi/3+pi/3#

#=(10pi+3pi)/12=(13pi)/12#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

2).#arcsin(-1/2)+arcsin(-sqrt3/2)#

#=arcsin(sin(-pi/6))+arcsin(sin(-pi/3))#

#=-pi/6-pi/3=-pi/2#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

3).#sin(arccos(-√3/2))#

#=sin(arccos(cos(pi-pi/6))#

#=sin(pi-pi/6)=sin(pi/6)=1/2#