Question

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`

Answer 1

(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`

Answer 2

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`