hard but not impossible complex q. (1 Viewer)

[Slidey]

hi! hey i know this is pretty pathetic of me to like ask you for help on this. Well to be askign help off a stranger even. But yeah im sorta stuck. Could you please try and help me. even direct me

okay so i need help with Algebra. the question is can i simplify >>>>

arccos(sin(5pi/4)) ??? man this would help me alot if you know it! Thanks!

Yeah you can. And don't worry about it! It's good to ask questions!

First of all, sin(5pi/4) = -sin(pi/4), correct?

Now, arccos(-x)=pi-arccos(x), so:
arccos(sin(5pi/4))=pi-arccos(sin(pi/4))
(because cos(a+(2k+1)*pi)=cos(a)*cos((2k+1)pi)=cos(a)*-1) - basically a negative number produces an angle displaced by some odd multiple of pi)

Recall from 2unit trig:
let x=sin(pi/4)
sin^2@+cos^2@=1 (alternatively, derive this yourself with a triangle of hypotenuse 1 and opposite side 1/sqrt(2))
cos(pi/4)=sqrt(1-x^2)
So arccos(sin(5pi/4))=pi - arccos(sqrt(1-x^2))
= pi - arccos(1/sqrt(2)) = about 2.35 or so

Alternatively:
arccos(sin(5pi/4))=arccos(cos(pi/2-5pi/4))=arccos(cos(-3pi/4))=arccos(cos(3pi/4))=3pi/4

Edit: Fixed a typo.