Inverse trigonometric functions
Transforming equations

The equations above can be transformed by using the reflection and shift identities:

These formulas imply, in particular, that the following hold:

sin θ = − sin (− θ) = − sin (π + θ) = − sin (π − θ) = − cos (π 2 + θ) = − cos (π 2 − θ) = − cos (− π 2 − θ) = − cos (− π 2 + θ) = − cos (3 π 2 − θ) = − cos (− 3 π 2 + θ) cos θ = − cos (− θ) = − cos (π + θ) = − cos (π − θ) = − sin (π 2 + θ) = − sin (π 2 − θ) = − sin (− π 2 − θ) = − sin (− π 2 + θ) = − sin (3 π 2 − θ) = − sin (− 3 π 2 + θ) tan θ = − tan (− θ) = − tan (π + θ) = − tan (π − θ) = − cot (π 2 + θ) = − cot (π 2 − θ) = − cot (− π 2 − θ) = − cot (− π 2 + θ) = − cot (3 π 2 − θ) = − cot (− 3 π 2 + θ) {\displaystyle {\begin{aligned}\sin \theta &=-\sin(-\theta)&&=-\sin(\pi +\theta)&&={\phantom {-}}\sin(\pi -\theta)\\&=-\cos \left({\frac {\pi }{2}}+\theta \right)&&={\phantom {-}}\cos \left({\frac {\pi }{2}}-\theta \right)&&=-\cos \left(-{\frac {\pi }{2}}-\theta \right)\\&={\phantom {-}}\cos \left(-{\frac {\pi }{2}}+\theta \right)&&=-\cos \left({\frac {3\pi }{2}}-\theta \right)&&=-\cos \left(-{\frac {3\pi }{2}}+\theta \right)\\\cos \theta &={\phantom {-}}\cos(-\theta)&&=-\cos(\pi +\theta)&&=-\cos(\pi -\theta)\\&={\phantom {-}}\sin \left({\frac {\pi }{2}}+\theta \right)&&={\phantom {-}}\sin \left({\frac {\pi }{2}}-\theta \right)&&=-\sin \left(-{\frac {\pi }{2}}-\theta \right)\\&=-\sin \left(-{\frac {\pi }{2}}+\theta \right)&&=-\sin \left({\frac {3\pi }{2}}-\theta \right)&&={\phantom {-}}\sin \left(-{\frac {3\pi }{2}}+\theta \right)\\\tan \theta &=-\tan(-\theta)&&={\phantom {-}}\tan(\pi +\theta)&&=-\tan(\pi -\theta)\\&=-\cot \left({\frac {\pi }{2}}+\theta \right)&&={\phantom {-}}\cot \left({\frac {\pi }{2}}-\theta \right)&&={\phantom {-}}\cot \left(-{\frac {\pi }{2}}-\theta \right)\\&=-\cot \left(-{\frac {\pi }{2}}+\theta \right)&&={\phantom {-}}\cot \left({\frac {3\pi }{2}}-\theta \right)&&=-\cot \left(-{\frac {3\pi }{2}}+\theta \right)\\\end{aligned}}}

where swapping sin ↔ csc, {\displaystyle \sin \leftrightarrow \csc,} swapping cos ↔ sec, {\displaystyle \cos \leftrightarrow \sec,} and swapping tan ↔ cot {\displaystyle \tan \leftrightarrow \cot } gives the analogous equations for csc, sec, and cot, {\displaystyle \csc,\sec,{\text{ and }}\cot,} respectively.

So for example, by using the equality sin (π 2 − θ) = cos θ, {\textstyle \sin \left({\frac {\pi }{2}}-\theta \right)=\cos \theta,} the equation cos θ = x {\displaystyle \cos \theta =x} can be transformed into sin (π 2 − θ) = x, {\textstyle \sin \left({\frac {\pi }{2}}-\theta \right)=x,} which allows for the solution to the equation sin φ = x {\displaystyle \;\sin \varphi =x\;} (where φ:= π 2 − θ {\textstyle \varphi :={\frac {\pi }{2}}-\theta }) to be used; that solution being: φ = (− 1) k arcsin (x) + π k for some k ∈ Z, {\displaystyle \varphi =(-1)^{k}\arcsin(x)+\pi k\;{\text{ for some }}k\in \mathbb {Z},} which becomes: π 2 − θ = (− 1) k arcsin (x) + π k for some k ∈ Z {\displaystyle {\frac {\pi }{2}}-\theta ~=~(-1)^{k}\arcsin(x)+\pi k\quad {\text{ for some }}k\in \mathbb {Z} } where using the fact that (− 1) k = (− 1) − k {\displaystyle (-1)^{k}=(-1)^{-k}} and substituting h:= − k {\displaystyle h:=-k} proves that another solution to cos θ = x {\displaystyle \;\cos \theta =x\;} is: θ = (− 1) h + 1 arcsin (x) + π h + π 2 for some h ∈ Z. {\displaystyle \theta ~=~(-1)^{h+1}\arcsin(x)+\pi h+{\frac {\pi }{2}}\quad {\text{ for some }}h\in \mathbb {Z}.} The substitution arcsin x = π 2 − arccos x {\displaystyle \;\arcsin x={\frac {\pi }{2}}-\arccos x\;} may be used express the right hand side of the above formula in terms of arccos x {\displaystyle \;\arccos x\;} instead of arcsin x. {\displaystyle \;\arcsin x.\;}
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