Though trigonometric functions are not bijective yet their inverse are defined explain?

Answer 1

Lets take sin x for example.

we know that sin is defined from R to [-1,1]
when we take R as domain, sin is not bijective

but note that if we take the domain as [-pi/2 , pi/2], the function becomes bijective. the inverse of this function (and not the main sin function) is taken as sin-inverse function. So, taking domain into account, sin-inverse is not the inverse of sine function. However, formula-wise and all, it is the inverse.

Similarly for other functions too.

cos-inverse assumes cos to be having a domain of [0, pi]
tan-inverse assumes tan to be having a domain of (-pi/2, pi/2)
cot-inverse assumes cot to be having a domain of (0, pi)
sec-inverse assumes sec to be having a domain of [0,pi]-{pi/2}
cosec-inverse assumes cosec to have a domain of [-pi/2,pi/2]-{0}

Answer 2

To define the inverse trigonometric function you must define the domain and limit is
so Arcsin x (A capital letter) is defined for -pi/2<=x<=pi/2
Arccosx 0<=x<=pi
Arctan x -pi/2 In all those intervals sin,cos and tan are monotonous functions

Answer 3

chill out! that is not as annoying as you think of it extremely is! permit's not start up off with something so complicated. provided that t = arcsin(a), then sin t = a. because of the fact working backwards to stumble on t gets you t= arcsin (a) that's comparable to what i've got in the commencing up. So, decrease back on your question, t = arcsin(x/3), sin t = x/3 and utilising the double attitude formulation, sin 2t is such as 2sintcost then you definitely could ask, how do you hit upon what cos t is. in actuality, you're able to truly locate cos t via drawing a appropriate attitude traingle. Referencing decrease back to sin t = x/3, it exhibits that the choice is of length x and the hyp is of length 3. Given so, the adjoining would be sq. root of 9 - x^2. consequently cos t would be adj/hyp = (root 9- x^2)/ 3 then 2sintcos t or sin2t would be 2 x t x (root 9- x^2)/ 3