Mathematics Advance discussion of JEE part-I :

Methodology to calculate the value of `sin^(-1)sinx` and `cos^(-1)cosx` etc

Writing `x` in form of `alpha`

If `x in I^(st) " quadrant " `.

`x = alpha`

`x= pi- alpha` if `x in II^(nd) " quadrant "`.

`x= pi+ alpha` if `x in III^(nd) " quadrant "`.

`x= 2pi- alpha` if `x in IV^(td) " quadrant "`.

Dealing with -ve sign :

(i) `sin^-1 (–x) = – sin^-1 x, x ∈ [– 1, 1]`

(ii) `tan^-1 (–x) = – tan^-1 x, x ∈ R`

(iii) `cosec^-1 (–x) = – cosec^-1 x, | x | ≥ 1`

(iv) `cos^-1 (–x) = π – cos^-1 x, x ∈ [– 1, 1]`

(v) `sec^-1 (–x) = π – sec^-1 x, | x | ≥ 1`

(vi) `cot^-1 (–x) = π – cot^-1 x, x ∈ R`

Also recall from elementary trigonometry

`sin(-x) = -sinx`

`tan(-x) = -tanx`

`cosec(-x) = -cosecx`

`cos(-x) = cosx`

`sec(-x) = secx`

`cot(-x) = cotx`

Calculation :

`text(Step 1 :)`

Dealing with value of `x :`

Case I : If `x ` is in `I^(st)` quadrant means `x in [0,pi/2]`

Let `x` be as it is.

Case II : If `x` is in `II^(nd)` quadrant `x in [pi/2, pi]`

then write `x= pi- alpha` where `0 le alpha le pi/2`

eg. `(2 pi)/3 = pi- pi/3`

`(5 pi)/6 = pi- pi/6`

Case III : If `x` is in `III^(rd)` quadrant `x in [pi, (3 pi)/2]`

then, write `x= pi+ alpha` where, `0 le alpha pi/2`

eg. `(4 pi)/3 = pi + pi/3`

`(7 pi)/6 = pi+pi/6`

Case IV : If `x` is in `IV^(th)` quadrant `x in [3/2 pi , 2pi]`

write `x= 2 pi- alpha` where `0 le alpha le pi/2`

eg., `(5 pi)/3 = 2 pi- pi/3`

`(7 pi)/4 = 2 pi - pi/4`

How to find value of `sin^(-1) sin x`

Case (I) `y= sin^(-1) sin (pi/4) \ \ \ \ \ \ \ \ x in I^(st) " quadrant"`

`y= x =pi/4`

Case (II) `y= sin^(-1) sin ((2 pi)/3) \ \ \ \ \ x in II^(nd) " quadrant"`

`= sin^(-1) sin (pi- pi/3)`

`= sin^(-1) sin (pi/3)` Now ` alpha in I^(st) " quadrant"`.

`y = pi/3`

Case (III) `y = sin^(-1) sin ((5 pi)/4) \ \ \ \ \ \ x in III^(rd) " quadrant"`

`= sin^(-1) sin (pi+ pi/4)`

`= sin ^(-1) (- sin (pi/4))`

using property, `sin^(-1) (-x) = -sin^(-1) x`

`= -sin ^(-1) sin (pi/4) \ \ \ \ \ \ \ alpha in I^(st) " quadrant"`

`=-pi/4`

Case (IV) `y=sin^(-1) sin ((11 pi)/6) \ \ \ \ \ \ x in IV^(th) " quadrant"`

`= sin^(-1) sin (2 pi - pi/6)`

`= sin^(-1) (-sin(pi/6))`

`=-sin ^(-1) sin (pi/6)`

`=- pi/6`

`=> ` for `cos^(-1) cos x`

(I) `y= cos^(-1) cos (pi/4) \ \ \ \ \ \ \ \ x in I^(st) " quadrant"`

`y= x =pi/4`

(II) `y= cos^(-1) cos ((2 pi)/3) \ \ \ \ \ x in II^(nd) " quadrant"`

`= cos^(-1) cos (pi- pi/3)`

`= cos^(-1) (-cos (pi/3))`

Now using property `cos^(-1)(-x) = pi - cos^(-1)x`

`y =pi - cos^(-1)cos( pi/3) = pi - pi/3 = 2pi/3`

(III) `y = cos^(-1) cos ((5 pi)/4) \ \ \ \ \ \ x in III^(rd) " quadrant"`

`= cos^(-1) cos (pi+ pi/4)`

`= cos ^(-1) (-cos (pi/4))`

using property, `cos^(-1) (-x) =pi - cos^(-1) x`

`= pi - cos ^(-1) cos (pi/4) \ \ \ \ \ \ \ x in I^(st) " quadrant"`

`=pi-pi/4=(3pi)/4`

(IV) `y=cos^(-1) cos ((11 pi)/6) \ \ \ \ \ \ x in IV^(th) " quadrant"`

`= cos^(-1) cos (2 pi - pi/6)`

`= cos^(-1) ( cos(pi/6))`

`=cos ^(-1) cos (pi/6)`

`=pi/6`

Note : `cos^(-1) cos x ge 0` (by observation) `x in N`

`=> ` for `tan^(-1) tan x`

(I) `y= tan^(-1) tan (pi/4) \ \ \ \ \ \ \ \ x in I^(st) " quadrant"`

`y= x =pi/4`

(II) `y= tan^(-1) tan ((2 pi)/3) \ \ \ \ \ x in II^(nd) " quadrant"`

`= tan^(-1) tan (pi- pi/3)`

`= tan^(-1) (-tan (pi/3))`

using property ` tan ^(-1) (-x)= -tan^(-1) x`

`y = -pi/3`

(III) `y = tan^(-1) tan ((5 pi)/4) \ \ \ \ \ \ x in III^(rd) " quadrant"`

`= tan^(-1) tan (pi+ pi/4)`

`= tan ^(-1) (tan (pi/4))`

`= tan ^(-1) tan (pi/4)`

`=pi/4`

(IV) `y=tan^(-1) tan ((11 pi)/6) \ \ \ \ \ \ x in IV^(th) " quadrant"`

`= tan^(-1) tan (2 pi - pi/6)`

`= tan^(-1) (- tan(pi/6))`

`=-tan ^(-1) tan (pi/6)`

`=-pi/6`

`=> ` for `cosec^(-1) cosec x`

(I) `y= cosec^(-1) cosec (pi/4) \ \ \ \ \ \ \ \ x in I^(st) " quadrant"`

`y= x =pi/4`

(II) `y= cosec^(-1) cosec ((2 pi)/3) \ \ \ \ \ x in II^(nd) " quadrant"`

`= cosec^(-1) cosec (pi- pi/3)`

`= cosec^(-1) cosec (pi/3)`

`y = pi/3`

(III) `y = cosec^(-1) cosec ((5 pi)/4) \ \ \ \ x in III^(rd) " quadrant"`

`= cosec^(-1) cosec (pi+ pi/4)`

`= cosec ^(-1) (- cosec (pi/4))`

using property, `cosec^(-1) (-x) = -cosec^(-1) x`

`= -cosec ^(-1) cosec (pi/4) `

`=-pi/4`

(IV) `y=cosec^(-1) cosec ((11 pi)/6) \ \ \ \ x in IV^(th) " quadrant"`

`= cosec^(-1) cosec (2 pi - pi/6)`

`= cosec^(-1) -( cosec(pi/6))`

`=-cosec ^(-1) cosec (pi/6)`

`= - pi/6`

`=> ` for `sec^(-1) sec x`

(I) `y= sec^(-1) sec (pi/4) \ \ \ \ \ \ \ \ x in I^(st) " quadrant"`

`y= x =pi/4`

(II) `y= sec^(-1) sec ((2 pi)/3) \ \ \ \ \ x in II^(nd) " quadrant"`

`= sec^(-1) sec (pi- pi/3)`

`= sec^(-1) (-sec (pi/3))`

Now using property `sec^(-1)(-x) = pi - sec^(-1)x`

`y =pi - sec^(-1)sec( pi/3) = pi - pi/3 = 2pi/3`

(III) `y = sec^(-1) sec ((5 pi)/4) \ \ \ \ \ \ x in III^(rd) " quadrant"`

`= sec^(-1) sec (pi+ pi/4)`

`= sec ^(-1) -(sec (pi/4))`

using property, `sec^(-1) (-x) =pi - sec^(-1) x`

`= pi - sec ^(-1) sec (pi/4) \ \ \ \ \ \ \ x in I^(st) " quadrant"`

`=pi-pi/4=3pi/4`

(IV) `y=sec^(-1) sec ((11 pi)/6) \ \ \ \ \ \ x in IV^(th) " quadrant"`

`= sec^(-1) sec (2 pi - pi/6)`

`= sec^(-1) ( sec(pi/6))`

`=-sec ^(-1) sec (pi/6)`

`=pi/6`

Note : `sec^(-1) sec x ge 0` (by observation)

`=> ` for `cot^(-1) cot x`

(I) `y= cot^(-1) cot (pi/4) \ \ \ \ \ \ \ \ x in I^(st) " quadrant"`

`y= x =pi/4`

(II) `y= cot^(-1) cot ((2 pi)/3) \ \ \ \ \ x in II^(nd) " quadrant"`

`= cot^(-1) cot (pi- pi/3)`

`= cot^(-1) (-cot (pi/3))`

Now using property `cot^(-1)(-x) = pi - cot^(-1)x`

`y =pi - cot^(-1)cot( pi/3) = pi - pi/3 = 2pi/3`

(III) `y = cot^(-1) cot ((5 pi)/4) \ \ \ \ \ \ x in III^(rd) " quadrant"`

`= cot^(-1) cot (pi+ pi/4)`

`= cot ^(-1) (cot (pi/4))`

`= pi/4`

(IV) `y=cot^(-1) cot ((11 pi)/6) \ \ \ \ \ \ x in IV^(th) " quadrant"`

`= cot^(-1) cot (2 pi - pi/6)`

`= cot^(-1) ( -cot(pi/6))`

`=pi - cot ^(-1) cot (pi/6)`

`=pi - pi/6= (5pi) /6`

Most Generalized Case :

(I) Take negative sign out .
(II) Write `x` as `2n pi + alpha \ \ \ \ n in I`
(III) Write `x = alpha`

General Case Illustration :

E.g 1 :

Find the value of

`sin^(-1)sin(-pi/4) + cos^(-1)cos((-3pi)/4)+tan^(-1)tan((-5pi)/4)+cosec^(-1)cosec((-7pi)/4)+sec^(-1)sec((-9pi)/4) +cot^(-1)cot((-37pi)/4)`

Sol :

Lets solve all trigonometry function one by one

=> `sin^(-1)sin(-pi/4) = sin^(-1)(-sin(pi/4)) = -(sin^(-1)sin(pi/4)) = -pi/4`

=> ` cos^(-1)cos((-3pi)/4) = cos^(-1)cos((3pi)/4) = cos^(-1)cos(pi - pi/4)`
` = cos^(-1)(-cos(pi/4)) = pi - cos^(-1)(cos(pi/4)) = pi - pi/4 = ((3pi)/4) `

=> `tan^(-1)tan((-5pi)/4) = tan^(-1)(-tan((5pi)/4)) = -tan^(-1)tan(pi + pi/4)`

`=- tan^(-1)tan(pi + pi/4) = -pi/4`

=> `cosec^(-1)cosec((-7pi)/4) = - cosec^(-1)cosec((7pi)/4) = -cosec^(-1)cosec(2pi -pi/4)`

`-cosec^(-1)(-cosec(pi/4)) = cosec^(-1)cosec(pi/4) = pi /4`

=> `sec^(-1)sec(-9pi/4) = sec^(-1)sec(9pi/4) = sec^(-1)sec(2pi + pi/4) = sec^(-1)sec(pi/4) = pi/4`

=> `cot^(-1)cot((-37pi)/4) = cot^(-1)(-cot((37pi)/4)) = pi - cot^(-1)cot((37pi)/4)`

`= pi - cot^(-1)cot(8pi + 5pi/4) = pi - cot^(-1)cot( 5pi/4) = pi - cot^(-1)cot(pi + pi/4))`

`= pi - cot^(-1)(cot(pi/4)) = pi - pi/4 = 3pi/4`

Now by putting all value solve it

` -pi/4 + (3pi)/4 + (-pi/4) + pi /4 + pi/4 + (3pi)/4 = pi/2`

E.g. 2 :

`sin^(-1)sin1 + cos^(-1)cos2 + tan^(-1)tan4+ cot^(-1)cot6`

Sol :

`=> sin^(-1)sin1 = 1` Here `1 in Ist \ \ Q u a d`

`=>cos^(-1)cos2 ` Here `2 in IInd \ \ Q u a d`

`(2 = pi - alpha)`

`= cos^(-1)cos(pi - alpha) = cos^(-1)(-cos(alpha))`

`pi -cos^(-1)cos(alpha) = pi -alpha = pi-pi+2= 2`

`=> tan^(-1)tan4` Here `4 in IIIrd \ \ Q u a d`

(`4 = pi + alpha`)

` = tan^(-1)tan(pi + alpha) = tan^(-1)tanalpha`

` = 4 - pi`

`=> cot^(-1)cot6` Here `6 in IVth \ \ Q u a d`

(`6 = 2pi - alpha`)

`= cot^(-1)cot(2pi - alpha) = cot^(-1)(-cotalpha)`

` = pi - cot^(-1)cotalpha = pi - alpha = pi -(6-2pi) `

`= pi - 6`

Now put all calculated value in question

`1+2+ (4-pi)+ (pi - 6) = 1`

Graph of `y = sin^(-1) (sinx) , x ∈ R , y ∈ [ - pi/2 , pi/2]` , Periodic with period `2pi`

`y= sin^(-1) sin x`

` ={ tt (( - (pi+x), ; pi le x < -pi/2), (x, ; -pi/2 le x le pi/2), ( pi-x , ; pi/2 < x le pi))`

Graph of `y = cos^(-1) (cosx) = x , x ∈ R , y ∈ [0 , pi] ` , periodic with period `2pi`

`y= cos^(-1) cos x { tt ((-x , ; -pi le x < 0),(x,; 0 le x le pi))`

Graph of `y = tan^(-1)(tanx) = x , x ∈ R - { (2n-1) \ \ pi/2 n ∈ 1 } , y ∈ (-pi/2 , pi/2) , ` periodic with period `pi`

`y=tan^(-1) tan x`

`={ tt (( x+pi, ; -pi le x < -pi/2), (x, ; -pi/2 < x < pi/2), ( x-pi , ; pi/2 < x le pi))`

Graph of `y = cosec^(-1) (cosec x) , x ∈ R - { npi , n ∈ 1 }`

`y= cosec^(-1) cosecx`

`= { tt ((-(pi+x),; -pi < x < -pi/2), (x,; -pi/2 le x le pi/2(x != 0)),( pi-x,; pi/2 < x < pi))`

Graph of `y = sec^(-1) (secx) = x , y` is periodic ;

`y = sec^(-1) sec x`

`= { tt ((-x,; -pi < x <0),( x,; 0 < x < pi))`

Graph of `y = cot^(-1) (cotx) = x , x ∈ R - { n pi } , y ∈ (0 , pi)` , periodic with `pi`

`y= cot^(-1) cot x`

`= { tt (( x+ pi , ; -pi < x < 0),(x,; 0 < x < pi))`