Mathematics TRIGONOMETRIC FUNCTIONS, THEIR PERIODICITY AND GRAPHS

Simplification & Transformation of Inverse Functions By Elementary Substitution and Their Graphs :

(1) `sin^(-1) 2x/(1+x^2)` `= [tt((2 tan^(-1) x , -1 le x le 1),(pi-2 tan^(-1) x , text(if) x ge 1),(-pi-2 tan^(-1) x , x le -1))`

Proof:

Let `x=tan theta , theta in (-pi/2 , pi/2) => theta =tan^(-1)x`
Now , `sin^(-1) (2x /(1+x^2)) =sin^(-1) (2 tan theta /(1+tan^2 theta))= sin^(-1) (sin 2 theta)=sin^(-1) (sin alpha)`, where `alpha in (-pi ,pi)`

`sin^(-1) (2x/(1+x^2))=sin^(-1) (sin alpha)`

`= {tt((-alpha -pi , -pi < alpha < -pi/2), (alpha, -pi/2 le alpha le -pi/2), (-alpha +pi , pi/2 < alpha < pi) ) = {tt( (-2 tan^(-1) x-pi, -pi < 2 tan^(-1) x < -pi/2) , (2 tan^(-1) x, -pi/2 le 2 tan^(-1) x le pi/2), (-2 tan^(-1) x +pi, pi/2 < 2 tan^(-1) x < pi) )`

`= {tt( (-2 tan^(-1) x -pi , -pi/2 < tan6(-1) x < -pi/4) ,(2 tan^(-1) x , -pi/4 le tan^(-1) x le pi/4) ,(-2 tan^(-1)x+pi, pi/4 < tan^(-1) x < pi/2 ) ) ={tt( (-2 tan^(-1) x-pi, x < -1 ), (2 tan^(-1) x, -1 le x le 1) ,(-2 tan^(-1) x + pi , x >1) )`

(2) `cos^(-1) (1-x^2)/(1+x^2)` `= [tt( (2 tan^(-1)x, x ge 0), (-2 tan^(-1) x, x <0) )`

Proof:

Let `x = tan theta , theta in ( -pi/2,pi/2) => theta = tan^(-1) x`
Now , `cos^(-1) ((1-x^2)/(1=x^2))=cos^(-1) ((1-tan^2 theta)/(1+tan^2 theta)) =cos^(-1) (cos 2 theta)= cos^(-1) (cos alpha)`, where `alpha in (-pi,pi)`

`cos^(-1) ((1-x^2)/(1+x^2))= cos^(-1) (cos alpha)`

`= {tt( (-alpha , -pi < alpha < 0 ) ,(alpha , 0 le alpha < pi) ) ={tt( (-2 tan^(-1) x , -pi< 2 tan^(-1) x < 0), (2 tan^(-1) x, 0 le 2 tan^(-1) x < pi) )`

`= {tt( (-2 tan^(-10 x , -pi/2 < tan^(-10 x < 0), (2 tan^(-1) x , 0 le tan^(-1) x < pi/2) ={tt( (-2tan6(-1) x , x < 0) ,(2 tan^(-1) x , x ge 0) )`

(3) `tan^(-1) 2x/ (1-x^2) ` `= [tt( ( pi+2 tan^(-1) x , x < -1), (2tan^(-1) x, -1 < x < 1), (2 tan^(-1)x -pi , x>1) )`

Proof:
Let `x= tan theta , theta in (-pi/2 ,pi/2) => theta = tan^(-1)x`

Now , `tan^(-1) (2x/ (1-x^2)) = tan^(-1) ((2tan theta)/(1- tan^2 theta)) =tan^(-1) (tan2 theta)= tan^(-1) (tan alpha)`, where `alpha in (-pi,pi)`

`tan^(-1) ((2x)/(1-x^2))=tan^(-1) (tan alpha)`

`={tt( (alpha+pi , -pi < alpha < -pi/2), (alpha, -pi/2 le alpha le pi/2), (alpha -pi , pi/2 < alpha < pi) ) ={tt( (2tan^(-1) x+pi, -pi < 2 tan^(-1) x < -pi/2), (2tan^(-1) x , -pi/2 le 2 tan^(-1) x le pi/2), (2 tan^(-1) x-pi , pi/2 < 2 tan^(-1) x < pi) )`

`= {tt( (2tan^(-1) x=pi , -pi/2 < tan^(-1) x < -pi/4), (2tan^(-1) x, -pi/4 le tan^(-1) x le pi/4), (2tan^(-1) x-pi, pi/4 < tan^(-1) x < pi/2) ) ={tt( (pi+2tan^(-1) x , x < -1 ), (2tan^(-1)x ,-1 le x le 1 ) , (2tan^(-1) x -pi , x >1) )`

Highlights : -

(a) `f(x) =sin ^(-1) ((2x)/(1+x^2)) +2 tan^(-1) x = pi ` if `x ge 1`

(b) `f(x) =sin^(-1) ((2x)/(1+x^2)) +2 tan^(-1) x = -pi ` if `x le -1`

(4) `sin^(-1) (3x -4x^3)` `= [tt( (- (pi +3sin^(-1) x) ,text(if) -1 le x le -1/2 ) , (3sin^(-1) x , text(if) -1 /2 le x le 1/2), (pi-3sin^(-1) x , text(if) 1/2 le x le1) )`

`sin^(-1) (3x -4x^3)` `= [tt( (- (pi +3sin^(-1) x) ,text(if) -1 le x le -1/2 ) , (3sin^(-1) x , text(if) -1 /2 le x le 1/2), (pi-3sin^(-1) x , text(if) 1/2 le x le1) )`

(5) `cos^(-1) (4x^3-3x)` `= [tt( (3cos^(-1) x -2 pi, text(if) -1 le x le -1/2), (2pi-3cos^(-1) x, text(if) -1 /2 le x le 1/2), (3 cos^(-1) x, text (if) 1/2 le x le 1) )`

`cos^(-1) (4x^3-3x)` `= [tt( (3cos^(-1) x -2 pi, text(if) -1 le x le -1/2), (2pi-3cos^(-1) x, text(if) -1 /2 le x le 1/2), (3 cos^(-1) x, text (if) 1/2 le x le 1) )`

(6) `tan^(-1) (3x-x^3)/(1-3x^2)` `= [tt( (3tan^(-1) x , text(if) -1/(sqrt3) < x < 1/(sqrt3)), (-pi+3tan^(-1) x text(if) x > 1/(sqrt3)), (pi+3tan^(-1) x , text (if) x < -1/(sqrt3)) )`

`tan^(-1) (3x-x^3)/(1-3x^2)` `= [tt( (3tan^(-1) x , text(if) -1/(sqrt3) < x < 1/(sqrt3)), (-pi+3tan^(-1) x text(if) x > 1/(sqrt3)), (pi+3tan^(-1) x , text (if) x < -1/(sqrt3)) )`