1. Mathematics

2. Inverse Trigonometric Functions Session

3. Session Objectives

4. 2.Properties of Inverse Trigonmetric Function 1.Basic Concepts of inverse trigonometric functions Definition Domain and Range 3.Conversion of one form of Inverse Trig. Fn. to other Forms 4.Identities containing inverse trigonometric functions

5. Basic Concepts - Definition Inverse of a Function : Function must be bijective Trigonometric functions are periodic. sin x is periodic with period equal to 2π Not Bijective Hence, inverse of sin x should not be valid ????

6. Basic Concepts - Definition However, trigonometric functions are bijective for particular value sets in the domain. sin -1 x is valid in these value sets Inverse trigonometric function - principal value set Smallest Numerical Angle sin x is bijective in [-π/2, π/2 ] and in [π/2,3π/2 ] ……. for x  R [-π/2, π/2 ]

7. Basic Concepts - Definition Inverse trigonometric functions  inverse circular function. arc sin x  sin -1 x ( principal value )

8. Inverse Trigonometric function - sin –1 x Domain of sin –1 x ( value which x can take ) is [-1,1] Range of sin –1 x (values which sin –1 x can take ) is [- /2, /2 ] 1  /2 -  /2 3  /2 X Y

9. Inverse Trigonometric function - cos –1 x Domain of cos –1 x ( value which x can take ) is [-1,1] Range of cos –1 x (values which cos –1 x can take ) is [0, ] X 1  /2 --  Y

10. Inverse Trigonometric function - tan –1 x Domain of tan –1 x ( value which x can take ) is (-, ) Range of tan –1 x (values which sec –1 x can take ) is (-/2, /2) X /2  -/2 --

11. Inverse Trigonometric function - sec –1 x Domain of sec –1 x ( value which x can take ) is (- -1] U [1, ) Range of sec –1 x (values which sec –1 x can take ) is [0, ] excl. x = /2 Y X /2  -/2 -- 1

12. Inverse Trigonometric function - cot –1 x Domain of cot –1 x ( value which x can take ) is (-, ) Range of cot –1 x (values which cot –1 x can take ) is (0, ) Y X /2  -/2 --

13. Inverse Trigonometric function - Domain and Range FunctionDomainRange sin -1 x[-1,1] [-/2, /2] cos -1 x[-1,1] [0, ] tan -1 x (-, )(-/2, /2) sec -1 x (-, -1] U [1, )[0, ] excl. /2 cosec -1 x (-, -1] U [1, )[-/2, /2] excl. 0 cot -1 x (-, )(0, ) FunctionDomainRange

14. Inverse Trigonometric function – Properties Always remember to keep the constraint of domain and range, while solving inverse trigonometric functions. sin( sin -1 x) = x and cos (cos -1 x) = x if x is in [-1,1] tan( tan -1 x) = x x if x is in ( -, ) sin -1 ( sin x) = x if x is in [-/2, /2] cos -1 ( cos x) = x if x is in [0, ] sec -1 ( sec x) = x if x is in [0, ] excl. x = /2

15. Inverse Trigonometric function – Properties cos -1 (-x) = - cos -1 x if x is in [-1,1] sin -1 (-x) = -sin -1 (x) if x is in [-1,1] tan -1 (-x) = - tan -1 x if x is in ( -, ) cot -1 (-x) =  - cot -1 x if x is in (-,) cosec -1 (-x) = - cosec -1 x if x is in (-,-1] U [1,) sec -1 (-x) =  - sec -1 x if x is in (-,-1] U [1,)

16. Inverse Trigonometric function – Properties Always remember to keep the constraint of domain and range, while solving inverse trigonometric functions. sin -1 (-x) = -sin -1 (x) if x is in [-1,1] Let y = sin -1 (-x) ; constraint : y is in [-/2, /2] sin y = - x  x = - sin y = sin ( -y ) sin -1 (-x) = sin -1 ( sin (-y)) sin -1 (-x) = -y  sin -1 (-x) = -sin -1 x

17. Class Exercise - 1 Find the principal value of Solution :

18. Class Exercise - 2 Find the principal value of sin –1 ( sin 5 ) Let y = sin -1 (sin 5).Hence y is in [-/2,/2] 5 Wrong Now, sin 5 = sin [(5/). ] = sin ( 1.59) = - sin (2 - 1.59) = sin ( 1.59  -2) in [-/2, /2] sin 5 = sin ( 5 - 2 ) sin -1 (sin 5) = sin -1 ( sin ( 5 - 2 )) = 5 - 2  Solution :

19. Other important properties If x > 0, y > 0 and xy < 1 If x > 0, y > 0 and xy > 1 If x<0,y<0 and xy < 1 sin -1 x+ cos -1 x = /2 ; if x is in [-1,1]

20. Class Exercise - 5 Find the value of Solution :

21. Class Exercise - 5 Find the value of Solution :

22. Class Exercise - 9 In triangle ABC if A = tan -1 2 and B = tan -1 3, prove that C = 45 0 Solution : For triangle ABC, A+B+C = 

23. Inverse Trigonometric function – Conversion To convert one inverse function to other inverse function : 1.Assume given inverse function as some angle ( say  ) 2. Draw a right angled triangle satisfying the angle. Find the third un known side 3.Find the trigonometric function from the triangle in step 2. Take its inverse and we will get  = desired inverse function

24. Conversion - Illustrative Problem The value of cot -1 3 + cosec -1  5 is (a) /3 (b) /2 ( c) /4 (d) none Step 1 Assume given inverse function as some angle ( say  ) Let cot -1 3 + cosec -1  5 = x + y, Where x = cot -1 3 ; cot x = 3 and y = cosec -1  5 ; cosec y =  5

25. Conversion - Illustrative Problem The value of cot -1 3 + cosec -1  5 is (a) /3 (b) /2 ( c) /4 (d) none Step 2 Draw a right angled triangle satisfying the angle. Find the third unknown side x 1 3  10 y  5 1 2 cot x = 3, tan x = 1/3 cosec y =  5, tan y = 1/2

26. Conversion - Illustrative Problem The value of cot -1 3 + cosec -1  5 is (a) /3 (b) /2 ( c) /4 (d) none Step 3 Find the trigonometric function from the triangle in step 2. Take its inverse and we will get  = desired inverse function tan ( x+ y) = 1  x + y = /4 tan x = 1/3,tan y = 1/2

27. Conversion - Illustrative Problem Prove that sin cot -1 tan cos -1 x = x Step 1 Assume given inverse function as some angle ( say  ) Let y = sin cot -1 tan cos -1 x And cos -1 x = , cos  = x Hence, y = sin cot -1 tan  Solution :

28. Conversion - Illustrative Problem Prove that sin cot -1 tan cos -1 x = x Step 2 Draw a right angled triangle satisfying the angle. Find the third unknown side  1 x cos  = x, tan  = Hence, y = sin cot -1 y = sin cot -1 tan 

29. Conversion - Illustrative Problem Prove that sin cot -1 tan cos -1 x = x Step 2 Draw a right angled triangle satisfying the angle. Find the third unknown side  1 x and y = sin  From the adjoining triangle, sin  = x Let cot -1 = , cot  = Hence y = x = R.H.S. y = sin cot -1

30. Class Exercise - 3 Find the value of Solution :

31. Class Exercise - 4 Find the value of Solution :

32. Class Exercise - 6 Prove that

33. Class Exercise - 6 Prove that Solution : And L.H.S. of the given identity is +

34. Class Exercise - 6 Prove that Solution : given identity is +

35. Class Exercise - 7 Solve the equation Solution :

36. Class Exercise - 7 Solve the equation Solution :

37. Class Exercise - 7 Solve the equation Solution :

38. Class Exercise - 8 If sin -1 x + sin -1 (1- x) = cos -1 x, the value of x could be (a) 1, 0 (b) 1,1/2 (c) 0,1/2 (d) 1, -1/2

39. Class Exercise - 8 If sin -1 x + sin -1 (1- x) = cos -1 x, the value of x could be (a) 1, 0 (b) 1,1/2 (c) 0,1/2 (d) 1, -1/2 Solution : and given equation is + = cos -1 x  cos (+) = x

40. Class Exercise - 8 If sin -1 x + sin -1 (1- x) = cos -1 x, the value of x could be (a) 1, 0 (b) 1,1/2 (c) 0,1/2 (d) 1, -1/2 Solution : cos (+) = x

41. Class Exercise - 8 If sin -1 x + sin -1 (1- x) = cos -1 x, the value of x could be (a) 1, 0 (b) 1,1/2 (c) 0,1/2 (d) 1, -1/2 Solution :

42. Class Exercise - 10 If cos -1 x + cos -1 y + cos -1 z = , Then prove that x 2 +y 2 +z 2 = 1 - 2xyz Solution : and given : A+B+C =  Now, L.H.S. = cos 2 A + cos 2 B +cos 2 C = cos 2 A + 1- sin 2 B +cos 2 C = 1+(cos 2 A - sin 2 B) +cos 2 C

43. Class Exercise - 10 If cos -1 x + cos -1 y + cos -1 z = , Then prove that x 2 +y 2 +z 2 = 1 - 2xyz Solution : Given : A+B+C =  L.H.S. = 1+(cos 2 A - sin 2 B) +cos 2 C = 1+ cos(A+B).cos(A–B) +cos 2 C = 1+ cos(  – C).cos(A–B) +cos 2 C = 1+ cosC [– cos(A–B) +cosC ] = 1+ cosC [– cos(A–B) – cos(A+B)]

44. Class Exercise - 10 If cos -1 x + cos -1 y + cos -1 z = , Then prove that x 2 +y 2 +z 2 = 1 - 2xyz Solution : Given : A+B+C =  L.H.S. = 1+ cosC [– cos(A–B) – cos(A+B)] = 1– cosC [ cos(A–B) + cos(A+B)] = 1– cosC [2cos A. cos B] = 1– 2xyz = R.H.S.

45. Thank you