Revision: Relations and Functions >> Inverse Trigonometric Functions Maths Commerce (English Medium) Class 12 CBSE

If

• sin θ = x ⟹ θ = sin⁻¹x...θ ∈ [−π/2, π/2]

• cos θ = x ⟹ θ = cos⁻¹x...θ ∈ [0, π]

• tan θ = x ⟹ θ = tan⁻¹x...θ ∈ (−π/2, π/2)

sin⁻¹x, cos⁻¹x, tan⁻¹x, etc. are called inverse trigonometric functions.

sin⁻¹ x = cos⁻¹ (√(1 − x²)), 0 ≤ x ≤ 1

cos⁻¹ x = sin⁻¹ (√(1 − x²)), 0 ≤ x ≤ 1

cos(sin⁻¹ x) = sin(cos⁻¹ x) = √(1 − x²), |x| ≤ 1

• sin⁻¹x = tan⁻¹( x / √(1−x²) ), |x| < 1

• cos⁻¹x = tan⁻¹( √(1−x²) / x ), x > 0

• tan⁻¹x = sin⁻¹( x / √(1+x²) ), ∀ x

• tan⁻¹x = cos⁻¹( 1 / √(1+x²) ), x ≥ 0

• cosec⁻¹ x = sin⁻¹1 (1/x), x ∈ R − (−1, 1)

• sec⁻¹ x = cos⁻¹ (1/x), x ∈ R − (−1, 1)

• cot⁻¹ x = tan⁻¹ (1/x), for x > 0

• cot⁻¹ x = π + tan⁻¹ (1/x), for x < 0
  [only if tan⁻¹ is taken in (−π/2, π/2)]

• sin⁻¹x + cos⁻¹x = π/2, |x| ≤1

• tan⁻¹x + cot⁻¹x = π/2, x ∈ ℝ

• sec⁻¹x + cosec⁻¹x = π/2, |x| ≥ 1

• sin⁻¹(−x) = −sin⁻¹x, |x| ≤1

• tan⁻¹(−x) = −tan⁻¹x, x ∈ ℝ

• cosec⁻¹(−x) = −cosec⁻¹x, |x| ≥ 1

• cos⁻¹(−x) = π − cos⁻¹x, |x| ≤ 1

• sec⁻¹(−x) = π − sec⁻¹x, |x| ≥ 1

• cot⁻¹(−x) = π − cot⁻¹x, x ∈ ℝ

2 sin⁻¹ x = sin⁻¹ (2x√(1 − x²))

3 sin⁻¹ x = sin⁻¹ (3x − 4x³)

2 cos⁻¹ x = cos⁻¹ (2x² − 1)

3 cos⁻¹ x = cos⁻¹ (4x³  − 3x)

3 tan⁻¹ x = tan⁻¹ ((3x − x³ )/(1 − 3x²))

(A) tan⁻¹ formulas

• tan⁻¹x + tan⁻¹y = tan⁻¹( (x+y)/(1−xy) ), if xy < 1

• tan⁻¹x + tan⁻¹y = π + tan⁻¹( (x+y)/(1−xy) ), if x,y > 0 & xy > 1

• tan⁻¹x − tan⁻¹y = tan⁻¹( (x−y)/(1+xy) ) if x,y> -1

(B) sin⁻¹ formulas  

• sin⁻¹x + sin⁻¹y
  = sin⁻¹( x√(1−y²) + y√(1−x²) )

• sin⁻¹x − sin⁻¹y
  = sin⁻¹( x√(1−y²) − y√(1−x²) )

(C) cos⁻¹ formulas

• cos⁻¹x + cos⁻¹y
  = cos⁻¹( xy − √(1−x²)√(1−y²) )

• cos⁻¹x − cos⁻¹y
  = cos⁻¹( xy + √(1−x²)√(1−y²) )

(A) Direct identities

• sin(sin⁻¹x) = x, |x| ≤ 1

• cos(cos⁻¹x) = x, |x| ≤ 1

• tan(tan⁻¹x) = x, x ∈ ℝ

• cot(cot⁻¹x) = x, x ∈ ℝ

• sec(sec⁻¹x) = x, |x| ≥ 1

• cosec(cosec⁻¹x) = x, |x| ≥ 1

(B) Inverse of trigonometric expressions

Valid ONLY in principal value range:

• sin⁻¹(sin θ) = θ, θ ∈ [−π/2, π/2]

• cos⁻¹(cos θ) = θ, θ ∈ [0, π]

• tan⁻¹(tan θ) = θ, θ ∈ (−π/2, π/2)

LHS = `(cos A)/a + (cos B)/b + (cos C)/c`

`= ((("b"^2 + "c"^2 - "a"^2)/"2bc"))/"a" + ((("c"^2 + "a"^2 - "b"^2)/"2ca"))/"b" + ((("a"^2 + "b"^2 - "c"^2)/"2ab"))/"c"`

`= ("b"^2 + "c"^2 - "a"^2)/"2abc" + ("c"^2 + "a"^2 - "b"^2)/"2abc" + ("a"^2 + "b"^2 - "c"^2)/"2abc"`

`= ("b"^2 + "c"^2 - "a"^2 + "c"^2 + "a"^2 - "b"^2 + "a"^2 + "b"^2 - "c"^2)/"2abc"`

`= ("a"^2 + "b"^2 + "c"^2)/"2abc"`

= RHS

LHS = `(cos A)/a + (cos B)/b + (cos C)/c`

= `(b cos A + a cos B)/(ab) + (cos C)/c`

= `c/(ab) + (cos C)/c`    ...(By projection rule)

= `c/(ab) + (a^2 + b^2 - c^2)/(2 abc)`    ...(By cosine rule)

= `(2c^2 + a^2 + b^2 - c^2)/(2 abc)`

= `(a^2 + b^2 + c^2)/(2 abc)` = R.H.S.

Put x = cos θ

∴ θ = cos^–1 x

L.H.S. = `tan^-1 ((sqrt(1 + x) - sqrt(1 - x))/(sqrt(1 + x) + sqrt(1 - x)))`

= `tan^-1 ((sqrt(1 + cos θ) - sqrt(1 - cos θ))/(sqrt(1 + cos θ) + sqrt(1 - cos θ)))`

= `tan^-1 [(sqrt(2 cos^2(θ/2)) - sqrt(2 sin^2 (θ/2)))/(sqrt(2 cos^2 (θ/2)) + sqrt(2 sin^2 (θ/2)))]`

= `tan^-1 [(sqrt(2) cos (θ/2) - sqrt(2) sin (θ/2))/(sqrt(2) cos (θ/2) + sqrt(2) sin (θ/2))]`

= `tan^-1 [((sqrt(2) cos (θ/2))/(sqrt(2) cos (θ/2)) - (sqrt(2) sin (θ/2))/(sqrt(2) cos (θ/2)))/((sqrt(2) cos (θ/2))/(sqrt(2) cos (θ/2)) + (sqrt(2) sin (θ/2))/(sqrt(2) cos (θ/2)))]`

= `tan^-1 [(1 - tan(θ/2))/(1 + tan (θ/2))]`

= `tan^-1 [(tan  pi/4 - tan (θ/2))/(1 + tan  pi/4. tan (θ/2))]  ....[∵ tan  pi/4 =1]`

= `tan^-1 [tan (pi/4 - θ/2)]`

= `pi/4 - θ/2`

= `pi/4 - 1/2 cos^-1`x  .....[∵ θ = cos^–1 x]

∴ L.H.S. = R.H.S.

`2sin^-1  3/5`

= `2tan^-1  3/sqrt(5^2 - 3^2)   ...[sin^-1  p/h = tan^-1  p/sqrt(h^2 - p^2)]`

= `2 tan^-1  3/sqrt(25 - 9)`

= `2 tan^-1  3/sqrt16`

= `2tan^-1  3/4`

= `tan^-1  (2 xx 3/4)/(1 - (3/4)^2)   ...[2tan^-1 x = tan^-1  (2x)/(1 - x^2)]`

= `tan^-1  (3/2)/(1 - 9/16)`

= `tan^-1  (3/2)/((16 - 9)/16)`

= `tan^-1  (3/2)/(7/16)`

= `tan^-1 (3/2 xx 16/7)`

= `tan^-1 (3/1 xx 8/7)`

= `tan^-1  24/7`

Let `cos^(-1)  4/5` = x

Then, cos x = `4/5`

⇒ sin x = `sqrt (1 - (4/5)^2)`

⇒ sin x = `sqrt(1 - 16/25)`

⇒ sin x = `sqrt(9/25)`

⇒ sin x = `3/5`

∴ tan x = `3/4` ⇒ x = `tan^(-1)  3/4`

∴ `cos^(-1)  4/5 =  tan^(-1)  3/4`   ...(1)

Now let `cos^(-1)  12/13` = y

Then cos y = `12/13`

⇒ sin y = `5/13`

∴ tan y = `5/12` ⇒ y = `tan^(-1)  5/12`

∴ `cos^(-1)  12/13 = tan^(-1)  5/12`  ....(2)

Let `cos^(-1)  33/65` = z

Then cos z = `33/65`

⇒ sin z = `56/65`

∴ tan z = `56/33` ⇒ z = `tan^(-1)  56/33`

∴ `cos^(-1)  33/65 = tan^(-1)  56/33`  ....(3)

Now, we will prove that:

L.H.S = `cos^(-1)  4/5 + cos^(-1)  12/13`

= `tan^(-1)  3/4 + tan^(-1)  5/12`  ....[Using (1) and (2)]

= `tan^(-1)  (3/4 + 5/12)/(1 - 3/4 * 5/12)    ....[tan^(-1) x + tan^(-1) y = tan^(-1)  (x + y)/(1 - xy)]`

= `tan^(-1)  (36+20)/(48-15)`

= `tan^(-1)  56/33`

= `cos^(-1)  33/65`   .....[by (3)]

= R.H.S.

Let x = cos θ

Then, cos⁻¹x =  θ

We have,

R.H.S = cos⁻¹(4x³ − 3x)

⇒ cos⁻¹(4 cos³θ − 3 cos θ)

⇒ cos⁻¹(cos 3θ) = cos⁻¹(4x³ − 3x)

⇒ 3θ =  cos⁻¹(4x³ − 3x)

⇒ 3 cos⁻¹x = cos⁻¹(4x³ − 3x)

R.H.S = L.H.S

`sin^-1  8/17 + sin^-1  3/5`

= `tan^-1  8/sqrt(17^2 - 8^2) + tan^-1  3/sqrt(5^2 - 3^2)  ...[sin^-1  p/h = tan^-1  p/sqrt(h^2 - p^2)]`

= `tan^-1  8/sqrt(289 - 64) + tan^-1  3/sqrt(25 - 9)`

= `tan^-1  8/sqrt225 + tan^-1  3/sqrt16`

= `tan^-1  8/15 + tan^-1  3/4`

= `tan^-1  ((8/15 + 3/4)/(1 - 8/15 xx 3/4))  ...[tan^-1x + tan^-1y = tan^-1((x + y)/(1 - x xx y))]`

= `tan^-1[((32 + 45)/60)/(1 - 24/60)]`

= `tan^-1  77/36`

Let x = `cos^(-1)  12/13` and y = `sin^(-1)  3/5`

or cos x = `12/13` and sin y = `3/5`

sin x = `sqrt (1 - cos^2 x)` and cos y = `sqrt(1 - sin^2 y)`

Now, sin x = `sqrt(1 - 144/169)` and cos y = `sqrt( 1 - 9/25)`

⇒ sin x = `5/13` and cos y = `4/5`

We know that,

sin (x + y) = sin x cos y + cos x sin y

= `5/13 xx 4/5 + 12/13 xx 3/5 `

= `20/65 + 36/65 `

= `56/65`

⇒ x + y = `sin ^-1(56/65)`

or, `cos^-1(12/13) + sin^-1 (3/5)`

= `sin^-1(56/65)`

Let `sin^(-1)  5/13` = x and `cos^(-1)  3/5` = y

⇒ sin x = `5/13 ` and cos y = `3/5`

or tan x = `5/12` and tan y = `4/3`

⇒ x = `tan^-1  5/12` and y = `tan^(-1)  4/3`

x + y = `tan^-1  5/12 + tan^-1  4/3`

= `tan^-1 ((5/12 + 4/3)/(1 - 5/12 xx 4/3))`

= `tan^(-1) ((15+48)/(36-20))`

= `tan^(-1)  63/16`

Let x = tan² θ

⇒ `sqrtx` = tan θ

⇒ θ = `tan^(-1) sqrtx`  ...(1)

∴ `(1-x)/(1+x)`

= `(1-tan^2 θ)/(1 + tan^2 θ)`

= cos 2θ

Now we have,

R.H.S = `1/2 cos^(-1)  (1-x)/(1+x)`

= `1/2 cos^(-1)(cos 2θ)`

= `1/2 xx 2θ`

= θ

= `tan^(-1) sqrtx`  ....[From (1)]

R.H.S. = L.H.S.

L.H.S. = `cot^(-1)  ((sqrt(1+sin x) + sqrt(1-sinx))/(sqrt(1+sin x) - sqrt(1- sinx)))`

= `cot^(-1)  (sqrt(1+sin x) + sqrt(1-sinx))/(sqrt(1+sin x) - sqrt(1 - sin x)) xx (sqrt(1+sin x) + sqrt(1-sinx))/(sqrt(1+sin x) - sqrt(1 - sin x))`

= `cot^(-1)  ((1+sinx) + (1-sinx) + 2sqrt(1 - sin^2 x))/((1+sinx) - (1 - sinx)`

= `cot^(-1)  (2(1 + cos x))/(2sin x)`

= `cot^(-1)  (1+ cosx)/sin x`

= `cot^(-1)  (2 cos^2  x/2)/(2sin  x/2 cos  x/2)`

= `cot^-1 (cot  x/2)`

= `x/2`

L.H.S. = R.H.S.

Let x = sin θ

Then, sin⁻¹ x = θ

We have

R.H.S = sin⁻¹ (3x − 4x³) = sin⁻¹ (3 sin θ − 4 sin³θ)

= sin⁻¹ (sin 3θ) = sin⁻¹ (3 sin θ − 4 sin³θ)

= 3θ = sin⁻¹ (3 sin θ − 4 sin³θ)

= 3 sin⁻¹ x = sin⁻¹ (3 sin θ − 4 sin³θ)

R.H.S = L.H.S