# Inverse Cosine

Cosine of a function is given by: cos θ = adjacent side / hypotenuse. The inverse of a cosine function is given by:
Θ = cos-1 (adjacent side / hypotenuse)

We use the inverse cosine function to find the unknown angles in a right-angled triangle.

• cos 0 = 1 ⇒ 0 = cos-1 (1)
• cos π/3 = 1/2 ⇒ π/3 = cos-1 (1/2)
• cos π/2 = 0 ⇒ π/2 = cos-1 (0)
• cos π = -1 ⇒ π = cos-1 (-1)

## Derivative of inverse cosine

We will derive the derivative of inverse cosine function using chain rule.

Let us assume y = cos-1x ⇒ cos y = x. Differentiating both sides, we get,
D (cos y) / dx = dx / dx
-sin y dy / dx = 1
dy / dx = -1 / sin y

We know, cos2 y + sin2 y = 1, we have,

sin y = √(1 - cos2 y) = √(1 - x2) [as cos y = x]

Substituting sin y = √(1 - x2) in (1), we have,

dy / dx = -1 / √(1 - x2)

Since x = -1, 1 makes the denominator √(1 - x2) equal to 0. Therefore, the derivative is not defined if x is -1 and 1.

We can conclude that the derivative of inverse cosine function is -1 / √(1 - x2), where -1 < x < 1.

## Integration of inverse cosine function

We will find the integration of the inverse cosine function using integration by parts, ILATE (inverse, logarithmic, algebraic, trigonometric, and exponent).

∫cos-1 x = ∫cos-1 x · 1 dx

Using integration by parts,

∫f (x) . g (x) dx = f (x) ∫g (x) dx − ∫(f′ (x) ∫g (x) dx) dx + C

Here f (x) = cos-1 x and g (x) = 1.

∫cos-1 x · 1 dx = cos-1 x ∫1 dx - ∫ [d (cos-1 x) / dx ∫1 dx] dx + C

∫cos-1 x dx = cos-1 x . (x) - ∫ [-1 / √(1 - x²)] x dx + C

We will evaluate this integral ∫ [-1 / √(1 - x²)] x dx using substitution method. Let us assume 1 - x2 = u. Then -2x dx = du (or) x dx = -1/2 du.

∫cos-1 x dx = x cos-1x - ∫(-1/√u) (-1/2) du + C

= x cos-1 x - 1/2 ∫u-1/2 du + C

= x cos-1 x - (1/2) (u1/2 / (1/2)) + C

= x cos-1x - √u + C

= x cos-1x - √(1 - x²) + C

Therefore, ∫cos-1x dx = x cos-1x - √(1 - x²) + C

## Properties of inverse cosines

• sin-1 x + cos-1 x = π/2, when x ∈ [-1, 1]
• cos (cos-1 x) = x only when x ∈ [-1, 1] (When x ∉ [-1, 1], cos (cos-1 x) is undefined)
• ∫cos-1 x dx = x cos-1 x - √(1 - x²) + C
• cos-1 (cos x) = x, only when x ∈ [0, π] (When x ∉ [0, π]. We can apply the trigonometric identities to find the equivalent angle of x that lies in [0, π])
• d (cos-1 x)/dx = -1/√(1 - x2), -1 < x < 1
• cos-1 (-x) = π - cos-1x
• cos-1 (1/x) = sec-1 x, when |x| ≥ 1

## Points to Ponder

• Inverse cosine is not the same as (cos x)-1 as (cos x)-1 = 1/(cos x) = sec x.
• cos-1 (-x) = π - cos-1x
• ∫cos-1 x dx = x cos-1 x - √(1 - x²) + C
• θ = cos-1[ (adjacent side) / (hypotenuse) ], θ ∈ [0, π]
• d (cos-1 x) / dx = -1/√(1 - x2), -1 < x < 1
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