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Inverse Cosine is considered to be an important inverse trigonometric function. As the name suggests, Inverse cosine is the inverse of the cosine function cos(x). It is a trigonometric function that reverses the cosine function. Mathematically, the inverse cosine function is written as cos-1(x). Also, one must understand that the inverse cosine is not the reciprocal of cos x. Every one of the six trigonometric functions has a corresponding inverse function which are sin-1x, cos-1x, tan-1x, cosec-1x, sec-1x, cot-1x. The inverse cosine function is also termed as the ‘arc cosine’ function. The Inverse cosine function is used to calculate the measure of angle by using the value of the trigonometric ratio cos x.
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Key Terms: Inverse Cosine Function, Domain, Range, Integral, Derivative, Inverse Trigonometric Functions, Cosine Function, Trigonometric Functions, Trigonometric Ratios
What is Inverse Cosine?
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Inverse cosine is the inverse function of trigonometric function cosine, i.e, cos(x). It is usually represented as cos-1(x). It does exactly the opposite of cos(x). It should be noted that inverse cosine is not the reciprocal of the cosine function. For all the trigonometric functions, there is an inverse function for it. Inverse cosine is used to find the corresponding angle by using the value of cosine value of that angle.
Let us assume y = cos(x)
Then x = cos-1(y)
Here are some examples to show how the inverse cosine function works
- cos (0) = 1 ⇒ cos-1(1) = 0
- cos(π/3) = 1/2 ⇒ cos-1(1/2) = π/3
- cos(π/2) = 0 ⇒ cos-1(0) = π/2
- cos(π) = -1 ⇒ cos-1(-1) = π
Using the definition from the right-angled triangle,
cos θ = [adjacent side / hypotenuse]
So as per the definition of inverse cosine
θ = cos-1( [ adjacent side / hypotenuse ] )
Inverse Cosine Function
Read More: Introduction to Trigonometry
Domain and Range of Inverse Cosine
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It is known to us that the domain of cosine function is R (the set of all real numbers) and the range is [-1, 1]. A function is invertible if and only if it is bijective (one-one and onto). But it is known that cos x is not an one-one function so the inverse cosine cannot have R as its range.
Hence inverse function for cos(x) cannot exist in the whole domain. The domain of the cosine function is restricted to [0, π] usually and its range remain as [-1, 1]. Hence the branch of cos inverse x with the range [0, π] is called principal branch. In inverse function the domain of cos becomes the range and range of cos becomes the domain. So the domain of the inverse cosine function is [-1, 1] and the range is [0, π] .
Hence, Cos-1x is a function from [-1, 1] → [0, π]
Read More: Domain and Range of Trigonometric Functions
Graph of Inverse Cosine Function
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Since the domain and range of the inverse cosine function are [-1, 1] and [0, π] respectively, we can use the values of cos-1x to plot the graph of cos-1x. For y = cos-1x, we get
- When x = 0 , y = π/2
- When X = ½ , y = π/3
- When X = 1 , y = 0
- When X = -1 , y= π
- When X = -½ , y = 2π/3
Inverse Cosine Graph
Read More: Graphs of Inverse Trigonometric Functions
Derivative of Inverse Cosine function
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To determine the derivative of inverse cosine function, we will be using some trigonometric identities and formulas.
Let us assume that y = cos-1x ⇒ cos y = x.
Now, differentiate both sides of the equation cos y = x with respect to x using the chain rule
cos y = x
d(cos y)/dx = dx/dx
-sin y dy/dx = 1
dy/dx = -1/sin y ---- (1)
As cos2y + sin2y = 1, we get sin y = √(1 - cos2y) = √(1 - x2) [Since cos y = x]
Substitute sin y = √(1 - x2) in equation (1), we will get
dy/dx = -1/√(1 - x2)
As x = -1, 1 will make the denominator √(1 - x2) equal to 0, and
Since the derivative is not defined, so x cannot be -1 and 1.
Therefore, the derivative of cos inverse x is given as -1/√(1 - x2), where -1 < x < 1
Read More: Derivative of Inverse Trigonometric Functions
Integration of Inverse Cosine Function
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The integration of the given inverse cosine function can be calculated by integration by parts principle.
∫cos-1x = ∫cos-1x · 1 dx
By using integration by parts principle,
∫f(x) . g(x) dx = f(x) ∫g(x) dx − ∫(f′(x) ∫g(x) dx) dx + C
Here, f(x) = cos-1x and g(x) = 1
∫cos-1x · 1 dx = cos-1x ∫1 dx - ∫ [d(cos-1x)/dx ∫1 dx]dx + C
∫cos-1x dx = cos-1x . (x) - ∫ [-1/√(1 - x²)] x dx + C
Now, 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-1x dx = x cos-1x - ∫(-1/√u) (-1/2) du + C
= x cos-1x - 1/2 ∫u-1/2 du + C
= x cos-1x - (1/2) (u1/2/(1/2)) + C
= x cos-1x - √u + C
= x cos-1x - √(1 - x2) + C
Hence, ∫cos-1x dx = x cos-1x - √(1 - x²) + C
Read More: Integrals- Definite, Indefinite and Methods of Integration
Properties of Inverse Cosine Function
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The properties of inverse cosine function are as follows:
- cos(cos-1x) = x only when x ∈ [-1, 1] (When x ∉ [-1, 1], cos(cos-1x) is not defined)
- cos-1(cos x) = x only when x ∈ [0, π] (When x ∉ [0, π], use trigonometric identities to find out equivalent angles between [0,π])
- cos-1(-x) = π - cos-1x
- cos-1(1/x) = sec-1x, when |x| ≥ 1
- sin-1x + cos-1x = π/2, when x ∈ [-1, 1]
- d(cos-1x)/dx = -1/√(1 - x2), -1 < x < 1
- ∫cos-1x dx = x cos-1x - √(1 - x2) + C
Read More: Properties of Inverse Trigonometric Functions
Things to Remember
- Inverse Cosine function is a trigonometric function which reverses the cosine function.
- Every one of the six trigonometric functions has a corresponding inverse function.
- Inverse cosine is not the reciprocal of cosine function.
- A function is invertible if and only if it is bijective (one-one and onto).
- For the inverse cosine function, the domain is [-1, 1] and the range is [0, π] .
- The branch of cos inverse x with the range [0, π] is called principal branch.
Sample Questions
Ques. Find out the value of x in cos-1(√3/2) = x using the inverse cosine formula. (3 Marks)
Ans. We know that cos x = adjacent / hypotenuse
We also know that cos-1x is inverse function of cos x
We know that cos (π/6) = √3/2
Since π/6 ∈ [0, π],
cos-1(√3/2) = π/6
Hence, the value of x in case when cos-1(√3/2) = x using the inverse cosine formula is π/6.
Ques. Evaluate cos(cos-15) and cos-1(cos 5π/3) using the properties of inverse cosine. (3 Marks)
Ans. Since 5 ∉ [-1, 1], cos(cos-15) is not defined.
Now we will evaluate the value of cos-1(cos 5π/3)
But we can see that 5π/3 ∉ [0, π],we will determine the equivalent value of 5π/3 that lies in [0, π]
As cos x = cos (2π - x), we have cos 5π/3 = cos (2π - 5π/3) = cos(π/3)
π/3 ∈ [0, π]
Therefore cos-1(cos 5π/3) = cos-1(cos π/3) = π/3
Thus, cos(cos-15) is not defined and cos-1(cos 5π/3) = π/3
Ques. Find the value of cos-1(-1/2 ) (3 Marks)
Ans. Assume that
y = cos-1( -1/2)
It can be written as
cos y = -1/2
cos y = cos (2π/3).
Therefore, the Range of the principal value of cos-1 is [0, π ]
Hence, the principal value of cos-1( -1/2) is 2π /3.
Ques. Find the value of i) cos-1(cos 3) ii) cos-1(cos 6) (3 Marks)
Ans. (i) Let us assume that
cos-1(cos 3) = θ
cos θ = cos 3
θ = 3
Hence, cos-1(cos 3) is 3.
(ii) Since, cos (2π-θ) = cos(θ)
= cos-1(cos 6)
= cos-1(cos (2π-θ))
= 2π - 6 as 0 ≤ (2π−6) ≤ π
Ques. Find the value of cos-1 (cos 14π/3) (3 Marks)
Ans. cos-1 (cos 14π/3) will be
= \(\cos^{-1} \left[ \cos(4\pi + \frac{2\pi}{3} ) \right]\)
= \(\cos^{-1} \left(\cos \frac{2\pi}{3} \right)\)
= \(2\pi/3\)
Ques. Find the principal value of sec-1(-2) (3 Marks)
Ans. Let y = sec-1(-2)
We get sec (y) = -2
Now we know that range of principal value branch is (0, π) - {π/2}
Thus,
sec y = 2 = sec (π/3)
\(Y = \pi/3 \in [0, \pi], y \neq \frac{\pi}{2} \)
Therefore, the principal value of sec-1(-2) is π/3
Ques. Find the value of sec-1(2/√3) (3 Marks)
Ans. Let x = sec-1(2/√3)
Thus, sec(x) = 2/√3
We know that
Cos π/6 = √3/2
∴ sec θ= 1/cosθ
∴ sec π/6 = 2/√3
Comparing with sec x = 2/√3
x = π/6
∴ sec-1(2/√3) = π/6
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