We will discuss here about Inverse trigonometric Functions or inverse circular functions.
The inverse of a function f: A → B exists if and only if f is one-one onto (i.e., bijection) and given by
f(x) = y⇔ f\(^{-1}\) (y) = x.
Consider the sine function. Clearly, sin: R → R given by sin θ = x for all θ ∈ R is a many-one into function. So, its inverse does not exist. If we restrict its domain to the interval [- \(\frac{π}{2}\), \(\frac{π}{2}\)] then we may have infinitely many values of the angle θ which satisfy the equation sin θ = x i.e., sine of any one of these angles is equal to x. Here angle θ is represented as sin\(^{-1}\)x which is read as sine inverse x or arc sin x. Therefore, the symbol sin\(^{-1}\)x represents an angle and the sine of this angle has the value x.
Note the difference between sin\(^{-1}\)x
and sin θ: sin\(^{-1}\)x represents an
angle while sin θ represents a
pure number; again, for a given value of x (- 1 ≤ x ≤ 1) we may have infinitely many vales of sin\(^{-1}\)x
i.e., sin\(^{-1}\)x is a multiple-valued
function; but a given value of θ gives a definite finite value of sin θ i.e.,
sin θ is a single-valued function. Thus, if x is a real number lying
between -1 and 1, then sin\(^{-1}\) x is an angle between - \(\frac{π}{2}\)
and \(\frac{π}{2}\) whose sine is x i.e.,
sin\(^{-1}\)x = θ
⇔ x = sin θ, where - \(\frac{π}{2}\) ≤ x ≤ \(\frac{π}{2}\) and - 1 ≤ x ≤ 1.
In the above discussion we have restricted the sine function to the interval [- \(\frac{π}{2}\), \(\frac{π}{2}\)] to ake it a bijection. In fact we restrict the domain of sin θ to any of the interval [- \(\frac{π}{2}\), \(\frac{π}{2}\)], [\(\frac{3π}{2}\), \(\frac{5π}{2}\)], [- \(\frac{5π}{2}\), -\(\frac{3π}{2}\)] etc. sin θ is one-one onto function with range [-1, 1]. We therefore conclude that each of these intervals we can define the inverse of sine function. Thus sin\(^{-1}\)x is a function with domain [-1, 1] = {x ∈ R: - 1 ≤ x ≤ 1} and range [- \(\frac{π}{2}\), \(\frac{π}{2}\)] or [\(\frac{3π}{2}\), \(\frac{5π}{2}\)] or [- \(\frac{5π}{2}\), -\(\frac{3π}{2}\)] and so on.
Similarly, if cos θ = x (- 1 ≤ x ≤ 1 ) then θ = cos\(^{-1}\)x i.e., cos\(^{-1}\)x (cos-inverse x) represents an angle and the cosine of this angle is equal to x. We have similar significances of the angles tan\(^{-1}\)x (tan-inverse x), cot\(^{-1}\)x (cot-inverse x), sec\(^{-1}\)x (sec-inverse x) and csc\(^{-1}\)x (csc-inverse x).
Therefore, if sin θ = x (- 1 ≤ x ≤ 1) then θ = sin\(^{-1}\)x;
if cos θ = x (- 1 ≤ x ≤ 1) then θ = cos\(^{-1}\)x ;
if tan θ = x (- ∞ < x < ∞) then θ = tan\(^{-1}\)x ;
if csc θ = x (I x I ≥ 1) then θ = csc\(^{-1}\)x.
if sec θ = x (I x I ≥ 1) then θ = sec\(^{-1}\)x ; and
if cot θ = x (- ∞ < x < ∞) then θ = cot\(^{-1}\)x ;
Conversely, sin\(^{-1}\)x = θ ⇒ sin θ = x;
cos\(^{-1}\)x = θ ⇒ cos θ = x
tan\(^{-1}\)x = θ ⇒ tan θ = x
csc\(^{-1}\)x = θ ⇒ csc θ = x
cot\(^{-1}\)x = θ ⇒ cot θ = x
The trigonometrical functions sin\(^{-1}\)x, cos\(^{-1}\)x, tan\(^{-1}\)x, cot\(^{-1}\)x, sec\(^{-1}\)x and csc\(^{-1}\)x are called Inverse Circular Functions.
Note: It should be noted that sin\(^{-1}\)x is not equal to (sin x)\(^{-1}\). Also noted that (sin x)\(^{-1}\)is an angle whose sin is x. Remember that sin\(^{-1}\)x is a circular function but (sin x )\(^{-1}\) is the reciprocal of sin x i.e., (sin x)\(^{-1}\) = 1/sin x and it represents a pure number.
● Inverse Trigonometric Functions
11 and 12 Grade Math
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