Inverse Trigonometric Functions

01 / UNDERSTAND & APPLY

1. Periodicity makes a branch choice unavoidable

Sine, cosine and tangent repeat their values, so none has an inverse on its full real domain. To obtain an inverse function, restrict the original function to an interval on which it is one-to-one and still attains its full range. The resulting inverse returns one principal angle, not every angle satisfying the original trigonometric equation.

The notation arcsin x or sin⁻¹x means inverse sine here. It does not mean 1/sin x, which is csc x. Likewise arctan x is an angle whose tangent is x, whereas cot x is the reciprocal of tan x where that reciprocal exists. Keep inverse-function notation separate from negative powers.

All angles in this lesson are measured in radians. An inverse-trigonometric output is a real angle, so it can be compared with π/2 or used inside another trigonometric function. Calculator mode changes numerical presentation but does not remove the need to know the principal range.

A branch choice is part of the function definition. The common arcsine, arccosine and arctangent conventions are widely shared. Arccotangent and inverse secant conventions can differ across sources, so the conventions used below are stated explicitly before any identities are applied.

02 / UNDERSTAND & APPLY

2. The domain is a ratio; the range is a selected angle

Arcsin maps [−1,1] to [−π/2,π/2], reversing sine restricted to that interval. It is increasing and includes both endpoints. Arccos maps [−1,1] to [0,π], reversing cosine on that interval. It is decreasing and also includes both endpoints.

Arctan maps all of ℝ to (−π/2,π/2). Its endpoint angles are excluded because tangent is undefined there; its outputs approach them only as the input tends to positive or negative infinity. Under the convention used here, arccot maps ℝ to (0,π), is decreasing, and satisfies arccot x=π/2−arctan x.

Arcsec is defined on (−∞,−1]∪[1,∞) with range [0,π] excluding π/2. Equivalently arcsec x=arccos(1/x). Arccsc has the same input domain and range [−π/2,π/2] excluding zero; equivalently arccsc x=arcsin(1/x). These reciprocal definitions make the sign and endpoint conventions explicit.

Before evaluating a nested inverse function, impose its argument domain. For arcsin(g(x)) and arccos(g(x)), require −1≤g(x)≤1. For arcsec(g(x)) or arccsc(g(x)), require |g(x)|≥1. Arctan and this arccot accept every real argument, though an inner rational formula can still have its own exclusions.

03 / UNDERSTAND & APPLY

3. Undoing works only on the chosen branch

The identity sin(arcsin x)=x holds for every x∈[−1,1]. But arcsin(sin θ)=θ only when θ lies in [−π/2,π/2]. Outside that interval, find the unique angle in the principal range with the same sine. The same distinction applies to cosine and tangent.

For θ∈[0,2π], arccos(cos θ) equals θ on [0,π] and 2π−θ on [π,2π]. For arctan(tan θ), subtract an integer multiple of π to place the result in (−π/2,π/2); values θ=π/2+kπ are excluded because the inner tangent does not exist.

A general arcsine folding rule is useful when its interval is stated: if θ∈[kπ−π/2,kπ+π/2] for an integer k, then arcsin(sin θ)=(−1)ᵏ(θ−kπ). At shared endpoints the neighboring formulas agree. This is a piecewise identity, not permission to cancel arcsin and sin globally.

For exact values, first determine the trigonometric ratio, then select the angle in the principal range. An angle with the right ratio but the wrong range is not the value of the inverse function, even though it solves a related trigonometric equation.

04 / UNDERSTAND & APPLY

4. A principal range decides the missing sign

If θ=arcsin x, then sin θ=x and θ∈[−π/2,π/2], so cos θ≥0. Therefore cos(arcsin x)=√(1−x²) for |x|≤1. The positive square root is justified by the angle range, not by a general rule that cosine is positive.

If θ=arccos x, then θ∈[0,π] and sin θ≥0, giving sin(arccos x)=√(1−x²). However cos θ=x may be negative. Thus tan(arccos x)=√(1−x²)/x for x≠0 within [−1,1], with the sign supplied by x.

For θ=arctan x, cosine is positive and tangent is x. A unit-circle or normalized-triangle calculation gives cos θ=1/√(1+x²) and sin θ=x/√(1+x²). The numerator x retains the sign; replacing it by |x| would incorrectly force sine positive.

Right-triangle reasoning is convenient for positive inputs, but a literal triangle with positive side lengths does not encode every quadrant. For negative inputs or an arccosine in the second quadrant, use the principal range and signed unit-circle coordinates to restore the correct signs.

05 / UNDERSTAND & APPLY

5. Complementary angles simplify without losing the branch

For x∈[−1,1], arcsin x+arccos x=π/2. If θ=arcsin x, then π/2−θ lies in [0,π] and has cosine x, so it is precisely arccos x. This argument proves both the ratio and the principal-range requirement.

Arcsin is odd: arcsin(−x)=−arcsin x. Arccos instead obeys arccos(−x)=π−arccos x. Arctan is odd, while the selected arccot branch obeys arccot(−x)=π−arccot x. A branch whose range is not symmetric about zero should not be assumed odd.

For x>0, arctan x+arctan(1/x)=π/2. For x<0, the sum is −π/2. The input x=0 is excluded from the reciprocal expression. A tangent calculation alone says the cosine of the sum is zero; the signs of the two principal angles select which half-turn boundary occurs.

Under the stated reciprocal definitions, arcsec x+arccsc x=π/2 for |x|≥1. This follows by applying the arcsine-arccosine complement identity to 1/x. Again, the identity depends on using the specified branches.

06 / UNDERSTAND & APPLY

6. A tangent value determines an angle only modulo π

Let A=arctan a and B=arctan b, so each lies in (−π/2,π/2) and their cosines are positive. The tangent addition formula gives tan(A+B)=(a+b)/(1−ab) when ab≠1. It does not by itself identify A+B as the principal arctangent of that ratio.

The cosine of A+B has the sign of 1−ab, while its sine has the sign of a+b. Also A+B∈(−π,π). These three facts locate the sum. If ab<1, its cosine is positive and the sum lies in the principal arctangent interval, so the uncorrected formula is valid.

If ab>1, the arguments have the same sign. For a,b>0 the sum lies in (π/2,π), so add π to the principal arctangent of the ratio. For a,b<0 the sum lies in (−π,−π/2), so subtract π. If ab=1, the sum is π/2 for positive arguments and −π/2 for negative arguments.

Differences can be treated as sums with −b, or analyzed through sin(A−B) and cos(A−B). Rather than memorizing an unqualified formula, determine the angle range before removing an arctangent. This is especially important when an equation involves several inverse tangents.

07 / UNDERSTAND & APPLY

7. Solve on the intersection of all domains

An inverse-trigonometric equation is an equation between principal values. Applying sine, cosine or tangent to both sides preserves a necessary equality, but may lose information because the outer trigonometric function is periodic or not injective over the relevant angle range.

Before applying a trigonometric function, bound the possible angles. If both sides lie in an interval where that function is injective, the transformation is reversible. Otherwise generate candidates and check them in the original principal-value equation.

Squaring a radical equation can introduce sign errors as well. For example, an equation 2x=√(1−x²) requires x≥0 before squaring. A negative algebraic root may satisfy the squared equation but fail the original sign requirement.

Endpoints need separate attention because arcsin and arccos include their endpoint ratios, while arctan excludes its limiting endpoint angles. A proposed equality to π/2 may be impossible for a single finite arctangent but possible for a sum of two.

08 / UNDERSTAND & APPLY

8. A double angle can leave the inverse range

Let θ=arctan x. Then sin2θ=2x/(1+x²), and this ratio always lies in [−1,1] because (|x|−1)²≥0. However 2θ ranges over (−π,π), wider than the principal arcsine interval.

If |x|≤1, then |θ|≤π/4 and arcsin(2x/(1+x²))=2arctan x. If x>1, the doubled angle lies in (π/2,π), so the principal arcsine is π−2arctan x. If x<−1, the doubled angle lies in (−π,−π/2), giving −π−2arctan x.

For cosine, cos(2θ)=(1−x²)/(1+x²). Since arccos returns an angle in [0,π], the corresponding identity is arccos((1−x²)/(1+x²))=2arctan|x| for every real x. The absolute value makes the selected angle nonnegative.

These formulas illustrate a general method: introduce the principal inner angle, express the ratio with a standard identity, and then fold the resulting angle into the outer inverse range. The branch correction is a geometric consequence of the chosen ranges.

09 / UNDERSTAND & APPLY

9. Ranges also control inequalities and derivative signs

Arcsin and arctan are increasing; arccos and the selected arccot are decreasing. Thus arcsin u≤arcsin v is equivalent to u≤v when u,v∈[−1,1], while arccos u≤arccos v reverses to u≥v. Domain restrictions remain necessary even when the inequality algebra looks simple.

On −1<x<1, the derivatives of arcsin and arccos are respectively 1/√(1−x²) and −1/√(1−x²). For all real x, arctan has derivative 1/(1+x²) and the selected arccot has derivative −1/(1+x²). These follow by differentiating the defining trigonometric identity on its branch.

With the stated conventions, arcsec has derivative 1/(|x|√(x²−1)) and arccsc has derivative −1/(|x|√(x²−1)) for |x|>1. The absolute value comes from the reciprocal definitions and √(x²)=|x|. The endpoint values ±1 are in the function domains but not in these finite-derivative domains.

Derivative formulas are useful checks on monotonicity within each connected domain interval. They do not erase discontinuities or gaps between domain components. A composite derivative also needs the chain rule and the original inner-domain restrictions.

10 / UNDERSTAND & APPLY

10. Original practice with branch-aware solutions

State the principal range used in each answer. For equations, verify the candidate in the original inverse-function statement.

11 / UNDERSTAND & APPLY

11. The ratio and the principal range must both match

An inverse-trigonometric value is determined by two conditions: its trigonometric ratio and its allowed angle range. Checking only the ratio leaves periodic and symmetry alternatives unresolved. This is the source of most incorrect cancellation and addition identities.

Before simplifying a composition, write the inner inverse range. Before using a square root, decide its sign from that range. Before replacing a sum by one arctangent, locate the sum’s quadrant. Before solving an equation, intersect the input domains and track whether applying a trigonometric function loses injectivity.

Remember the endpoint distinctions: arcsin and arccos accept ±1, arctan never returns ±π/2 for finite input, and reciprocal inverse functions exclude inputs between −1 and 1. Derivative domains can be narrower than function domains.

The final check is simple and powerful: does the proposed angle lie in the principal interval, and does its sine, cosine or tangent recover the requested ratio? If both conditions hold, the branch has been selected correctly.