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Inverse trigonometric compositions

Simplify trig functions of arcsin, arccos and arctan using signed triangles. Check domains, reversed compositions and piecewise inverse-trig identities.

Before you startPrincipal inverse trig ranges, Pythagoras and exact trig values

01 / Name the inner angle

Its principal range fixes the signs of the other ratios.

For an expression such as sin(arccos x), begin with θ = arccos x. You know cos θ = x, and you also know 0 ≤ θ ≤ π. That second fact determines which square root to choose.

Name the angle → state its range → find the other ratios → retain domains

The circle model shows why an obtuse principal arccos angle has negative cosine but positive sine. Change the inner inverse and the input to see which signs and exclusions change.

Let the branch choose the signsExplore

θ = arccos(3/5) ≈ 0.927 radians.

sin θ = √(1 − x²) = 0.800.

cos θ = x = 0.600.

tan θ = √(1 − x²)/x = 1.333.

θ lies in [0, π]; sine is nonnegative and cosine keeps the sign of x.

For x ≥ 0, arccos x = arcsin(√(1 − x²)).

Green is the signed horizontal coordinate; blue is the signed vertical coordinate. The radius is 1. Read signs from the principal angle, not from an unsigned triangle sketch. Values are rounded to 3 decimal places.

Inner inverseInput x

Watch equal sine values give different arccos angles

Pause, replay or seek freely. The notes explain the same idea and stay in view.

02 / Compose with arcsin

Cosine is nonnegative on the arcsin branch.

cos(arcsin x) = √(1 − x²), for −1 ≤ x ≤ 1
tan(arcsin x) = x/√(1 − x²), for −1 < x < 1

Find cos(arcsin(−3/5)) and tan(arcsin(−3/5))Worked example

Let θ = arcsin(−3/5), so −π/2 ≤ θ ≤ π/2

Since its sine is negative, θ lies in Quadrant IV or the equivalent negative-angle interval.

cos θ = √(1 − 9/25) = 4/5

Use the nonnegative cosine on the principal branch.

tan θ = (−3/5)/(4/5) = −3/4

The sign of the numerator is retained.

At x = ±1 the cosine formula still works and gives 0, but the tangent is undefined. The outer function can narrow the domain even when the inner inverse exists.

03 / Compose with arccos

Sine is nonnegative, but cosine can have either sign.

sin(arccos x) = √(1 − x²), for −1 ≤ x ≤ 1
tan(arccos x) = √(1 − x²)/x, for −1 ≤ x ≤ 1 and x ≠ 0

Find sin(arccos(−3/5)) and tan(arccos(−3/5))Worked example

Let θ = arccos(−3/5)

It lies in Quadrant II, inside [0, π].

sin θ = √(1 − 9/25) = 4/5

The sine is positive.

tan θ = (4/5)/(−3/5) = −4/3

The negative cosine stays in the denominator.

The tangent formula is defined at x = ±1: it gives 0. It is undefined at x = 0 because arccos 0 = π/2. Check the actual denominator instead of copying the arcsin case.

04 / Compose with arctan

Use a positive adjacent coordinate and a signed opposite coordinate.

sin(arctan x) = x/√(1 + x²)
cos(arctan x) = 1/√(1 + x²)
Both hold for every real x.

Find sin(arctan(−2)) and cos(arctan(−2))Worked example

Let θ = arctan(−2), so −π/2 < θ < π/2

The cosine is positive and the sine negative.

Use horizontal coordinate 1 and vertical coordinate −2

The distance from the origin is √5.

sin θ = −2/√5 and cos θ = 1/√5

Divide each signed coordinate by the positive radius.

The quantity √(1 + x²) is always positive. These two compositions have no extra excluded real input.

05 / Keep the domain of the inner expression

A formula is incomplete without its allowed inputs.

Simplify cos(arcsin(2x − 1))Worked example

Require −1 ≤ 2x − 1 ≤ 1

This gives 0 ≤ x ≤ 1.

Use the arcsin composition with input 2x − 1

The result is √(1 − (2x − 1)²).

Keep 0 ≤ x ≤ 1

The square root is nonnegative on this domain.

Simplify tan(arccos(3x))Worked example

Require −1 ≤ 3x ≤ 1

Initially −1/3 ≤ x ≤ 1/3.

The tangent also requires 3x ≠ 0

Exclude x = 0.

Result: √(1 − 9x²)/(3x)

Domain [−1/3, 1/3] with 0 removed.

06 / Reverse compositions can fold an angle

The output must return to the principal interval.

For −π ≤ θ ≤ π:
arcsin(sin θ) = −π − θ when −π ≤ θ < −π/2
= θ when −π/2 ≤ θ ≤ π/2
= π − θ when π/2 < θ ≤ π

Evaluate arcsin(sin(−5π/6))Worked example

The angle is in [−π, −π/2)

Use −π − θ.

−π − (−5π/6) = −π/6

This has the same sine and lies in the arcsin range.

Compare cosine and tangent foldingWorked example

arccos(cos θ) = |θ| on −π ≤ θ ≤ π

Cosine is even and arccos returns [0, π].

arctan(tan θ) = θ − π on π/2 < θ < 3π/2

Subtract one tangent period to return to (−π/2, π/2).

The tangent endpoints are excluded

The inner tangent is undefined there.

07 / Use complementary principal angles

Arcsin and arccos add to a fixed half-turn fraction.

arcsin x + arccos x = π/2, for −1 ≤ x ≤ 1

Explain why the identity works even for negative xWorked example

Let α = arcsin x, so −π/2 ≤ α ≤ π/2

Then β = π/2 − α lies in [0, π].

cos β = cos(π/2 − α) = sin α = x

β has the correct cosine and the correct arccos range.

Therefore β = arccos x

Rearrange to obtain the identity.

Simplify arcsin(cos θ) for 0 ≤ θ ≤ πWorked example

cos θ = sin(π/2 − θ)

The angle π/2 − θ lies in [−π/2, π/2].

arcsin(cos θ) = π/2 − θ

Here the inverse can undo sine because its input angle lies on the chosen branch.

08 / A square root loses the cosine sign

An acute angle cannot replace an obtuse arccos output.

Let α = arcsin(√(1 − x²)). For −1 ≤ x ≤ 1:
arccos x = α if x ≥ 0
arccos x = π − α if x < 0

Test x = −3/5Worked example

√(1 − x²) = 4/5, so α = arcsin(4/5)

This is an acute angle.

arccos(−3/5) is obtuse

Its sine is also 4/5, but its cosine is negative.

The correct result is π − arcsin(4/5)

Using α alone would choose the wrong quadrant.

At x = 0, both branches meet at π/2. At x = −1, the square-root expression gives α = 0 while arccos(−1) = π. Endpoint tests quickly reveal a missing branch correction.

09 / An arctan quotient needs a branch correction

The quotient is undefined at zero, and its principal angle can be negative.

For −1 ≤ x ≤ 1:
arccos x = arctan(√(1 − x²)/x) if x > 0
= π + arctan(√(1 − x²)/x) if x < 0
= π/2 if x = 0

Use x = −3/5 againWorked example

√(1 − x²)/x = −4/3

The quotient is the tangent of the required obtuse angle.

arctan(−4/3) lies between −π/2 and 0

It is not an arccos output for this input.

Add π to reach Quadrant II

arccos(−3/5) = π + arctan(−4/3).

No single constant added to the arctan quotient works across both positive and negative x. Also, assigning x = 0 to the quotient is invalid; evaluate arccos 0 separately.

10 / Check the inverse range before applying trig

Applying a function can hide an impossible target angle.

Solve arcsin(2x) = π/6Worked example

The target π/6 lies in the arcsin range

The inner input also requires −1/2 ≤ x ≤ 1/2.

2x = sin(π/6) = 1/2

So x = 1/4.

Check the original inverse equation

Arcsin(1/2) is indeed π/6.

Solve arctan x = 3π/4Worked example

3π/4 is outside (−π/2, π/2)

Solve arcsin x = 5π/6. Why does taking sine not settle the question?

Hint

Compare 5π/6 with the arcsin range.

Worked solution

No real solution. Taking sine gives x = 1/2, but arcsin(1/2) = π/6. The target 5π/6 is outside the principal range.

12 / Recap

A radical gives a magnitude; the branch supplies its meaning.

Name the inner principal angle and state its range.
Use signed coordinates when deriving the other ratios.
Check the inner domain and the outer denominator.
Reversed compositions return a principal angle, not necessarily the original angle.
Arcsin x + arccos x = π/2 on [−1, 1].
Square-root and arctan representations of arccos need piecewise branch corrections.
Check the inverse range before solving an inverse equation.

Section 1 of 12 · Name the inner angle