Inverse trigonometric identities combine familiar trigonometric relationships with the principal branches of arcsine, arccosine and arctangent. The algebra can look simple, but the branch conditions are part of the mathematics.

A formula that is correct for ordinary tangent may need a ±π correction after arctangent is applied. A square root may need a sign justified by a principal range. A composition may undo exactly in one order but fold angles back into a restricted interval in the other.

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Inverse identities are branch-sensitive

An inverse identity must return a value in the chosen principal range. The conditions on the input and output are part of the result, not optional notes.

The identities below use real inputs and radian outputs. Arcsine returns [−π/2,π/2][-\pi/2,\pi/2], arccosine returns [0,π][0,\pi], and arctangent returns (−π/2,π/2)(-\pi/2,\pi/2).

Direct compositions

A trig function applied after its inverse returns the input on the inverse function’s domain:

sin(arcsinx)=x(−1≤x≤1)\sin(\arcsin x)=x\quad(-1\le x\le1)
cos(arccosx)=x(−1≤x≤1)\cos(\arccos x)=x\quad(-1\le x\le1)
tan(arctanx)=x(x∈ℝ)\tan(\arctan x)=x\quad(x\in\mathbb R)

The reverse compositions equal x only on the principal ranges: arcsin(sin x) = x for x in [−π/2, π/2], arccos(cos x) = x for x in [0, π], and arctan(tan x) = x for x in (−π/2, π/2).

For a second explanation, see OpenStax’s treatment of inverse compositions.

What reverse composition returns

A reverse composition returns the principal angle with the same trigonometric value. It need not return an angle coterminal with the original one. For example,

arcsin(sin(5π/6))=π/6\arcsin(\sin(5\pi/6))=\pi/6

The two angles differ by 2π/32\pi/3, not by a whole turn, so they are not coterminal. Their sines agree because they are supplementary.

On the interval −π/2≤x≤3π/2-\pi/2\le x\le3\pi/2, the folding can be written exactly:

arcsin(sinx)={x,−π/2≤x≤π/2π−x,π/2<x≤3π/2\arcsin(\sin x)=\begin{cases}x,&-\pi/2\le x\le\pi/2\\\pi-x,&\pi/2<x\le3\pi/2\end{cases}

Extend this result periodically for other inputs. For arctan(tanx)\arctan(\tan x), first exclude x=π/2+kπx=\pi/2+k\pi; tangent is undefined there, so the composition cannot assign a value.

Complement and parity identities

For −1 ≤ x ≤ 1, arcsine and arccosine are complementary:

arcsinx+arccosx=π2\arcsin x+\arccos x=\frac\pi2

Arcsine and arctangent are odd:

arcsin(−x)=−arcsinx,arctan(−x)=−arctanx\arcsin(-x)=-\arcsin x,\qquad \arctan(-x)=-\arctan x

Arccosine instead satisfies arccos(−x) = π − arccos x because its principal outputs lie in [0, π].

Radical composition identities

The Pythagorean identity gives several useful compositions. Let θ = arcsin x. Since θ is in [−π/2, π/2], cosine is non-negative, so

cos(arcsinx)=1−x2\cos(\arcsin x)=\sqrt{1-x^2}

Similarly,

sin(arccosx)=1−x2\sin(\arccos x)=\sqrt{1-x^2}

For |x| < 1,

tan(arcsinx)=x1−x2\tan(\arcsin x)=\frac{x}{\sqrt{1-x^2}}

and for every real x,

sin(arctanx)=x1+x2,cos(arctanx)=11+x2\sin(\arctan x)=\frac{x}{\sqrt{1+x^2}},\qquad \cos(\arctan x)=\frac1{\sqrt{1+x^2}}

Converting between inverse functions

A radical composition can convert one inverse function to another. For |x| < 1,

arcsinx=arctan(x1−x2)\arcsin x=\arctan\left(\frac{x}{\sqrt{1-x^2}}\right)

Both sides lie in (−π/2, π/2), so the branch choice agrees. At x = ±1, the arcsine side is defined while the displayed arctangent expression has a zero denominator, so the endpoints require separate treatment.

The reciprocal arctangent identity

The familiar identity involving arctan x and arctan(1/x) depends on the sign of x:

arctanx+arctan(1/x)={π/2,x>0−π/2,x<0\arctan x+\arctan(1/x)=\begin{cases}\pi/2,&x>0\\-\pi/2,&x<0\end{cases}

There is no x = 0 case because 1/x is undefined. The sign condition is not optional; it is what keeps the result inside the correct branch geometry.

Arctangent addition needs branch corrections

Let α=arctanu\alpha=\arctan u and β=arctanv\beta=\arctan v, with real u,vu,v. Both angles lie between −π/2-\pi/2 and π/2\pi/2, so their sum lies between −π-\pi and π\pi. When uv≠1uv\ne1, define

R=u+v1−uvR=\frac{u+v}{1-uv}

The principal value arctanR\arctan R must sometimes be adjusted:

arctanu+arctanv={arctanR,uv<1arctanR+π,uv>1,u>0arctanR−π,uv>1,u<0π/2,uv=1,u>0−π/2,uv=1,u<0\arctan u+\arctan v=\begin{cases}\arctan R,&uv<1\\\arctan R+\pi,&uv>1,\ u>0\\\arctan R-\pi,&uv>1,\ u<0\\\pi/2,&uv=1,\ u>0\\-\pi/2,&uv=1,\ u<0\end{cases}

Why these cases are necessary

sin(α+β)=u+v(1+u2)(1+v2)\sin(\alpha+\beta)=\frac{u+v}{\sqrt{(1+u^2)(1+v^2)}}
cos(α+β)=1−uv(1+u2)(1+v2)\cos(\alpha+\beta)=\frac{1-uv}{\sqrt{(1+u^2)(1+v^2)}}

The denominator is positive. When uv<1uv<1, the cosine of the sum is positive; since the sum is between −π-\pi and π\pi, it is already in the principal arctangent range. When uv>1uv>1, the cosine is negative, and the sign of u+vu+v chooses the required upper or lower quadrant. When uv=1uv=1, the cosine is zero and the sine sign chooses π/2\pi/2 or −π/2-\pi/2. This establishes the cases for all real inputs, not merely for a few numerical examples.

For u=2u=2 and v=3v=3, R=−1R=-1. Its principal arctangent is −π/4-\pi/4, but the sum of the two positive angles is 3π/43\pi/4, so adding π\pi is essential.

The branch chosen by arctangent

Arctangent stays strictly between minus pi over two and pi over two.

The graph’s restricted output range explains why a sum of inverse tangents may need a π\pi correction.

Arctangent.svg · Geek3; SVG correction by Tiger66 · CC BY-SA 4.0.

Diagram explained

For positive inputs the curve lies above zero; for negative inputs it lies below zero. It approaches the horizontal limits minus pi/2 and pi/2 but never reaches them at a finite input. A sum of two outputs can leave this interval even though each output lies inside it.

Chart values and boundaries
Chart values and boundaries
InputArctangent
−1-1−π/4-\pi/4
0000
11π/4\pi/4
u→+∞u\to+\inftyApproaches π/2\pi/2; never reaches it

No simple distribution over sums

Inverse trigonometric functions do not distribute over addition. In general, arcsin(u + v) is not arcsin u + arcsin v, and arccos(uv) is not arccos u · arccos v. These functions are non-linear.

If an expression contains a sum or product inside an inverse trig function, use a valid angle formula or a geometric substitution rather than inventing a distribution rule.

Using triangles to derive identities

For θ=arctanx\theta=\arctan x with unrestricted real xx, use the signed coordinate (1,x)(1,x). Its distance from the origin is 1+x2\sqrt{1+x^2}. The horizontal coordinate is positive, so cosine is positive; the sign of sine is the sign of xx.

An ordinary reference triangle uses side lengths 11, |x||x| and 1+x2\sqrt{1+x^2}. Restore the sign from the coordinate quadrant after using its lengths. At x=0x=0 the reference triangle is degenerate, but the coordinate formulas still give sine zero and cosine one.

sin(arctanx)=x1+x2\sin(\arctan x)=\frac{x}{\sqrt{1+x^2}}
cos(arctanx)=11+x2\cos(\arctan x)=\frac1{\sqrt{1+x^2}}

For example, sin(arctan(−3))=−3/10\sin(\arctan(-3))=-3/\sqrt{10}. Treating the opposite side as a negative physical length would hide the distinction between geometry and signed coordinates.

Related topics

Use these branch-aware relationships when simplifying inverse-trig expressions, solving equations or converting coordinate directions. For the non-inverse identity families, return to Trigonometric Identities.

Practice questions

  1. Find sin(arctan3)\sin(\arctan3).
  2. Find cos(arcsin(−4/5))\cos(\arcsin(-4/5)).
  3. Evaluate arctan2+arctan(1/2)\arctan2+\arctan(1/2).
Check your answers
  1. 3/103/\sqrt{10}, using the coordinate (1,3)(1,3).
  2. 1−16/25=3/5\sqrt{1-16/25}=3/5. Cosine is non-negative on the arcsine range.
  3. π/2\pi/2. Both inputs are positive and their product is one.

Sources and further reading

Further explanations of the definitions and conventions used here: