Each inverse trigonometric function inherits properties from its parent. Monotonicity flips orientation only when the parent is decreasing, and the graph is always a reflection of the restricted parent across the line y=x. This subtopic catalogues the six graphs and the symmetries you should know on sight.
The six graphs
y=sin−1x. Domain [−1,1], range [−π/2,π/2]. Increasing. Passes through (−1,−π/2),(0,0),(1,π/2). Odd function: sin−1(−x)=−sin−1x.
y=cos−1x. Domain [−1,1], range [0,π]. Decreasing. Passes through (−1,π),(0,π/2),(1,0). Reflection of sin−1 about the line y=π/4: in fact cos−1x=π/2−sin−1x.
y=cot−1x. Domain R, range (0,π). Decreasing. Asymptotes y=0 (as x→∞) and y=π (as x→−∞). cot−1(−x)=π−cot−1x.
y=sec−1x. Domain ∣x∣≥1, range [0,π]∖{π/2}. Two branches: one on [1,∞) increasing from 0 to π/2, one on (−∞,−1] increasing from π/2 to π.
y=csc−1x. Domain ∣x∣≥1, range [−π/2,π/2]∖{0}. Decreasing on each branch, symmetric to sec−1.
Symmetry identities
Identity
Reason
sin−1(−x)=−sin−1x
sin is odd on its branch
cos−1(−x)=π−cos−1x
cos is even but the branch shifts the angle
tan−1(−x)=−tan−1x
tan is odd on its branch
cot−1(−x)=π−cot−1x
cot shifts to the other branch
Monotonicity by inheritance
Since sin is increasing on [−π/2,π/2], sin−1 is increasing on [−1,1]. Since cos is decreasing on [0,π], cos−1 is decreasing on [−1,1]. The same logic applies to the other four.
Derivatives (preview)
The derivatives are computed in Chapter 5, but it is useful to know:
Function
Derivative
sin−1x
1−x21
cos−1x
−1−x21
tan−1x
1+x21
cot−1x
−1+x21
sec−1x
∥x∥x2−11
csc−1x
−∥x∥x2−11
Reading the graph: an example
Suppose you are asked: for which x is sin−1x=tan−1x? From the graphs, both start at 0 for x=0 and both are increasing. sin−1 rises faster (its derivative 1/1−x2 exceeds 1/(1+x2) for x∈(0,1)). So the only intersection is at x=0.
Worked examples
Example 1. Without computing, decide whether sin−1(0.6)>sin−1(0.5).
Yes, because sin−1 is strictly increasing on [−1,1].
Example 2. Decide whether cos−1(0.4)>cos−1(0.5).
Yes, because cos−1 is strictly decreasing.
Example 3. Sketch y=sin−1(sinx) for x∈[−2π,2π].
Inside [−π/2,π/2], the function is just x. Outside, fold using sin(π−x)=sinx. The graph is a triangular wave with peaks at ±π/2 and slope alternating between +1 and −1.
Example 4. Sketch y=cos−1(cosx) on [−2π,2π].
Inside [0,π], it is x. Use cos(−x)=cosx and periodicity to fold. The graph is a triangular wave with peaks at π and troughs at 0, 2π.
Example 5. Show that sin−1x+cos−1x=π/2 for all x∈[−1,1].
Let θ=sin−1x, so x=sinθ and θ∈[−π/2,π/2]. Then cos(π/2−θ)=sinθ=x, and π/2−θ∈[0,π]. So cos−1x=π/2−θ, giving the identity.
Example 6. Plot tan−1x and cot−1x on the same axes. Confirm that tan−1x+cot−1x=π/2 for all x.
The two curves are reflections of each other across y=π/4, summing to a constant π/2 at every x.
Try it yourself
Sketch y=sin−1x and mark its monotonicity.
Find cos−1(cos(2.5)) given π≈3.14.
Show sin−1(−x)=−sin−1x using the principal branch.
Why is the graph of cot−1x continuous at x=0 but not differentiable in elementary ways?
Sketch y=sin−1x+cos−1x for x∈[−1,1] , note the constant value.
Compute sec−1(2).
Compute sec−1(−2).
Find the points where y=sin−1(sinx) has its peaks for x∈[−3π,3π].
Sketch y=tan−1(tanx) for x∈[−π,π].
State the horizontal asymptotes of tan−1x.
Why does cos−1 have no points of inflection?
Verify the identity tan−1x+cot−1x=π/2 by differentiating both sides.
For which x is sin−1x=cos−1x?
Show that sin−1x+sin−1(−x)=0 for x∈[−1,1].
Pitfalls / Tricks
cos−1 is not odd. cos−1(−x)=π−cos−1x, not −cos−1x.
The graph of sin−1(sinx) is piecewise linear , a sawtooth, not a smooth sine.
Conventions for sec−1 vary; in some books the range includes [π/2,π) on the left branch instead of (π/2,π].
The derivative of sin−1x blows up at x=±1 , the graph has vertical tangent at the endpoints.
Next we collect the standard inverse trigonometric identities.