A small set of identities lets you rewrite almost any inverse trigonometric expression. This subtopic catalogues the foundational identities , those that do not require addition formulas. They follow from the principal range conventions and from basic trigonometric identities like sin2θ+cos2θ=1.
Complementary identities
The most important identity in the chapter:
sin−1x+cos−1x=2π,x∈[−1,1].
Proof. Let θ=sin−1x, so sinθ=x and θ∈[−π/2,π/2]. Then cos(π/2−θ)=sinθ=x, with π/2−θ∈[0,π]. Hence cos−1x=π/2−θ, and addition gives π/2. ■
By the same argument:
tan−1x+cot−1x=2π,x∈R.
sec−1x+csc−1x=2π,∣x∣≥1.
These three identities are used in nearly every problem of the chapter.
Reciprocal identities
For ∣x∣≥1 we have csc−1x=sin−1(1/x) and sec−1x=cos−1(1/x).
Reason. If θ=csc−1x then cscθ=x, so sinθ=1/x. Since θ∈[−π/2,π/2]∖{0}, we have θ=sin−1(1/x).
For x>0 we also have cot−1x=tan−1(1/x), but this fails for x<0 , there one must write cot−1x=π+tan−1(1/x).
Negative argument identities
Identity
Domain
sin−1(−x)=−sin−1x
x∈[−1,1]
cos−1(−x)=π−cos−1x
x∈[−1,1]
tan−1(−x)=−tan−1x
x∈R
cot−1(−x)=π−cot−1x
x∈R
sec−1(−x)=π−sec−1x
$
csc−1(−x)=−csc−1x
$
Cross identities
Often we need to convert one inverse function into another. For x∈[0,1]:
sin−1x=cos−11−x2=tan−11−x2x.
Reason. If sinθ=x with θ∈[0,π/2], then cosθ=1−x2, hence cos−1(1−x2)=θ. And tanθ=x/1−x2.
For x≥0, tan−1x=sin−1(x/1+x2)=cos−1(1/1+x2).
These conversions are essential for combining inverse trig terms into a single one.
Worked examples
Example 1. Simplify sin−1x+cos−1x for x=0.3.
By the identity, the sum is π/2, regardless of the value of x (as long as ∣x∣≤1).
Example 2. Simplify sin−1(53)+cos−1(53).
Same identity: answer π/2.
Example 3. Compute tan−1(31)+cot−1(31).
Answer π/2.
Example 4. Show sin−1(sin32π)=π/3.
sin(2π/3)=sin(π−2π/3)=sin(π/3)=3/2. The principal value answer is sin−1(3/2)=π/3.
Example 5. Express sin−1(53) in terms of cos−1.
If sinθ=3/5 with θ∈[0,π/2], then cosθ=4/5. So sin−1(3/5)=cos−1(4/5).
Example 6. Show cos−1(−1/2)=π−cos−1(1/2)=π−π/3=2π/3.
Direct: cos(2π/3)=−1/2 and 2π/3∈[0,π]. So the answer is 2π/3.
Try it yourself
Evaluate sin−1(2/2)+cos−1(2/2).
Evaluate tan−1(3)+cot−1(3).
Show sec−1(2)+csc−1(2)=π/2.
Express sin−1(4/5) as cos−1 of something.
Express sin−1(5/13) as tan−1 of something.
Simplify sin−1(−1/2)+cos−1(−1/2).
Evaluate tan−1(−1)+cot−1(−1) , careful with branches.
Show cos−1(−x)=π−cos−1x by direct verification at x=1/2.
Show sin−1(−3/5)+cos−1(3/5)=π/2−2sin−1(3/5).
Find the value of sin−1(1)+cos−1(1).
Evaluate tan−1(2)+cot−1(2).
Verify the identity sec−1x=cos−1(1/x) at x=2.
For x>0 show cot−1x=tan−1(1/x).
For x<0 find a correct identity replacing cot−1x=tan−1(1/x).
Pitfalls / Tricks
The identity sin−1x+cos−1x=π/2 holds for all x∈[−1,1], including the boundary values.
cot−1x=tan−1(1/x) only for x>0; for x<0 add π.
Always restrict x to the natural domain before applying any identity. Asking for sin−1(2) is meaningless in real analysis.
For symbolic manipulations, sin−1(sinθ)=θ only when θ∈[−π/2,π/2].
The identity cos−1(−x)=π−cos−1x is asymmetric (compare to sin): the cosine inverse is not an odd function.
Next we tackle the additive identities for sums of two inverse trig terms.