Concept understanding — Composite and Sum/Difference Identities of Inverse Trigonometric Functions
These are the working-formula properties (Properties VI–X) for combining or composing inverse trig functions, together with the reference-triangle technique for a raw composite like tan(sin−1x).
Composing a trig function with an unrelated inverse trig function (reference triangle). To evaluate f(g−1(x)) where f=g (e.g. cot(sin−1x)), let θ=g−1(x), build a right triangle encoding θ from the definition of g−1 (e.g. sinθ=x gives opposite =x, hypotenuse =1, so adjacent =1−x2 by Pythagoras — taking the adjacent side non-negative since θ∈[−2π,2π] keeps cosine ≥0 there), then read f(θ) straight off the triangle. This proves, e.g., tan(sin−1x)=1−x2x, −1<x<1.
Property VI (addition/subtraction formulas).
sin−1x+sin−1y=sin−1(x1−y2+y1−x2),if x2+y2≤1 or xy<0
with analogous subtraction forms for sine and cosine. Extending the tangent addition formula to three terms gives tan−1x+tan−1y+tan−1z=tan−1[1−xy−yz−zxx+y+z−xyz]; setting the left side equal to π and taking the tangent of both sides (which is 0) proves the classical identity x+y+z=xyz whenever tan−1x+tan−1y+tan−1z=π.
Property VII (double-angle-style formulas, from setting y=x in Property VI).
Property VIII.sin−1(2x1−x2)=2sin−1x for ∣x∣≤21, and sin−1(2x1−x2)=2cos−1x for 21≤x≤1 (the SAME left side splits into two different right sides depending on which half of [−1,1], i.e. which principal-range piece, x falls in). …
Combine the three arctangents pairwise using the addition formula, reduce to a single equation, and solve the resulting cubic — checking each root against the original equation. …
Combining tan−1(x−1)+tan−1(x+1) first (their sum is symmetric) and then adding tan−1x, matching against tan−1(3x), reduces the equation to a cubic with three real roots — and all three check out directly in the original equation.
Step 1. Combine the outer two terms. For ∣x∣<2 (so that (x−1)(x+1)=x2−1<1), tan−1(x−1)+tan−1(x+1)=tan−11−(x−1)(x+1)(x−1)+(x+1)=tan−12−x22x.
Step 2. Add the middle term.tan−12−x22x+tan−1x=tan−11−2−x22x⋅x2−x22x+x=tan−12−3x2x(4−x2) (multiplying numerator and denominator by 2−x2 and simplifying).
Step 3. Equate with the right-hand side. Matching tan−12−3x2x(4−x2)=tan−1(3x) gives 2−3x2x(4−x2)=3x (for 2−3x2=0).
Step 4. Clear the denominator and simplify.x(4−x2)=3x(2−3x2)⇒4x−x3=6x−9x3⇒8x3−2x=0⇒2x(4x2−1)=0.
Step 5. Solve the cubic.x=0 or x2=41⇒x=±21 — three real roots. …