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). …
Each part is a distinct equation-solving technique built from the chapter's sum/double-angle identities; we solve each in turn and check the resulting value against any stated restriction.
Step 1. (i) Use the complementary-angle idea. If x=13, then 135 and 1312 are the legs-over-hypotenuse of a 5–12–13 right triangle, so sin−1135+sin−11312=2π exactly (since (135)2+(1312)2=1 means the two angles are complementary).
Step 2. (i) Confirm by direct algebra. Let θ=sin−1x5. The equation forces cosθ=sin(2π−θ)=x12, so 1−x225=x12⇒1−x225=x2144⇒x2=169⇒x=±13. Only x=13 keeps both arguments positive so the sum of two positive arcsines can equal +2π (at x=−13 the sum would be −2π). So x=13.
Step 3. (ii) Apply the double-angle cosine-inverse identity. For t≥0, cos−11+t21−t2=2tan−1t. With a,b>0: cos−11+a21−a2=2tan−1a and cos−11+b21−b2=2tan−1b.
Step 4. (ii) Substitute and simplify.2tan−1x=2tan−1a−2tan−1b⇒tan−1x=tan−1a−tan−1b=tan−11+aba−b (valid since a,b>0⇒ab>−1). Hence x=1+aba−b.
Step 5. (iii) Apply the double-angle tangent identity to the left side. With t=cosx (so ∣t∣<1 generically), 2tan−1(cosx)=tan−11−cos2x2cosx=tan−1sin2x2cosx.
Step 6. (iii) Equate the arguments.sin2x2cosx=2cosecx=sinx2⇒sin2xcosx=sinx1⇒cosx=sinx⇒tanx=1. …