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All chaptersMaths · Unit 1

Inverse Trigonometric Functions

6 articles26 formulas32 ways the board asks it
MATProperties & Identities

Properties and Identities of Inverse Trig Functions

This is the toolbox of standard identities: complementary relations, negative-argument rules, the arctangent/arcsine/arccosine addition formulae, and the double-angle conversions of 2tan⁡−1x2\tan^{-1}x. The method is to recognise which identity matches the structure of an expression and to apply it within its stated validity interval.

These identities are the building blocks for every proof, simplification, and equation in the chapter.

Complementary identities
sin⁡−1x+cos⁡−1x=π2,tan⁡−1x+cot⁡−1x=π2,sec⁡−1x+cosec⁡−1x=π2\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2},\quad \tan^{-1}x+\cot^{-1}x=\dfrac{\pi}{2},\quad \sec^{-1}x+\operatorname{cosec}^{-1}x=\dfrac{\pi}{2}
First holds for x∈[−1,1]x\in[-1,1], second for all real xx, third for ∣x∣≥1|x|\ge1.
Reciprocal identities
tan⁡−1x+tan⁡−11x={π2,x>0−π2,x<0\tan^{-1}x+\tan^{-1}\dfrac{1}{x}=\begin{cases}\dfrac{\pi}{2}, & x>0\\[4pt] -\dfrac{\pi}{2}, & x<0\end{cases}
Also cot⁡−1x=tan⁡−11x\cot^{-1}x=\tan^{-1}\dfrac{1}{x} for x>0x>0 (add π\pi if x<0x<0).
Arctangent addition (worked)
tan⁡−112+tan⁡−113=tan⁡−112+131−16=tan⁡−11=π4\tan^{-1}\dfrac{1}{2}+\tan^{-1}\dfrac{1}{3}=\tan^{-1}\dfrac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}=\tan^{-1}1=\dfrac{\pi}{4}
Uses tan⁡−1x+tan⁡−1y=tan⁡−1x+y1−xy\tan^{-1}x+\tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy} with xy=16<1xy=\dfrac{1}{6}<1.
Double-angle conversions of 2tan⁡−1x2\tan^{-1}x
2tan⁡−1x=sin⁡−12x1+x2=cos⁡−11−x21+x2=tan⁡−12x1−x22\tan^{-1}x=\sin^{-1}\dfrac{2x}{1+x^{2}}=\cos^{-1}\dfrac{1-x^{2}}{1+x^{2}}=\tan^{-1}\dfrac{2x}{1-x^{2}}
Arcsin form: −1≤x≤1-1\le x\le1; arccos form: x≥0x\ge0; arctan form: −1<x<1-1<x<1.
Composition (drawing a right triangle)
cos⁡(sin⁡−1x)=1−x2,sin⁡ ⁣(cos⁡−1x)=1−x2\cos(\sin^{-1}x)=\sqrt{1-x^{2}},\qquad \sin\!\left(\cos^{-1}x\right)=\sqrt{1-x^{2}}
Let θ=sin⁡−1x\theta=\sin^{-1}x so sin⁡θ=x\sin\theta=x; then read off the other ratio, e.g. cos⁡ ⁣(sin⁡−135)=45\cos\!\left(\sin^{-1}\dfrac{3}{5}\right)=\dfrac{4}{5}.
  • The complementary identity is the single most-used tool: it lets you swap cos⁡−1x\cos^{-1}x for π2−sin⁡−1x\dfrac{\pi}{2}-\sin^{-1}x to unify an expression.
  • Negative-argument rules: sin⁡−1(−x)=−sin⁡−1x\sin^{-1}(-x)=-\sin^{-1}x and tan⁡−1(−x)=−tan⁡−1x\tan^{-1}(-x)=-\tan^{-1}x (odd), but cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x)=\pi-\cos^{-1}x and cot⁡−1(−x)=π−cot⁡−1x\cot^{-1}(-x)=\pi-\cot^{-1}x.
  • The three forms of 2tan⁡−1x2\tan^{-1}x each have a DIFFERENT validity interval — choose the form whose interval contains your xx.
  • For sin⁡−1x+sin⁡−1y\sin^{-1}x+\sin^{-1}y and cos⁡−1x+cos⁡−1y\cos^{-1}x+\cos^{-1}y, apply the addition formula only when the sum stays in range (check x2+y2≤1x^{2}+y^{2}\le1 or x+y≥0x+y\ge0); otherwise add/subtract a correction.
  • To evaluate compositions like cos⁡ ⁣(sin⁡−135)\cos\!\left(\sin^{-1}\dfrac{3}{5}\right), set the inverse equal to an angle, build a right triangle, and read the required ratio.
  • The arctangent sum needs xy<1xy<1 for the clean form; for xy>1xy>1 use the ±π\pm\pi correction (see Proving Identities).
  • These identities work in both DIRECTIONS — expanding a single inverse term into a sum, or collapsing a sum into one term.
Where the marks go
  • Applying 2tan⁡−1x=cos⁡−11−x21+x22\tan^{-1}x=\cos^{-1}\dfrac{1-x^{2}}{1+x^{2}} for x<0x<0 — the arccos form requires x≥0x\ge0 (for x<0x<0 the right side equals −2tan⁡−1x-2\tan^{-1}x).
  • Treating cos⁡−1\cos^{-1} as odd: writing cos⁡−1(−x)=−cos⁡−1x\cos^{-1}(-x)=-\cos^{-1}x instead of π−cos⁡−1x\pi-\cos^{-1}x.
  • Using tan⁡−1x+tan⁡−11x=π2\tan^{-1}x+\tan^{-1}\dfrac{1}{x}=\dfrac{\pi}{2} for negative xx, where it should be −π2-\dfrac{\pi}{2}.
  • Forgetting the domain x∈[−1,1]x\in[-1,1] on sin⁡−1x+cos⁡−1x=π2\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2} and applying it to an out-of-range value.
How the board asks it
  • Numericaldrawing a right triangle for compositions and the arccosine/arcsine addition formula
    Find the value of cos⁡ ⁣(sin⁡−135+cos⁡−11213)\cos\!\left(\sin^{-1}\dfrac{3}{5}+\cos^{-1}\dfrac{12}{13}\right).
  • Derive / provethe arctangent addition formula with xy<1xy<1
    Prove that tan⁡−112+tan⁡−113=π4\tan^{-1}\dfrac{1}{2}+\tan^{-1}\dfrac{1}{3}=\dfrac{\pi}{4}.
  • Numericalthe arctangent addition formula reduced to a single equation
    Solve for xx: tan⁡−1(x+1)+tan⁡−1(x−1)=tan⁡−1831\tan^{-1}(x+1)+\tan^{-1}(x-1)=\tan^{-1}\dfrac{8}{31}.
  • Numericalthe substitution x=tan⁡θx=\tan\theta with half-angle identities
    Express tan⁡−11+x2−1x\tan^{-1}\dfrac{\sqrt{1+x^{2}}-1}{x} in its simplest form.
  • Numericalnegative-argument rules within the principal range
    Find the principal value of sin⁡−1 ⁣(−12)+cos⁡−1 ⁣(−32)\sin^{-1}\!\left(-\dfrac{1}{2}\right)+\cos^{-1}\!\left(-\dfrac{\sqrt{3}}{2}\right).

Practise this topic

PreviousEvaluating Principal ValuesNextProving Identities

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.