Sublevo
ISC 2027
All chaptersMaths · Unit 1

Inverse Trigonometric Functions

6 articles26 formulas32 ways the board asks it
MATSimplifying & Solving

Solving Equations involving Inverse Trig Functions

These problems require finding the value(s) of xx satisfying an equation built from inverse-trig terms. The standard method is to combine the inverse terms into a single one using the addition formulae (or to take tan⁡\tan/sin⁡\sin/cos⁡\cos of both sides), reduce to an algebraic equation, solve it, and finally reject any root for which a term leaves its principal-value branch.

The compulsory last step — checking each root — is what earns full marks.

Combine then solve
tan⁡−1(x+1)+tan⁡−1(x−1)=tan⁡−1831 ⇒ tan⁡−12x2−x2=tan⁡−1831\tan^{-1}(x+1)+\tan^{-1}(x-1)=\tan^{-1}\dfrac{8}{31}\ \Rightarrow\ \tan^{-1}\dfrac{2x}{2-x^{2}}=\tan^{-1}\dfrac{8}{31}
Apply tan⁡−1a+tan⁡−1b=tan⁡−1a+b1−ab\tan^{-1}a+\tan^{-1}b=\tan^{-1}\dfrac{a+b}{1-ab}; equating arguments gives 4x2+31x−8=04x^{2}+31x-8=0, so x=14x=\dfrac{1}{4} (reject the extraneous x=−8x=-8).
Halving-type equation
tan⁡−11−x1+x=12tan⁡−1x, x>0 ⇒ x=13\tan^{-1}\dfrac{1-x}{1+x}=\dfrac{1}{2}\tan^{-1}x,\ x>0\ \Rightarrow\ x=\dfrac{1}{\sqrt{3}}
Use tan⁡−11−x1+x=π4−tan⁡−1x\tan^{-1}\dfrac{1-x}{1+x}=\dfrac{\pi}{4}-\tan^{-1}x, so π4=32tan⁡−1x\dfrac{\pi}{4}=\dfrac{3}{2}\tan^{-1}x, giving tan⁡−1x=π6\tan^{-1}x=\dfrac{\pi}{6}.
Sum equal to π4\dfrac{\pi}{4}
tan⁡−1(2x)+tan⁡−1(3x)=π4 ⇒ 5x1−6x2=1\tan^{-1}(2x)+\tan^{-1}(3x)=\dfrac{\pi}{4}\ \Rightarrow\ \dfrac{5x}{1-6x^{2}}=1
Take tan⁡\tan of both sides; solve 6x2+5x−1=06x^{2}+5x-1=0 to get x=16x=\dfrac{1}{6} (reject x=−1x=-1, since both terms are then negative and cannot sum to π4\dfrac{\pi}{4}).
Take the trig function of both sides
2tan⁡−1(cos⁡x)=tan⁡−1(2cosec⁡x) ⇒ tan⁡−12cos⁡x1−cos⁡2x=tan⁡−1(2cosec⁡x)2\tan^{-1}(\cos x)=\tan^{-1}(2\operatorname{cosec}x)\ \Rightarrow\ \tan^{-1}\dfrac{2\cos x}{1-\cos^{2}x}=\tan^{-1}(2\operatorname{cosec}x)
2cos⁡xsin⁡2x=2sin⁡x⇒cos⁡x=sin⁡x⇒x=π4\dfrac{2\cos x}{\sin^{2}x}=\dfrac{2}{\sin x}\Rightarrow\cos x=\sin x\Rightarrow x=\dfrac{\pi}{4} on (0,π)(0,\pi).
  • Strategy 1: collapse two inverse terms into one with the addition formula, then equate the inner arguments (since the outer functions are equal and one-to-one on their range).
  • Strategy 2: take tan⁡\tan (or sin⁡\sin/cos⁡\cos) of both sides to clear the inverses, but remember this can introduce extraneous roots — those MUST be checked.
  • Use tan⁡−11−x1+x=π4−tan⁡−1x\tan^{-1}\dfrac{1-x}{1+x}=\dfrac{\pi}{4}-\tan^{-1}x for the common 'halving' equations.
  • When the equation involves sin⁡−1x+cos⁡−1x\sin^{-1}x+\cos^{-1}x or tan⁡−1x+cot⁡−1x\tan^{-1}x+\cot^{-1}x, substitute π2\dfrac{\pi}{2} immediately to simplify.
  • After solving the algebraic equation, substitute each root back into the ORIGINAL equation and confirm every inverse term stays inside its principal branch and the two sides agree in sign.
  • Respect any stated domain (e.g. 0<x<π0<x<\pi) and discard roots outside it.
  • When two arctangents must sum to a value >π2>\dfrac{\pi}{2}, the xy>1xy>1 correction term (±π\pm\pi) can change which root is valid — keep it in mind when equating.
Where the marks go
  • Skipping the verification step and reporting an extraneous root introduced by taking tan⁡\tan of both sides (e.g. keeping x=−1x=-1 when the LHS is then negative).
  • Equating arguments without checking ab<1ab<1 in tan⁡−1a+tan⁡−1b\tan^{-1}a+\tan^{-1}b, missing a needed ±π\pm\pi shift and getting a wrong root.
  • Ignoring the given interval (such as 0<x<π0<x<\pi) and listing roots that violate it.
  • Squaring or cross-multiplying carelessly, creating spurious solutions whose inverse terms fall outside the principal-value range.
How the board asks it
  • Numericalcombining two inverse terms with the addition formula, then equating inner arguments
    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 halving identity tan⁡−11−x1+x=π4−tan⁡−1x\tan^{-1}\dfrac{1-x}{1+x}=\dfrac{\pi}{4}-\tan^{-1}x
    Find xx if tan⁡−11−x1+x=12tan⁡−1x\tan^{-1}\dfrac{1-x}{1+x}=\dfrac{1}{2}\tan^{-1}x, where x>0x>0.
  • Numericalequation mixing cos⁡−1\cos^{-1} and sin⁡−1\sin^{-1}, with a rejected root
    Solve cos⁡−1x+sin⁡−1x2=π6\cos^{-1}x+\sin^{-1}\dfrac{x}{2}=\dfrac{\pi}{6}, stating clearly any rejected root.
  • Derive / provetaking sin⁡\sin of both sides, then checking each root in the principal branch
    Solve sin⁡−1(1−x)−2sin⁡−1x=π2\sin^{-1}(1-x)-2\sin^{-1}x=\dfrac{\pi}{2} and verify which value of xx actually satisfies the original equation.
  • Give reasonsextraneous root introduced when two arctangents sum past π2\dfrac{\pi}{2}
    While solving tan⁡−12x+tan⁡−13x=π4\tan^{-1}2x+\tan^{-1}3x=\dfrac{\pi}{4}, a student obtains x=16x=\dfrac{1}{6} and x=−1x=-1. State which root must be rejected and give reasons.

Practise this topic

PreviousSimplification of Expressions

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.