Simplification of Expressions

MATSimplifying & Solving

These questions hand you a complicated inverse-trig expression (often with a surd or a half-angle structure) and ask for its simplest form. The standard method is a trigonometric substitution — put $x=\tan\theta$ , $x=\sin\theta$ , or $x=\cos\theta$ — so the inner expression collapses to a single trig ratio of a multiple/sub-multiple angle, then cancel with the inverse.

Picking the right substitution turns a page of algebra into one line.

Surd form (put

x=\tan\theta

)

\tan^{-1}\dfrac{\sqrt{1+x^{2}}-1}{x}=\dfrac{1}{2}\tan^{-1}x,\quad x\ne0

x=\tan\theta

gives

\sqrt{1+x^{2}}=\sec\theta

and the fraction becomes

\tan\dfrac{\theta}{2}

Half-angle (cosine) form

\tan^{-1}\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac{x}{2},\quad 0<x<\pi

Uses

\dfrac{1-\cos x}{1+\cos x}=\tan^{2}\dfrac{x}{2}

; the radical equals

\tan\dfrac{x}{2}

since it is positive on

(0,\pi)

Cosine-over-(1+sine) form

\tan^{-1}\dfrac{\cos x}{1+\sin x}=\dfrac{\pi}{4}-\dfrac{x}{2},\quad -\dfrac{\pi}{2}<x<\dfrac{\pi}{2}

Write

\cos x

and

1+\sin x

in half-angle form, divide, then recognise

\tan\!\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)

Double-angle arcsine (put

x=\sin\theta

)

\sin^{-1}\!\left(2x\sqrt{1-x^{2}}\right)=2\sin^{-1}x,\quad -\dfrac{1}{\sqrt{2}}\le x\le\dfrac{1}{\sqrt{2}}

x=\sin\theta

gives

2x\sqrt{1-x^{2}}=\sin2\theta

; the range keeps

2\theta

in the arcsine branch.

Choose the substitution by the algebraic shape: $\sqrt{1+x^{2}}$ or $\dfrac{2x}{1-x^{2}}$ suggests $x=\tan\theta$ ; $\sqrt{1-x^{2}}$ suggests $x=\sin\theta$ (or $\cos\theta$ ).
After substituting, simplify the inner expression to a single ratio of $\theta$ , $\dfrac{\theta}{2}$ , or $2\theta$ , then cancel the outer inverse function.
Convert half-angle radicals with $1-\cos x=2\sin^{2}\dfrac{x}{2}$ and $1+\cos x=2\cos^{2}\dfrac{x}{2}$ .
Always carry the validity interval forward — the final answer (like $\dfrac{x}{2}$ or $\dfrac{\pi}{4}-\dfrac{x}{2}$ ) is only valid where the inverse cancellation is allowed.
A radical $\sqrt{(\cdot)^{2}}$ equals $|\cdot|$ ; on the given interval check the sign so you drop the modulus correctly (on the standard intervals it is positive).
Resubstitute $\theta=\tan^{-1}x$ (or $\sin^{-1}x$ ) at the very end to express the answer back in $x$ where required.
These simplifications often feed straight into differentiation questions, so reducing to $\dfrac{1}{2}\tan^{-1}x$ or $\dfrac{x}{2}$ first makes $\dfrac{dy}{dx}$ trivial.

Where the marks go

Dropping the modulus when taking $\sqrt{(\cdot)^{2}}$ — failing to check the sign on the given interval gives a wrong-signed answer.
Omitting the validity interval, so a result like $\sin^{-1}\!\left(2x\sqrt{1-x^{2}}\right)=2\sin^{-1}x$ is quoted where it actually fails (outside $|x|\le\dfrac{1}{\sqrt{2}}$ ).
Choosing the wrong substitution (e.g. $x=\sin\theta$ for a $\sqrt{1+x^{2}}$ expression), which produces a $\sqrt{1+\sin^{2}\theta}$ that won't simplify.
Stopping at $\dfrac{\theta}{2}$ without resubstituting back to $\dfrac{1}{2}\tan^{-1}x$ , leaving the answer in terms of an undefined $\theta$ .

How the board asks it

Numericalthe multiple-angle form (put $x=\tan\theta$ )
Express $\tan^{-1}\left(\dfrac{3x-x^{3}}{1-3x^{2}}\right)$ in the simplest form.
Derive / provedouble-angle arcsine (put $x=\sin\theta$ )
Prove that $\sin^{-1}\left(2x\sqrt{1-x^{2}}\right)=2\sin^{-1}x$ , stating the interval of $x$ for which it is valid.
Numericalreducing the surd form to $\dfrac{1}{2}\tan^{-1}x$ first
If $y=\tan^{-1}\left(\dfrac{\sqrt{1+x^{2}}-1}{x}\right)$ , simplify $y$ and hence find $\dfrac{dy}{dx}$ .
Derive / provethe half-angle (cosine) form
Prove that $\tan^{-1}\left(\dfrac{\cos x}{1+\sin x}\right)=\dfrac{\pi}{4}-\dfrac{x}{2}$ for $x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ .
Give reasonsthe modulus from $\sqrt{(\cdot)^{2}}$ and the validity interval
Simplify $\cos^{-1}\left(\dfrac{1-x^{2}}{1+x^{2}}\right)$ and give reasons for the sign of your answer over the interval $x>0$ .

Practise this topic

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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.