Proving Identities

MATProperties & Identities

These problems ask you to prove a target value (often $\dfrac{\pi}{4}$ , $\dfrac{\pi}{2}$ or $\pi$ ) by combining several inverse-trig terms using the addition formulae. The core method is to add two terms at a time with $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy}$ (adjusting by $\pm\pi$ when $xy>1$ ) and to convert $\sin^{-1}$ / $\cos^{-1}$ terms into a common function first.

They are examined because they test whether you can pick the right identity AND respect its principal-value condition.

Sum of two arctangents

\tan^{-1}x+\tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy},\quad xy<1

Valid when

xy<1

so the sum stays in

\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)

; typically used with

x>0,\,y>0

Arctangent sum when

xy>1

\tan^{-1}x+\tan^{-1}y=\pi+\tan^{-1}\dfrac{x+y}{1-xy},\quad x>0,\,y>0,\,xy>1

Add

\pi

because the true sum exceeds

\dfrac{\pi}{2}

; subtract

\pi

instead if

x<0,\,y<0,\,xy>1

Sum of two arcsines

\sin^{-1}x+\sin^{-1}y=\sin^{-1}\!\left(x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}\right)

Holds when

x^{2}+y^{2}\le1

, or when

xy<0

; here

x,y\in[-1,1]

so the result stays in the principal range.

Sum of two arccosines

\cos^{-1}x+\cos^{-1}y=\cos^{-1}\!\left(xy-\sqrt{1-x^{2}}\sqrt{1-y^{2}}\right)

Valid when

x+y\ge0

; if

x+y<0

the right side becomes

2\pi-\cos^{-1}(\cdots)

Worked target

\tan^{-1}\dfrac{1}{2}+\tan^{-1}\dfrac{1}{5}+\tan^{-1}\dfrac{1}{8}=\dfrac{\pi}{4}

Add

\tan^{-1}\dfrac{1}{2}+\tan^{-1}\dfrac{1}{5}=\tan^{-1}\dfrac{7}{9}

, then add

\tan^{-1}\dfrac{1}{8}

to get

\tan^{-1}1=\dfrac{\pi}{4}

Combine the terms two at a time; after using the formula, simplify the single fraction before bringing in the next term.
Before applying $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy}$ , always check $xy$ : use the plain formula only when $xy<1$ .
If $xy>1$ with both $x,y>0$ , add $\pi$ ; if both $x,y<0$ , subtract $\pi$ . This is exactly why $\tan^{-1}1+\tan^{-1}2+\tan^{-1}3=\pi$ , not $\dfrac{\pi}{4}$ .
To prove a value like $\dfrac{\pi}{4}$ , reduce the left side to a single inverse function whose value is standard, e.g. $\tan^{-1}1=\dfrac{\pi}{4}$ .
Mixed $\sin^{-1}$ and $\cos^{-1}$ terms: convert one type to the other (using $\cos^{-1}t=\sin^{-1}\sqrt{1-t^{2}}$ for $t\in[0,1]$ ) so a single addition formula applies.
In application problems (a frame/billboard subtending angle $\theta$ , or angle of elevation), write $\theta$ as a difference of two arctangents and collapse it: $\tan^{-1}\dfrac{b}{d}-\tan^{-1}\dfrac{a}{d}=\tan^{-1}\dfrac{(b-a)d}{d^{2}+ab}$ .
For elevation $E=\tan^{-1}\dfrac{h}{\text{horizontal distance}}$ : as the distance $\to0$ , $E\to\dfrac{\pi}{2}$ ( $90^{\circ}$ ); a pole of height $h$ with shadow $h\sqrt3$ gives $E=\tan^{-1}\dfrac{1}{\sqrt3}=30^{\circ}$ .
Verify the final value by taking $\tan$ (or $\sin$ / $\cos$ ) of both sides and confirming the quadrant — a proof is incomplete without checking the answer lies in the correct range.

Where the marks go

Blindly using $\tan^{-1}\dfrac{x+y}{1-xy}$ when $xy>1$ (e.g. with $2$ and $3$ ): forgetting the $+\pi$ correction gives a wrong negative angle instead of the correct positive sum.
Quoting the arccosine/arcsine sum formula without checking the $x+y\ge0$ or $x^{2}+y^{2}\le1$ condition, so the result falls outside the principal range.
Stating the answer numerically (a decimal) instead of an exact value like $\dfrac{\pi}{4}$ — board proofs require the exact form.
In word problems, mixing up which edge is $a$ (lower) and which is $b$ (upper), or putting the larger arctangent second, flipping the sign of $\theta$ .

How the board asks it

Derive / provearctangent addition formula, two terms at a time
Prove that $\tan^{-1}\dfrac{1}{2}+\tan^{-1}\dfrac{1}{5}+\tan^{-1}\dfrac{1}{8}=\dfrac{\pi}{4}$ by combining the terms two at a time and simplifying each fraction before adding the next.
Derive / prove $\pm\pi$ correction when $xy>1$
Prove that $\tan^{-1}1+\tan^{-1}2+\tan^{-1}3=\pi$ , justifying the $+\pi$ correction since $xy>1$ when adding $\tan^{-1}2$ and $\tan^{-1}3$ .
Derive / proveconverting $\sin^{-1}$ / $\cos^{-1}$ to a common function
Prove that $\sin^{-1}\dfrac{4}{5}+\sin^{-1}\dfrac{5}{13}+\sin^{-1}\dfrac{16}{65}=\dfrac{\pi}{2}$ by converting to a single inverse function before adding.
Numericalreducing an equation to a single inverse function
Solve for $x$ : $\tan^{-1}\dfrac{x-1}{x-2}+\tan^{-1}\dfrac{x+1}{x+2}=\dfrac{\pi}{4}$ .
Applicationangle written as a difference of two arctangents
A billboard whose lower and upper edges are $2\,\text{m}$ and $5\,\text{m}$ above eye level is viewed from a horizontal distance $d$ . Show that the angle $\theta$ it subtends is $\tan^{-1}\dfrac{3d}{d^{2}+10}$ , and find $\theta$ when $d=5\,\text{m}$ .
Give reasonsprincipal-value and quadrant check via $\tan$ of both sides
Explain why $\tan^{-1}2+\tan^{-1}3$ equals $\dfrac{3\pi}{4}$ and not $-\dfrac{\pi}{4}$ , justifying from $xy>1$ that the sum lies in $\left(\dfrac{\pi}{2},\pi\right)$ .

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.