Reciprocal and inverse trig functions, compound angle identities – IB Maths exam questions with mark scheme

1HardNon-calculatorHL only7 marks

The angle $A$ is obtuse with $\sin A=\dfrac5{13}$, and the angle $B$ is acute with $\cos B=\dfrac35$.

(a) Find $\cos A$ and $\sin B$. [3]

(b) Find the exact value of $\sin(A+B)$. [2]

$\cos^2A=1-\left(\tfrac5{13}\right)^2$ (or 5-12-13 triangle)M1
$\cos A=-\dfrac{12}{13}$ (negative as $A$ is obtuse)A1
$\sin B=\dfrac45$A1
$\sin(A+B)=\sin A\cos B+\cos A\sin B$ with values substitutedM1
$-\dfrac{33}{65}$A1
$\cos(A+B)=\cos A\cos B-\sin A\sin B$ with values substitutedM1
$-\dfrac{56}{65}$A1
Answer: (a) $\cos A=-\tfrac{12}{13}$, $\sin B=\tfrac45$ (b) $-\tfrac{33}{65}$ (c) $-\tfrac{56}{65}$
Examiner tip: The sign of $\cos A$ is the usual lost mark: obtuse means cosine is negative.

2HardCalculatorHL only6 marks

Solve $2\tan^2x+\sec x=1$ for $0\le x\le2\pi$. Give non-exact answers to 3 significant figures.

Uses $\tan^2x=\sec^2x-1$M1
$2\sec^2x+\sec x-3=0$A1
Factorises/solves: $(2\sec x+3)(\sec x-1)=0$M1
$\sec x=1\Rightarrow x=0,\ 2\pi$ (both needed)A1
$\cos x=-\tfrac23\Rightarrow x=2.30$A1
$x=3.98$ (from $2\pi-2.301$)A1
Answer: $x=0,\ 2.30,\ 3.98,\ 2\pi$
Examiner tip: Convert $\sec x=-\tfrac32$ to $\cos x=-\tfrac23$ before using the inverse function, and remember the end-point solutions.

3HardNon-calculatorHL only4 marks

Show that $\arctan\dfrac12+\arctan\dfrac13=\dfrac\pi4$.

Lets $\alpha=\arctan\tfrac12$, $\beta=\arctan\tfrac13$ and uses $\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$M1
$\dfrac{\frac12+\frac13}{1-\frac12\cdot\frac13}$A1
$=\dfrac{5/6}{5/6}=1$A1
Both $\alpha,\beta\in\left(0,\tfrac\pi4\right)$ (as $0<\tan<1$), so $0<\alpha+\beta<\tfrac\pi2$, hence $\alpha+\beta=\dfrac\pi4$ AGR1
Answer: Shown
Examiner tip: Evaluating $\tan(\alpha+\beta)=1$ is not enough on its own: say why the angle sum is in the range where $\tan^{-1}$ gives $\pi/4$.

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4CoreNon-calculatorHL only4 marks

The angle $\theta$ satisfies $0<\theta<\dfrac{\pi}{2}$ and $\sin\theta=p$.

Find, in terms of $p$,

(a) $\sin(\pi-\theta)$ [1]

(b) $\cos\left(\dfrac{\pi}{2}+\theta\right)$ [1]

(a) $p$A1
(b) $-p$ (accept from compound angle expansion $\cos\frac{\pi}{2}\cos\theta-\sin\frac{\pi}{2}\sin\theta$ or from the symmetry of the graphs)A1
(c) Uses $\cos\theta=\sqrt{1-p^2}$ (positive root, as $\theta$ is acute), or uses $1+\cot^2\theta=\operatorname{cosec}^2\theta=\dfrac1{p^2}$M1
(c) $\cot\theta=\dfrac{\sqrt{1-p^2}}{p}$ (accept $\sqrt{\dfrac{1-p^2}{p^2}}$; do not accept $\pm$ in the final answer)A1
Answer: (a) $p$; (b) $-p$; (c) $\dfrac{\sqrt{1-p^2}}{p}$
Examiner tip: Leaving $\pm$ in part (c) loses the final mark: $\theta$ is acute, so $\cot\theta$ is positive.

5CoreNon-calculatorHL only4 marks

The function $f$ is defined by $f(x)=\arccos(2x-1)$.

(a) Find the largest possible domain of $f$. [2]

(b) Write down the range of $f$. [1]

(a) Uses the domain of $\arccos$: $-1\le 2x-1\le 1$M1
(a) $0\le x\le 1$ (accept $[0,1]$; strict inequalities A0)A1
(b) $0\le f(x)\le\pi$ (accept $[0,\pi]$; accept $y$ in place of $f(x)$; do not accept degrees)A1
(c) $\arccos\left(-\dfrac12\right)=\dfrac{2\pi}{3}$A1
Answer: (a) $0\le x\le1$; (b) $0\le f(x)\le\pi$; (c) $\dfrac{2\pi}{3}$
Examiner tip: $\arccos$ gives answers in $[0,\pi]$, so $\arccos(-\tfrac12)$ is $\tfrac{2\pi}{3}$, not $-\tfrac{\pi}{3}$ or $\tfrac{4\pi}{3}$.

6StandardNon-calculatorHL only5 marks

The angle $\theta$ is obtuse and $\operatorname{cosec}\theta=\dfrac54$.

(a) Use an appropriate Pythagorean identity to find the value of $\cot\theta$. [2]

(b) Write down the value of $\sec\theta$. [1]

(a) Uses $1+\cot^2\theta=\operatorname{cosec}^2\theta$ to get $\cot^2\theta=\dfrac{25}{16}-1=\dfrac9{16}$M1
(a) $\cot\theta=-\dfrac34$, negative root chosen because $\theta$ is in the second quadrant (A0 for $\pm\dfrac34$ or $\dfrac34$)A1
(b) $\sec\theta=-\dfrac53$ (follow through from their $\cot\theta$ with $\sin\theta=\frac45$)A1
(c) Substitutes $\tan\theta=-\dfrac43$ into $\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}$: $\dfrac{-\frac83}{1-\frac{16}9}$ (follow through their $\tan\theta$)M1
(c) $\tan2\theta=\dfrac{24}7$A1
Answer: (a) $-\dfrac34$; (b) $-\dfrac53$; (c) $\dfrac{24}{7}$
Examiner tip: The question says obtuse: cot, sec and tan are all negative in the second quadrant, and a wrong sign in (a) carries through to (c).

7StandardNon-calculatorHL only6 marks

The diagram shows the graph of $y=\operatorname{cosec}x$ for $0

(a) Write down the coordinates of $A$ and of $B$. [2]

(b) Write down the equation of the vertical asymptote that lies between $A$ and $B$. [1]

(a) $A\left(\dfrac{\pi}{2},\,1\right)$A1
(a) $B\left(\dfrac{3\pi}{2},\,-1\right)$A1
(b) $x=\pi$ (must be an equation)A1
(c) Rearranges to $\sin^2x=\dfrac34$ (or $\operatorname{cosec}x=\pm\dfrac{2}{\sqrt3}$)M1
(c) $\sin x=\pm\dfrac{\sqrt3}{2}$, both signs considered, leading to at least two correct valuesA1
(c) $x=\dfrac{\pi}{3},\ \dfrac{2\pi}{3},\ \dfrac{4\pi}{3},\ \dfrac{5\pi}{3}$ — all four and no others in range (degrees A0)A1
Answer: (a) $A\left(\frac{\pi}{2},1\right)$, $B\left(\frac{3\pi}{2},-1\right)$; (b) $x=\pi$; (c) $x=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}$
Examiner tip: Square roots give two signs: forgetting the negative case loses the solutions in $(\pi,2\pi)$, where the graph is below the axis.

8HardNon-calculatorHL only6 marks

(a) Show that $\tan\left(x+\dfrac{\pi}{4}\right)=\dfrac{1+\tan x}{1-\tan x}$, for $\tan x\ne1$. [2]

(b) Hence solve $\tan\left(x+\dfrac{\pi}{4}\right)=6\tan x$ for $0\le x<\pi$, giving your answers in the form $\arctan k$, where $k\in\mathbb{Q}$. [4]

(a) Uses compound angle formula: $\dfrac{\tan x+\tan\frac{\pi}{4}}{1-\tan x\tan\frac{\pi}{4}}$M1
(a) Uses $\tan\dfrac{\pi}{4}=1$ to obtain $\dfrac{1+\tan x}{1-\tan x}$ (AG)A1
(b) Uses (a) and multiplies out: $1+\tan x=6\tan x(1-\tan x)$M1
(b) $6\tan^2x-5\tan x+1=0$ (or equivalent three-term quadratic)A1
(b) $\tan x=\dfrac12$ or $\tan x=\dfrac13$ (allow factorised form $(2\tan x-1)(3\tan x-1)=0$ followed by both values)A1
(b) $x=\arctan\dfrac12,\ \arctan\dfrac13$ only (no extra solutions: $\tan x>0$ gives first-quadrant angles only in $[0,\pi)$)A1
Answer: (b) $x=\arctan\dfrac12$ or $x=\arctan\dfrac13$
Examiner tip: 'Hence' means use part (a): expanding with the compound angle formula again wastes time, and decimal answers are not in the requested form.

9HardCalculatorHL only7 marks

A rectangular sign $BT$ is fixed to a vertical wall. Its lower edge $B$ is $2\text{ m}$ above, and its upper edge $T$ is $5\text{ m}$ above, the point $F$ on the wall that is level with an observer's eye $E$. The observer stands so that $EF=x$ metres, and the angle $BET$ is $\theta$, as shown in the diagram.

(a) Show that $\tan\theta=\dfrac{3x}{x^2+10}$. [3]

(b) Hence find the value of $x$ for which $\theta$ is a maximum, justifying why maximising $\tan\theta$ maximises $\theta$. [3]

(a) Writes $\theta=\angle TEF-\angle BEF$ with $\tan\angle TEF=\dfrac5x$ and $\tan\angle BEF=\dfrac2x$ (accept $\theta=\arctan\frac5x-\arctan\frac2x$)M1
(a) Uses $\tan(A-B)$ formula: $\tan\theta=\dfrac{\frac5x-\frac2x}{1+\frac{10}{x^2}}$M1
(a) Multiplies numerator and denominator by $x^2$ to obtain $\dfrac{3x}{x^2+10}$ (AG)A1
(b) $0<\theta<\dfrac{\pi}{2}$ and $\tan$ is increasing on this interval, so $\theta$ is greatest when $\tan\theta$ is greatestR1
(b) Attempts to differentiate $\dfrac{3x}{x^2+10}$ and set equal to zero, e.g. $\dfrac{3(x^2+10)-6x^2}{(x^2+10)^2}=0$ (accept GDC maximum of graph of $\tan\theta$ or $\theta$ against $x$)M1
(b) $x=\sqrt{10}\ (=3.16)$A1
(c) $\theta=0.443$ (rad) (accept $\arctan\frac{3\sqrt{10}}{20}$; $25.4^\circ$ A0)A1
Answer: (b) $x=\sqrt{10}\approx3.16$; (c) $\theta\approx0.443$ rad
Examiner tip: In (b) the reason that $\tan$ is increasing for acute angles earns its own mark; leaving it out costs the R1 even with the right $x$.

10StretchNon-calculatorHL only7 marks

Let $\alpha=\arctan\dfrac15$ and $\beta=\arctan\dfrac1{239}$.

(a) Show that $\tan2\alpha=\dfrac5{12}$. [2]

(b) Find the exact value of $\tan4\alpha$. [2]

(a) Uses $\tan2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha}=\dfrac{\frac25}{1-\frac1{25}}$M1
(a) $=\dfrac{2}{5}\times\dfrac{25}{24}=\dfrac5{12}$ (AG) — intermediate step must be seenA1
(b) Applies double angle formula again: $\tan4\alpha=\dfrac{2\cdot\frac5{12}}{1-\frac{25}{144}}$M1
(b) $\tan4\alpha=\dfrac{120}{119}$A1
(c) Uses $\tan(4\alpha-\beta)=\dfrac{\frac{120}{119}-\frac1{239}}{1+\frac{120}{119}\cdot\frac1{239}}$M1
(c) Simplifies to $\dfrac{120\times239-119}{119\times239+120}=\dfrac{28561}{28561}=1$A1
(c) Justifies the angle: $\tan\alpha=\frac15<\tan\frac{\pi}{8}=\sqrt2-1$, so $0<4\alpha<\frac{\pi}{2}$, and $0<\beta<\frac{\pi}{4}$, so $-\frac{\pi}{4}<4\alpha-\beta<\frac{\pi}{2}$, hence $4\alpha-\beta=\dfrac{\pi}{4}$ (AG). Accept any valid argument that $4\alpha-\beta$ lies in an interval containing only one solution of $\tan\phi=1$R1
Answer: (a) shown; (b) $\dfrac{120}{119}$; (c) shown
Examiner tip: Getting $\tan(4\alpha-\beta)=1$ is not enough: you must show the angle lies in an interval where $\frac{\pi}{4}$ is the only possibility.

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