Radians, arcs and sectors, unit circle and trig identities – IB Maths exam questions with mark scheme

1CoreCalculator4 marks

The diagram shows a sector of a circle with centre $O$, radius $12$ cm and angle $1.4$ radians.

(a) Find the length of the arc. [2]

(b) Find the area of the sector. [2]

Uses $s=r\theta$: $12\times1.4$M1
$16.8$ cmA1
Uses $A=\tfrac12r^2\theta$: $\tfrac12\times12^2\times1.4$M1
$100.8$ cm$^2$ (accept $101$)A1
Answer: (a) $16.8$ cm (b) $100.8$ cm$^2$
Examiner tip: Both formulas only work in radians; check the calculator mode and never use $\tfrac12r^2\theta$ with degrees.

2HardCalculator6 marks

A circle has centre $O$ and radius $9$ cm. The chord $AB$ subtends an angle of $2.2$ radians at $O$. The shaded segment is the region between the chord $AB$ and the minor arc $AB$.

(a) Find the length of the minor arc $AB$. [2]

(b) Find the length of the chord $AB$. [2]

$s=9\times2.2$M1
$19.8$ cmA1
Chord via cosine rule or $2r\sin\frac\theta2$: $2(9)\sin1.1$M1
$AB=16.0$ cm (3 s.f.)A1
Segment = sector $-$ triangle: $\tfrac12(9)^2(2.2)-\tfrac12(9)^2\sin2.2$M1
$56.4$ cm$^2$A1
Answer: (a) $19.8$ cm (b) $16.0$ cm (c) $56.4$ cm$^2$
Examiner tip: For the segment use $\tfrac12r^2(\theta-\sin\theta)$ and evaluate $\sin\theta$ in radian mode.

3HardCalculator6 marks

A sector of a circle has radius $r$ cm and angle $\theta$ radians. The perimeter of the sector is $24$ cm and its area is $32$ cm$^2$.

(a) Show that $r^2-12r+32=0$. [3]

(b) Hence find the two possible pairs of values of $r$ and $\theta$. [3]

Perimeter: $2r+r\theta=24$A1
Area: $\tfrac12r^2\theta=32$A1
Eliminates $\theta$, e.g. $r\theta=24-2r$ so $\tfrac12r(24-2r)=32$, leading to $12r-r^2=32\Rightarrow r^2-12r+32=0$ AG (must see the substitution step)M1
Factorises/solves: $(r-4)(r-8)=0$M1
$r=4,\ \theta=4$A1
$r=8,\ \theta=1$A1
Answer: $r=4,\theta=4$ or $r=8,\theta=1$
Examiner tip: Do not forget the two straight edges ($2r$) in the perimeter of a sector.

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4StandardNon-calculator6 marks

The angle $\theta$ is acute and $\sin\theta=\dfrac35$.

(a) Find the exact value of $\cos\theta$. [2]

(b) Find the exact value of $\sin2\theta$. [2]

$\cos^2\theta=1-\left(\tfrac35\right)^2$ (or a 3-4-5 triangle)M1
$\cos\theta=\dfrac45$ (positive since acute)A1
$\sin2\theta=2\sin\theta\cos\theta=2\times\tfrac35\times\tfrac45$M1
$\dfrac{24}{25}$A1
$\cos2\theta=1-2\sin^2\theta$ (or $\cos^2\theta-\sin^2\theta$) with values substitutedM1
$\dfrac{7}{25}$A1
Answer: (a) $\tfrac45$ (b) $\tfrac{24}{25}$ (c) $\tfrac7{25}$
Examiner tip: State that $\cos\theta$ is positive because $\theta$ is acute, and use the double-angle formulae: $\sin2\theta$ is not $2\sin\theta$.

5StandardNon-calculator4 marks

Prove that $\dfrac{1-\cos2x}{\sin2x}\equiv\tan x$ for $0

Uses a double-angle identity for $\cos2x$, e.g. $\cos2x=1-2\sin^2x$M1
Numerator $1-\cos2x=2\sin^2x$A1
Denominator $\sin2x=2\sin x\cos x$A1
$\dfrac{2\sin^2x}{2\sin x\cos x}=\dfrac{\sin x}{\cos x}$ (cancelling $2\sin x\neq0$ on the interval) $=\tan x$ AGA1
Answer: Proved: LHS $=\tan x$
Examiner tip: Start from one side only and write every line of the working; the final line must be the target expression.

6CoreNon-calculator6 marks

(a) Express $135^\circ$ in radians, giving your answer in terms of $\pi$. [1]

(b) Express $\dfrac{7\pi}{12}$ radians in degrees. [1]

(d) Find the exact value of $\tan\dfrac{11\pi}{6}$. [2]

$\dfrac{3\pi}{4}$ (accept $0.75\pi$)A1
$105^\circ$ (the degree sign or the word degrees is not required)A1
Use of reference angle $\dfrac{\pi}{3}$ with the angle placed in the third quadrant, e.g. $-\cos\dfrac{\pi}{3}$ or a unit-circle sketchM1
$-\dfrac12$A1
Valid approach using exact values, e.g. $\dfrac{\sin(11\pi/6)}{\cos(11\pi/6)}=\dfrac{-1/2}{\sqrt3/2}$ or $-\tan\dfrac{\pi}{6}$M1
$-\dfrac{1}{\sqrt3}$ (accept $-\dfrac{\sqrt3}{3}$; A0 for a decimal)A1
Answer: (a) $\frac{3\pi}{4}$ (b) $105^\circ$ (c) $-\frac12$ (d) $-\frac{1}{\sqrt3}$
Examiner tip: Most lost marks are sign errors: place the angle on the unit circle first, then attach the reference-angle value with the sign for that quadrant.

7StandardNon-calculator6 marks

The diagram shows the unit circle with centre $O$. The point $P\left(-\dfrac{12}{13},\ \dfrac{5}{13}\right)$ lies on the circle, and $\theta$ is the angle, in radians, measured anticlockwise from the positive $x$-axis to $[OP]$, where $0<\theta<\pi$.

(a) Write down the value of $\cos\theta$ and the value of $\sin\theta$. [1]

(b) Find the value of $\tan\theta$. [1]

(d) Find the exact value of $\sin2\theta$. [2]

(e) Find the exact value of $\cos2\theta$. [1]

$\cos\theta=-\dfrac{12}{13}$ and $\sin\theta=\dfrac{5}{13}$ (both required)A1
$\tan\theta=-\dfrac{5}{12}$ (FT their (a) as $\sin\theta/\cos\theta$)A1
$Q\left(\dfrac{12}{13},\ -\dfrac{5}{13}\right)$ (accept $x=\frac{12}{13}, y=-\frac{5}{13}$)A1
Substitution into $\sin2\theta=2\sin\theta\cos\theta$, e.g. $2\left(\dfrac{5}{13}\right)\left(-\dfrac{12}{13}\right)$M1
$-\dfrac{120}{169}$A1
$\dfrac{119}{169}$ (from $\cos^2\theta-\sin^2\theta=\frac{144-25}{169}$ or equivalent; FT their (a))A1
Answer: (a) $\cos\theta=-\frac{12}{13}$, $\sin\theta=\frac{5}{13}$ (b) $-\frac{5}{12}$ (c) $\left(\frac{12}{13},-\frac{5}{13}\right)$ (d) $-\frac{120}{169}$ (e) $\frac{119}{169}$
Examiner tip: On the unit circle the $x$-coordinate is $\cos\theta$ and the $y$-coordinate is $\sin\theta$; students who swap them lose every follow-on mark's accuracy.

8StandardCalculator6 marks

In triangle $ABC$, $AB=11$ cm, $BC=8$ cm and $B\hat{A}C=40^\circ$. The diagram shows that two different triangles, $ABC_1$ and $ABC_2$, satisfy this information.

(a) Find the two possible values of $A\hat{C}B$. [3]

(b) Given that $A\hat{C}B$ is obtuse, find the area of triangle $ABC$. [3]

Correct substitution into the sine rule, e.g. $\dfrac{\sin C}{11}=\dfrac{\sin40^\circ}{8}$M1
$A\hat{C}B=62.1^\circ$ (accept $62.1083\ldots$; accept $1.08$ radians)A1
$A\hat{C}B=117.9^\circ$ (accept $117.892\ldots$; accept $2.06$ radians)A1
Attempt to find $A\hat{B}C=180-40-117.9\ (=22.1^\circ)$ (or equivalent valid route, e.g. finding $AC=4.68$ cm for the obtuse case)M1
Correct substitution into the area formula, e.g. $\dfrac12\times11\times8\times\sin22.1^\circ$ or $\dfrac12\times11\times4.68\times\sin40^\circ$M1
$16.6$ cm$^2$ (accept $16.5597\ldots$)A1
Answer: (a) $62.1^\circ$ or $117.9^\circ$ (b) $16.6$ cm$^2$
Examiner tip: Your calculator only gives the acute angle from $\sin^{-1}$; the second solution is $180^\circ$ minus it, and you must check it still leaves room for the third angle.

9HardCalculator7 marks

A windscreen wiper consists of an arm $[OB]$ of length $15$ cm and a blade $[BA]$ of length $30$ cm, so that $O$, $B$ and $A$ are collinear. The wiper rotates about $O$ through an angle of $110^\circ$. The blade sweeps out the shaded region shown in the diagram.

(a) Express $110^\circ$ in radians, giving your answer correct to $4$ significant figures. [1]

(b) Find the area of the shaded region. [3]

$1.920$ (accept $\dfrac{11\pi}{18}$; $1.91986\ldots$)A1
Use of sector area formula with $r=45$ or $r=15$, e.g. $\dfrac12(45)^2(1.920)$M1
Subtracting the two sectors, e.g. $\dfrac12(45)^2\theta-\dfrac12(15)^2\theta$M1
$1730$ cm$^2$ (accept $1727.88\ldots$; accept $550\pi$)A1
Use of arc length formula for either arc, e.g. $45\times1.920\ (=86.4)$ or $15\times1.920\ (=28.8)$M1
Adding both arcs and both straight edges, e.g. $86.39+28.80+30+30$M1
$175$ cm (accept $175.19\ldots$; accept $60+\dfrac{110\pi}{3}$)A1
Answer: (a) $1.920$ (b) $1730$ cm$^2$ (c) $175$ cm
Examiner tip: Arc and sector formulae need the angle in radians: substituting $110$ directly is the most common reason the whole of (b) and (c) scores zero.

10StretchNon-calculator9 marks

(a) Show that $(\sin x+\cos x)^2\equiv1+\sin2x$. [2]

It is given that $\sin x+\cos x=\dfrac13$, where $0

(b) Find the value of $\sin2x$. [2]

(d) By first finding the value of $\cos x-\sin x$, or otherwise, find the exact value of $\cos2x$. [4]

Expanding: $\sin^2x+2\sin x\cos x+\cos^2x$M1
Use of both $\sin^2x+\cos^2x=1$ and $2\sin x\cos x=\sin2x$ to reach $1+\sin2x$ (AG)A1
Substituting into (a): $\left(\dfrac13\right)^2=1+\sin2x$M1
$\sin2x=-\dfrac89$A1
Valid reasoning, e.g. $2\sin x\cos x<0$ and $\sin x>0$ for $0R1
$(\cos x-\sin x)^2=1-\sin2x=\dfrac{17}{9}$M1
$\cos x-\sin x=-\dfrac{\sqrt{17}}{3}$, negative root chosen with reason, e.g. $\cos x<0<\sin x$R1
Use of $\cos2x=(\cos x-\sin x)(\cos x+\sin x)$, e.g. $-\dfrac{\sqrt{17}}{3}\times\dfrac13$M1
$\cos2x=-\dfrac{\sqrt{17}}{9}$ (otherwise route: $\cos2x=\pm\sqrt{1-\frac{64}{81}}$ with a correct argument that $\pi<2x<\frac{3\pi}{2}$ also earns full marks; $\pm\frac{\sqrt{17}}{9}$ with no sign decision scores M1R0M1A0)A1
Answer: (a) shown (b) $-\frac89$ (c) $\cos x<0$ so $x$ obtuse (d) $-\frac{\sqrt{17}}{9}$
Examiner tip: In (d) squaring loses the sign: the R1 is for using the quadrant from (c) to choose the negative root, and a bare $\pm$ answer cannot earn the final A1.

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