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Trigonometric Functions

Practice exam-style IB Math AA questions for Trigonometric Functions, aligned with the syllabus and grouped by topic.

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Question 1
SL • Paper 1
Easy
Non Calculator
SL • Paper 1
Easy
Non Calculator

The point PP lies on the unit circle and corresponds to the angle θ\theta, where π<θ<3π2\pi<\theta<\dfrac{3\pi}{2}. Given that tan⁡θ=3\tan\theta=\sqrt{3}:

A

the value of θ\theta.

[2]
B

the coordinates of PP.

[2]
C

A line through the origin makes an angle 2π−θ2\pi-\theta with the positive xx-axis. the gradient of the line.

[1]
Question 2
SL • Paper 1
Easy
Non Calculator
SL • Paper 1
Easy
Non Calculator

Let xx be an angle such that π2<x<π\dfrac{\pi}{2}<x<\pi and cos⁡x=−35\cos x=-\dfrac{3}{5}.

A

sin⁡x\sin x.

[2]
B

cos⁡2x\cos 2x.

[1]
C

sin⁡2x\sin 2x.

[2]
Question 3
SL • Paper 1
Easy
Non Calculator
SL • Paper 1
Easy
Non Calculator

A

2sin⁡(x−π4)=12\sin\left(x-\dfrac{\pi}{4}\right)=1 for 0≤x≤2π0\le x\le 2\pi.

[4]
Question 4
SL • Paper 2
Easy
Calculator Permitted
SL • Paper 2
Easy
Calculator Permitted

A line ll passes through the origin and makes an angle θ\theta with the positive xx-axis, where π<θ<3π2\pi<\theta<\dfrac{3\pi}{2}. A point P(−4,k)P(-4,k) lies on ll. Given that sin⁡θ=−513\sin\theta=-\dfrac{5}{13}, answer the following.

A

cos⁡θ\cos\theta.

[1]
B

the exact value of kk.

[2]
C

the exact value of cos⁡2θ\cos 2\theta.

[2]

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Question 5
SL • Paper 2
Easy
Calculator Permitted
SL • Paper 2
Easy
Calculator Permitted

Let xx be an angle such that π2<x<π\dfrac{\pi}{2}<x<\pi and cos⁡x=−0.28\cos x=-0.28.

A

sin⁡x\sin x.

[2]
B

sin⁡2x\sin 2x and cos⁡2x\cos 2x.

[3]
Question 6
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

In triangle ABCABC, A=30∘A=30^\circ, a=5a=5 and b=52b=5\sqrt{2}, where aa and bb are opposite AA and BB respectively.

A

the possible values of angle CC.

[5]
Question 7
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

A periodic quantity is modelled by f(t)=asin⁡(bt)+df(t)=a\sin(bt)+d, where b>0b>0. The maximum value of ff is 1010, the minimum value is 22, and the period is π\pi. Also, for small positive values of tt, f(t)>f(0)f(t)>f(0).

A

the amplitude and the value of dd.

[2]
B

the value of bb.

[2]
C

the values of aa, bb and dd.

[2]
Question 8
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

A

2cos⁡2x−cos⁡x−1=02\cos^2 x-\cos x-1=0 for 0≤x≤2π0\le x\le 2\pi.

[5]

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Question 9
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let π<θ<3π2\pi<\theta<\dfrac{3\pi}{2} and tan⁡θ=2\tan\theta=2.

A

sec⁡θ\sec\theta.

[3]
B

csc⁡θ\csc\theta.

[2]
Question 10
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

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

A

the domain of ff.

[2]
B

f(x)=2π3f(x)=\dfrac{2\pi}{3}.

[3]
Question 11
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let 0<θ<π40<\theta<\dfrac{\pi}{4} and tan⁡θ=13\tan\theta=\dfrac{1}{3}.

A

tan⁡2θ\tan 2\theta.

[3]
B

tan⁡(2θ+π4)\tan\left(2\theta+\dfrac{\pi}{4}\right).

[2]
Question 12
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Consider the equation sin⁡(π−x)=cos⁡(−x)\sin(\pi-x)=\cos(-x) for −π≤x≤π-\pi\le x\le \pi.

A

Use symmetry properties to rewrite the equation in the form sin⁡x=cos⁡x\sin x=\cos x.

[2]
B

the equation for −π≤x≤π-\pi\le x\le \pi.

[2]

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Question 13
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

Consider the equation

2sin⁡2x−3cos⁡x=02\sin^2x-3\cos x=0

for 0≤x≤2π0\le x\le 2\pi.

A

the equation can be written as 2cos⁡2x+3cos⁡x−2=02\cos^2x+3\cos x-2=0.

[1]
B

the equation.

[3]
Question 14
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Let θ\theta be an angle such that π2<θ<π\dfrac{\pi}{2}<\theta<\pi and tan⁡θ=−125\tan\theta=-\dfrac{12}{5}.

A

the exact values of sec⁡θ\sec\theta and csc⁡θ\csc\theta.

[3]
B

sec⁡2x−3tan⁡x=5\sec^2x-3\tan x=5 for 0≤x<2π0\le x<2\pi.

[3]
Question 15
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The function ff is defined by

f(x)=arccos⁡(2x−1)f(x)=\arccos(2x-1)
A

the domain and range of ff.

[2]
B

exactly f(x)=2arctan⁡(34)f(x)=2\arctan\left(\dfrac{3}{4}\right).

[3]
Question 16
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The graph of y=2csc⁡(x−0.4)+1y=2\csc(x-0.4)+1 is considered for 0≤x≤2π0\le x\le 2\pi.

A

the equations of the vertical asymptotes in this interval.

[2]
B

2csc⁡(x−0.4)+1=52\csc(x-0.4)+1=5 in this interval.

[2]
C

the range of y=2csc⁡(x−0.4)+1y=2\csc(x-0.4)+1.

[1]

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Question 17
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

This question uses the compound angle identities.

A

sin⁡π12=6−24\sin\dfrac{\pi}{12}=\dfrac{\sqrt{6}-\sqrt{2}}{4}.

[3]
B

sin⁡(x+π12)=6−24\sin\left(x+\dfrac{\pi}{12}\right)=\dfrac{\sqrt{6}-\sqrt{2}}{4} for 0≤x≤2π0\le x\le 2\pi.

[3]
Question 18
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A

sec⁡x−tan⁡x=13\sec x-\tan x=\dfrac{1}{3} for 0≤x<π20\le x<\dfrac{\pi}{2}.

[6]
Question 19
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

In triangle ABCABC, AB=9 cmAB=9\ \text{cm}, AC=7 cmAC=7\ \text{cm} and ∠ABC=0.620\angle ABC=0.620 radians. This information gives two possible triangles.

A

the two possible values of ∠ACB\angle ACB.

[3]
B

the two possible areas of triangle ABCABC.

[3]
Question 20
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The height hh metres of a floating marker above the sea bed is modelled by

h(t)=asin⁡(b(t−c))+dh(t)=a\sin(b(t-c))+d

where tt is the time in hours after midnight and a>0a>0. The maximum height is 5.85.8 metres at t=2t=2, the minimum height is 1.41.4 metres at t=8t=8, and the motion has period 1212 hours, with b>0b>0 and −6<c≤6-6<c\le 6.

A

the values of aa, bb and dd.

[3]
B

the value of cc.

[2]
C

the first time after midnight when the height is 4.54.5 metres.

[2]

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Question 21
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The equation

2tan⁡(3(x−1))=12\tan(3(x-1))=1

is to be solved for 0≤x≤30\le x\le 3.

A

Let u=3(x−1)u=3(x-1). the corresponding interval for uu.

[1]
B

the equation for xx.

[4]
Question 22
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Let AA and BB be acute angles such that tan⁡A=2\tan A=2 and tan⁡B=13\tan B=\dfrac{1}{3}.

A

the exact value of tan⁡(A+B)\tan(A+B).

[2]
B

A+BA+B, giving your answer to 33 significant figures.

[1]
C

tan⁡(2x)=7\tan(2x)=7 for 0≤x≤π0\le x\le \pi.

[3]
Question 23
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

For 0≤x≤2π0\le x\le 2\pi, define

f(x)=sin⁡x+sin⁡(π−x)+cos⁡(π+x)f(x)=\sin x+\sin(\pi-x)+\cos(\pi+x)
A

Use symmetry properties of trigonometric functions to simplify f(x)f(x).

[2]
B

Write f(x)f(x) in the form Rsin⁡(x−α)R\sin(x-\alpha), where R>0R>0 and 0<α<π20<\alpha<\dfrac{\pi}{2}.

[2]
C

f(x)=1f(x)=1.

[2]
Question 24
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

Let θ\theta be an angle such that π2<θ<π\dfrac{\pi}{2}<\theta<\pi and cos⁡θ=−513\cos\theta=-\dfrac{5}{13}. A point PP on the unit circle corresponds to the angle θ\theta.

A
I.

sin⁡θ\sin\theta.

[2]
II.

tan⁡θ\tan\theta.

[2]
B
I.

A line ll passes through the origin and is perpendicular to OPOP. the gradient of ll.

[2]
II.

an equation of ll.

[1]
C

The point PP is reflected in the line y=xy=x. The image is QQ. the angle α\alpha, 0≤α<2π0\leq \alpha<2\pi, corresponding to QQ on the unit circle, in terms of θ\theta. tan⁡α\tan\alpha.

[3]

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Question 25
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

In triangle ABCABC, side lengths aa, bb and cc are opposite angles AA, BB and CC respectively. It is given that A=45∘A=45^\circ, a=4a=4 and b=26b=2\sqrt{6}.

A
I.

Use the sine rule to sin⁡B=32\sin B=\dfrac{\sqrt{3}}{2}.

[2]
II.

the two possible values of BB.

[2]
B
I.

the corresponding two possible values of CC.

[2]
II.

the two possible exact values of cc.

[2]
C

the two possible exact areas of triangle ABCABC.

[3]
Question 26
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

The function ff is defined by

f(x)=3−2sin⁡(2x+π3),0≤x≤πf(x)=3-2\sin\left(2x+\frac{\pi}{3}\right),\quad 0\leq x\leq \pi
A
I.

the amplitude and period of ff.

[2]
II.

the horizontal and vertical translations from y=−2sin⁡(2x)y=-2\sin(2x) to y=f(x)y=f(x).

[2]
B
I.

the maximum and minimum values of ff.

[2]
II.

the values of xx in the given interval for which f(x)=5f(x)=5.

[2]
C

f(x)=4f(x)=4 for 0≤x≤π0\leq x\leq \pi.

[3]
Question 27
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The brightness BB of a rotating warning light, in arbitrary units, is modelled by

B(t)=asin⁡(b(t−c))+d,0≤t≤16B(t)=a\sin(b(t-c))+d,\quad 0\leq t\leq 16

where tt is the time in seconds after the light is switched on and a>0a>0. The brightness has a maximum value of 7272 at t=1.5t=1.5 and a minimum value of 1818 at t=5.5t=5.5. The motion is periodic.

A
I.

the values of aa and dd.

[2]
II.

the value of bb, taking b>0b>0.

[2]
III.

one possible value of cc for this choice of bb.

[2]
B

Using your model, all times in the interval 0≤t≤160\leq t\leq 16 when B(t)=60B(t)=60. If you did not obtain a model in part (a), use B(t)=27sin⁡(π4(t+0.5))+45B(t)=27\sin\left(\dfrac{\pi}{4}(t+0.5)\right)+45.

[3]
C

the total length of time, in seconds, during 0≤t≤160\leq t\leq 16 for which the brightness is greater than 6060.

[3]
Question 28
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

In triangle ABCABC, the side lengths opposite AA, BB and CC are aa, bb and cc respectively. It is given that A=0.650A=0.650 radians, a=10.0 cma=10.0\ \text{cm} and b=13.0 cmb=13.0\ \text{cm}.

A
I.

Use the sine rule to the two possible values of angle BB.

[4]
B
I.

For each possible triangle, the corresponding value of CC.

[2]
II.

For each possible triangle, the corresponding length of cc.

[2]
C

A design condition states that the area of triangle ABCABC must exceed 20 cm220\ \text{cm}^2. which of the possible triangles satisfies this condition.

[2]

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Question 29
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A machine component moves vertically. Its height hh cm above a fixed level is modelled by

h(x)=acos⁡(b(x−c))+dh(x)=a\cos(b(x-c))+d

where xx is measured in seconds and b>0b>0. The graph of hh has consecutive maximum points at (2,9)(2,9) and (12,9)(12,9), and a minimum point at (7,1)(7,1).

A
I.

the amplitude and the value of dd.

[2]
II.

the value of bb.

[2]
III.

a possible model for h(x)h(x).

[3]
B

A second component has height g(x)=h(x−3)+2g(x)=h(x-3)+2. the transformation from the graph of hh to the graph of gg.

[2]
C

the first time after x=0x=0 when the height of the second component is 88 cm. If you did not obtain a model in part (a), use h(x)=4cos⁡(π5(x−2))+5h(x)=4\cos\left(\dfrac{\pi}{5}(x-2)\right)+5.

[3]
Question 30
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Consider the equation

arctan⁡x+arctan⁡(2x)=π4\arctan x+\arctan(2x)=\frac{\pi}{4}
A

any solution must satisfy 2x2+3x−1=02x^2+3x-1=0.

[2]
B

the exact solution of the original equation.

[3]
Question 31
SL • Paper 1
Hard
Non Calculator
SL • Paper 1
Hard
Non Calculator

The height hh metres of a point on a rotating wheel above the ground is modelled by

h(t)=acos⁡(b(t−c))+dh(t)=a\cos(b(t-c))+d

where tt is the time in seconds, a>0a>0 and b>0b>0. The maximum height is 1414 metres at t=1t=1, and the minimum height is 22 metres at t=5t=5. The maximum at t=1t=1 and the minimum at t=5t=5 are consecutive extrema.

A
I.

aa and dd.

[2]
II.

bb.

[2]
III.

a possible value of cc.

[1]
B

a model for h(t)h(t).

[2]
C
I.

the first time after t=1t=1 when h(t)=8+32h(t)=8+3\sqrt2.

[3]
II.

the values of tt in 1≤t≤91\leq t\leq 9 for which h(t)=5h(t)=5.

[2]
Question 32
SL • Paper 1
Hard
Non Calculator
SL • Paper 1
Hard
Non Calculator

Consider the equation

2sin⁡x=cos⁡2x2\sin x=\cos 2x

for −π≤x≤π-\pi\leq x\leq \pi.

A
I.

the equation can be written as 2sin⁡2x+2sin⁡x−1=02\sin^2x+2\sin x-1=0.

[3]
II.

the possible values of sin⁡x\sin x.

[2]
B

Let α=arcsin⁡(3−12)\alpha=\arcsin\left(\dfrac{\sqrt3-1}{2}\right), where 0<α<π20<\alpha<\dfrac{\pi}{2}. the equation for −π≤x≤π-\pi\leq x\leq \pi.

[3]
C
I.

the exact value of cos⁡2x\cos 2x for each solution.

[2]
II.

whether the equation has any solution in −π≤x<0-\pi\leq x<0. Give a reason.

[2]

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Question 33
SL • Paper 1
Hard
Non Calculator
SL • Paper 1
Hard
Non Calculator

Consider the equation

2tan⁡(2x−π4)=22\tan\left(2x-\frac{\pi}{4}\right)=2

where 0≤x≤2π0\leq x\leq 2\pi.

A
I.

Let u=2x−π4u=2x-\dfrac{\pi}{4}. the corresponding interval for uu.

[2]
II.

the equation for uu in this interval.

[2]
B

the original equation for xx.

[3]
C
I.

the values of xx in 0≤x≤2π0\leq x\leq 2\pi for which tan⁡(2x−π4)\tan\left(2x-\dfrac{\pi}{4}\right) is undefined.

[3]
II.

the number of branches of the graph of y=tan⁡(2x−π4)y=\tan\left(2x-\dfrac{\pi}{4}\right) in 0≤x≤2π0\leq x\leq 2\pi.

[2]
Question 34
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let 0≤x<2π0\leq x<2\pi. Consider the equation

sec⁡x+tan⁡x=3\sec x+\tan x=3
A
AI.

(sec⁡x+tan⁡x)(sec⁡x−tan⁡x)=1(\sec x+\tan x)(\sec x-\tan x)=1.

[2]
AII.

sec⁡x−tan⁡x\sec x-\tan x.

[2]
B
BI.

sec⁡x\sec x and tan⁡x\tan x.

[2]
BII.

sin⁡x\sin x and cos⁡x\cos x.

[2]
C

sec⁡x+tan⁡x=3\sec x+\tan x=3 for 0≤x<2π0\leq x<2\pi.

[4]
Question 35
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

The function ff is defined by

f(x)=arcsin⁡(2x−1)f(x)=\arcsin(2x-1)
A
AI.

the domain of ff.

[2]
AII.

the range of ff.

[2]
B
BI.

f(14)f\left(\dfrac{1}{4}\right).

[2]
BII.

f(x)=π3f(x)=\dfrac{\pi}{3}.

[2]
C

Let g(x)=arccos⁡(2x−1)g(x)=\arccos(2x-1) on the same domain as ff. f(x)+g(x)=π2f(x)+g(x)=\dfrac{\pi}{2} for 0≤x≤10\leq x\leq1.

[4]
Question 36
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

Consider the function

f(x)=2sin⁡x−cos⁡2x,0≤x≤2πf(x)=2\sin x-\cos 2x,\quad 0\leq x\leq 2\pi
A
I.

f(x)=2sin⁡2x+2sin⁡x−1f(x)=2\sin^2x+2\sin x-1.

[2]
B
I.

f(x)=0f(x)=0 for 0≤x≤2π0\leq x\leq 2\pi.

[4]
C

Use a GDC to the maximum and minimum values of ff on 0≤x≤2π0\leq x\leq 2\pi.

[3]
D

the set of values of xx for which f(x)>1.5f(x)>1.5.

[2]

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Question 37
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

A narrow laser beam is emitted from the origin in the coordinate plane. At time tt seconds, the beam makes an angle

θ=0.25t+0.300\theta=0.25t+0.300

with the positive xx-axis. A vertical screen is placed along the line x=4x=4. For times before the first parallel position, the point where the beam meets the screen has coordinates (4,y)(4,y).

A
I.

y=4tan⁡(0.25t+0.300)y=4\tan(0.25t+0.300).

[2]
II.

the first time at which the beam is parallel to the screen. Give your answer to three significant figures.

[2]
B

the time in the interval 0≤t<π2−0.3000.250\leq t<\dfrac{\dfrac{\pi}{2}-0.300}{0.25} when y=6y=6.

[3]
C

At the time found in part (b), the distance from the origin to the point where the beam meets the screen.

[2]
D

why the model cannot give a finite value of yy at the first parallel time, t=π2−0.3000.25t=\dfrac{\dfrac{\pi}{2}-0.300}{0.25} (which is 5.085.08 s to three significant figures).

[1]
Question 38
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

The function

f(x)=atan⁡(b(x−c))+df(x)=a\tan(b(x-c))+d

has vertical asymptotes at x=1x=1 and x=5x=5. The centre of one branch is the point (3,2)(3,2), and the point (4,5)(4,5) lies on the graph. Use the convention that b>0b>0 and that cc is the xx-coordinate of the centre of the branch between the two given asymptotes.

A
I.

the values of bb, cc and dd.

[3]
II.

the value of aa.

[3]
B

f(x)=0f(x)=0 for −1≤x≤9-1\leq x\leq 9. Give your answers to 3 significant figures. If you did not obtain ff, use f(x)=3tan⁡(π4(x−3))+2f(x)=3\tan\left(\dfrac{\pi}{4}(x-3)\right)+2.

[3]
C

the equations of all vertical asymptotes of ff in the interval −1≤x≤9-1\leq x\leq 9.

[1]
Question 39
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Consider the function

f(x)=3sec⁡(2x)−2,0≤x≤2πf(x)=3\sec(2x)-2,\quad 0\leq x\leq 2\pi
A
I.

the equations of the vertical asymptotes of the graph of ff.

[2]
II.

the range of ff.

[3]
B

f(x)=4f(x)=4 for 0≤x≤2π0\leq x\leq 2\pi.

[3]
C

the area enclosed by the curve y=f(x)y=f(x), the xx-axis, the yy-axis and the line x=π6x=\dfrac{\pi}{6}.

[2]
Question 40
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Consider the function

f(x)=5sin⁡x+12cos⁡xf(x)=5\sin x+12\cos x

where 0≤x≤2π0\leq x\leq 2\pi.

A
I.

Write f(x)f(x) in the form Rsin⁡(x+α)R\sin(x+\alpha), where R>0R>0 and 0<α<π20<\alpha<\dfrac{\pi}{2}.

[4]
II.

the maximum value of f(x)f(x).

[1]
B

f(x)=6f(x)=6 for 0≤x≤2π0\leq x\leq 2\pi. If you did not obtain part (a), use f(x)=13sin⁡(x+1.18)f(x)=13\sin(x+1.18).

[3]
C

Let AA be the area between the graph of y=f(x)y=f(x) and the xx-axis over the interval from the smaller solution in part (b) to the larger solution in part (b). AA.

[4]

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Question 41
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Define

f(x)=sin⁡(π−x)+cos⁡(−x),−π≤x≤3πf(x)=\sin(\pi-x)+\cos(-x),\quad -\pi\leq x\leq 3\pi
A
I.

Use symmetry properties of trigonometric functions to simplify f(x)f(x).

[2]
II.

, for any xx such that both xx and π2−x\dfrac{\pi}{2}-x lie in the stated domain, f(π2−x)=f(x)f\left(\dfrac{\pi}{2}-x\right)=f(x). the formula has reflection symmetry about the line x=π4x=\dfrac{\pi}{4} on the portions of the graph for which reflected points are in the domain.

[2]
B

Write f(x)f(x) in the form Rsin⁡(x+α)R\sin(x+\alpha), where R>0R>0 and 0<α<π20<\alpha<\dfrac{\pi}{2}.

[2]
C

f(x)=0.500f(x)=0.500 for −π≤x≤3π-\pi\leq x\leq 3\pi.

[4]
Question 42
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

This question investigates the graph of the reciprocal trigonometric function fa,d(x)=asec⁡(x−π6)+df_{a,d}(x)=a\sec\left(x-\frac{\pi}{6}\right)+d, where a>0a>0, for 0≤x≤2π0\le x\le 2\pi.

A

Consider first the function f(x)=2sec⁡(x−π6)−1f(x)=2\sec\left(x-\frac{\pi}{6}\right)-1.

I.

the equations of the vertical asymptotes of the graph of ff in the interval 0≤x≤2π0\le x\le 2\pi.

[2]
II.

the range of ff.

[2]
B

2sec⁡(x−π6)−1=32\sec\left(x-\frac{\pi}{6}\right)-1=3 for 0≤x≤2π0\le x\le 2\pi.

[3]
C

For the general function fa,df_{a,d}, its range is y≤d−ay\le d-a or y≥d+ay\ge d+a.

[2]
D

A member of the family has range y≤−4y\le -4 or y≥2y\ge 2. aa and dd.

[2]
Question 43
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For 0<a<π20<a<\frac{\pi}{2}, define ga(x)=sin⁡(x+a)+sin⁡(x−a)g_a(x)=\sin(x+a)+\sin(x-a) for 0≤x≤2π0\le x\le 2\pi.

A

Use compound angle identities to rewrite ga(x)g_a(x).

I.

ga(x)=2cos⁡asin⁡xg_a(x)=2\cos a\sin x.

[3]
II.

the amplitude and period of gag_a.

[2]
B

For a=π6a=\frac{\pi}{6}, ga(x)=1g_a(x)=1 for 0≤x≤2π0\le x\le 2\pi.

[3]
C

the value of aa for which the equation ga(x)=1g_a(x)=1 has exactly one solution in the interval 0≤x≤2π0\le x\le 2\pi.

[3]
Question 44
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

In triangle ABCABC, side lengths aa, bb and cc are opposite angles AA, BB and CC respectively. Let A=40∘A=40^\circ and a=8 cma=8\ \text{cm}. The side bb may vary.

A

When b=10 cmb=10\ \text{cm}, the two possible values of angle BB.

[3]
B

For b=10 cmb=10\ \text{cm}, the two possible values of cc.

[3]
C

the values of bb for which two different triangles are possible.

[3]
D

what happens geometrically when b=8sin⁡40∘b=\frac{8}{\sin40^\circ}.

[2]

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Question 45
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The temperature TT degrees Celsius in a greenhouse during one day is modelled by T(t)=d+acos⁡(b(t−c))T(t)=d+a\cos(b(t-c)), where 0≤t≤240\le t\le 24 is the time in hours after midnight and a>0a>0, b>0b>0, and 0≤c<240\le c<24. The maximum temperature is 28∘C28^\circ\text{C} at t=15t=15, and the minimum temperature is 16∘C16^\circ\text{C} at t=3t=3. The model has period 2424 hours.

A

three parameters in the model.

I.

aa and dd.

[3]
II.

bb.

[2]
B

cc for this model.

[3]
C

the first time after midnight when the model predicts 25∘C25^\circ\text{C}.

[2]
D

the rate of change of temperature at the time found in part (c), and its sign.

[2]
Question 46
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

This question uses compound angle identities. Let

F(x)=3cos⁡x−sin⁡xF(x)=\sqrt3\cos x-\sin x
A
I.

F(x)=2cos⁡(x+π6)F(x)=2\cos\left(x+\dfrac{\pi}{6}\right).

[3]
II.

the maximum value of F(x)F(x).

[2]
B

F(x)=1F(x)=1 for 0≤x≤2π0\leq x\leq2\pi.

[4]
C
I.

the exact value of F(7π12)F\left(\dfrac{7\pi}{12}\right).

[2]
II.

the values of kk for which F(x)=kF(x)=k has no solution. Give your answer in interval notation.

[2]
Question 47
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let 0<x<π20<x<\dfrac{\pi}{2} and let t=tan⁡xt=\tan x.

A
I.

Use the compound angle formula for tangent to derive tan⁡2x=2t1−t2\tan2x=\dfrac{2t}{1-t^2}, where t≠1t\ne1.

[3]
II.

why t>0t>0.

[2]
B

tan⁡2x=3tan⁡x\tan2x=3\tan x for 0<x<π20<x<\dfrac{\pi}{2}, excluding values for which tan⁡2x\tan2x is undefined.

[4]
C
I.

For x=π6x=\dfrac{\pi}{6}, the exact value of tan⁡(2x+π4)\tan\left(2x+\dfrac{\pi}{4}\right).

[2]
II.

why x=π4x=\dfrac{\pi}{4} is not a solution to the equation in part (b).

[2]
Question 48
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

For 0≤x≤2π0\leq x\leq2\pi, define

f(x)=sin⁡(π+x)+sin⁡(π−x)−2cos⁡(−x)f(x)=\sin(\pi+x)+\sin(\pi-x)-2\cos(-x)
A
I.

Use symmetry properties to simplify f(x)f(x).

[3]
II.

the range of ff.

[1]
B

f(x)=1f(x)=1 for 0≤x≤2π0\leq x\leq2\pi.

[3]
C
I.

that f(2π−x)=f(x)f(2\pi-x)=f(x) for all xx in the domain.

[3]
II.

this result as a symmetry of the graph of y=f(x)y=f(x) on 0≤x≤2π0\leq x\leq2\pi.

[2]

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Question 49
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Consider the function

g(x)=2sec⁡(x−π3)−1g(x)=2\sec\left(x-\frac{\pi}{3}\right)-1

for 0≤x≤2π0\leq x\leq2\pi.

A
I.

the equations of the vertical asymptotes of the graph of y=g(x)y=g(x) in the given interval.

[3]
II.

the range of gg on its domain.

[2]
B

g(x)=3g(x)=3 for 0≤x≤2π0\leq x\leq2\pi.

[4]
C
I.

the exact value of g(π3)g\left(\dfrac{\pi}{3}\right).

[1]
II.

On the interval π3≤x<5π6\dfrac{\pi}{3}\leq x<\dfrac{5\pi}{6}, define the inverse branch g−1g^{-1}. g−1(5)g^{-1}(5).

[3]
Question 50
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Let

g(x)=arctan⁡x+arctan⁡(1−x),x∈Rg(x)=\arctan x+\arctan(1-x),\quad x\in\mathbb{R}
A
I.

tan⁡(g(x))=1x2−x+1\tan(g(x))=\frac{1}{x^2-x+1}
[3]
II.

why g(x)>0g(x)>0 for all real xx.

[1]
B

g(x)=π4g(x)=\dfrac{\pi}{4}.

[3]
C

Use a GDC to the maximum value of g(x)g(x) and the value of xx at which it occurs.

[2]
D

the two solutions of g(x)=0.800g(x)=0.800.

[2]
Question 51
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Let

F(x)=tan⁡(2x)−3tan⁡xF(x)=\tan(2x)-3\tan x

where 0≤x<2π0\leq x<2\pi and both terms are defined.

A
I.

, for u=tan⁡xu=\tan x,

F(x)=u(3u2−1)1−u2F(x)=\frac{u(3u^2-1)}{1-u^2}
[4]
B

F(x)=0F(x)=0 for 0≤x<2π0\leq x<2\pi.

[4]
C

all values of xx in 0≤x<2π0\leq x<2\pi for which F(x)=4F(x)=4.

[3]
Question 52
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

An alternating voltage is modelled by

V(t)=8sin⁡(50t)−6cos⁡(50t),0≤t≤0.200V(t)=8\sin(50t)-6\cos(50t),\quad 0\leq t\leq0.200

where tt is measured in seconds and VV is measured in volts.

A
I.

V(t)V(t) can be written in the form

V(t)=10sin⁡(50t−α)V(t)=10\sin(50t-\alpha)

where 0<α<π20<\alpha<\dfrac{\pi}{2}.

[4]
II.

the period of the voltage. Give your answer in seconds.

[1]
B

all times in 0≤t≤0.2000\leq t\leq0.200 for which V(t)=7V(t)=7. If you did not obtain the form in part (a), use V(t)=10sin⁡(50t−0.644)V(t)=10\sin(50t-0.644).

[4]
C

the total length of time during 0≤t≤0.2000\leq t\leq0.200 for which V(t)>7V(t)>7.

[3]
D

the first time after t=0t=0 at which the voltage is increasing at its greatest possible rate.

[1]

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Question 53
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For k>0k>0, define Fk(x)=arctan⁡x+arctan⁡(kx)F_k(x)=\arctan x+\arctan(kx) for x≥0x\ge 0. This question investigates equations of the form Fk(x)=αF_k(x)=\alpha.

A

Consider FkF_k for a fixed positive value of kk.

I.

the range of FkF_k.

[2]
II.

FkF_k is increasing on x≥0x\ge 0.

[2]
B

F3(x)=π3F_3(x)=\frac{\pi}{3}, giving your answer exactly.

[4]
C

that for every α\alpha such that 0<α<π0<\alpha<\pi, the equation Fk(x)=αF_k(x)=\alpha has exactly one solution for x≥0x\ge 0.

[2]
D

For 0<α<π20<\alpha<\frac{\pi}{2}, let t=tan⁡αt=\tan\alpha. the solution of Fk(x)=αF_k(x)=\alpha in terms of kk and tt.

[2]
Question 54
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A line through the origin makes an angle θ\theta with the positive xx-axis, where tan⁡θ=14\tan\theta=\frac{1}{4}. A sequence of gradients is defined by m0=tan⁡θm_0=\tan\theta and mn+1=2mn1−mn2m_{n+1}=\frac{2m_n}{1-m_n^2} when mn2≠1m_n^2\ne 1.

A

the first two gradients in the sequence.

I.

m1m_1 exactly.

[2]
II.

m2m_2 exactly.

[2]
B

by induction that mn=tan⁡(2nθ)m_n=\tan(2^n\theta) whenever all the terms involved are defined.

[3]
C

Using technology, the least positive integer nn for which mn>10m_n>10.

[2]
D

the result in part (c) geometrically.

[2]
Question 55
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For real kk, consider the equation 2sin⁡x=k+cos⁡2x2\sin x=k+\cos 2x on the interval 0≤x≤2π0\le x\le 2\pi.

A

the equation can be written as 2sin⁡2x+2sin⁡x+k−1=02\sin^2x+2\sin x+k-1=0.

[2]
B

the equation for k=0k=0.

[3]
C

the set of values of kk for which the equation has at least one solution.

[3]
D

the value of kk for which the equation has exactly three solutions in the interval 0≤x≤2π0\le x\le 2\pi.

[3]
Question 56
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For 0<r≤10<r\le 1, define fr(x)=arcsin⁡x+arccos⁡(rx)f_r(x)=\arcsin x+\arccos(rx). This question investigates how the chosen ranges of inverse trigonometric functions affect equations involving frf_r.

A

the ranges of the two inverse functions used in frf_r.

[2]
B

f1(x)=π2f_1(x)=\frac{\pi}{2} for −1≤x≤1-1\le x\le 1.

[2]
C

f12(x)=2π3f_{\frac{1}{2}}(x)=\frac{2\pi}{3} exactly.

[4]
D

the value of rr for which fr(1)=2π3f_r(1)=\frac{2\pi}{3}.

[2]

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Question 57
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
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This question investigates the equation 4cos⁡3x−3cos⁡x=124\cos^3x-3\cos x=\frac{1}{2} using compound angle identities.

A

Derive a cubic expression for cos⁡3x\cos 3x in terms of cos⁡x\cos x.

[4]
B

4cos⁡3x−3cos⁡x=124\cos^3x-3\cos x=\frac{1}{2} for 0≤x≤2π0\le x\le 2\pi.

[4]
C

the number of distinct real roots of 8t3−6t−1=08t^3-6t-1=0.

[2]
D

the three roots of 8t3−6t−1=08t^3-6t-1=0 to 33 significant figures.

[2]
Question 58
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Let f(x)=tan⁡x+tan⁡(π3−x)f(x)=\tan x+\tan\left(\frac{\pi}{3}-x\right) on the interval −π6<x<π2-\frac{\pi}{6}<x<\frac{\pi}{2}, excluding any value where the function is undefined.

A

the graph of ff is symmetric about the vertical line x=π6x=\frac{\pi}{6}.

[1]
B

Let t=tan⁡xt=\tan x. f(x)=3(1+t2)1+3tf(x)=\frac{\sqrt{3}(1+t^2)}{1+\sqrt{3}t}.

[3]
C

the value of xx in 0≤x≤π30\le x\le \frac{\pi}{3} for which f(x)f(x) is a minimum, and this minimum value.

[3]
D

why f(x)f(x) is undefined at x=−π6x=-\frac{\pi}{6}.

[2]
E

f(x)=4f(x)=4 on −π6<x<π2-\frac{\pi}{6}<x<\frac{\pi}{2}.

[3]
Question 59
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A simplified sound wave is modelled by S(t)=sin⁡(5t)+sin⁡(7t)S(t)=\sin(5t)+\sin(7t) for 0≤t≤π0\le t\le \pi. This question investigates the zeros of this type of combined wave.

A

Use compound angle identities to S(t)=2sin⁡(6t)cos⁡tS(t)=2\sin(6t)\cos t.

[3]
B

all zeros of S(t)S(t) in the interval 0≤t≤π0\le t\le \pi.

[3]
C

For an integer n≥2n\ge 2, define Sn(t)=sin⁡((n−1)t)+sin⁡((n+1)t)S_n(t)=\sin((n-1)t)+\sin((n+1)t). Sn(t)=2sin⁡(nt)cos⁡tS_n(t)=2\sin(nt)\cos t.

[3]
D

the number of distinct zeros of Sn(t)S_n(t) in 0≤t≤π0\le t\le \pi.

[3]
Question 60
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Consider f(x)=sec⁡x+csc⁡xf(x)=\sec x+\csc x on the interval 0<x<π20<x<\frac{\pi}{2}.

A

Rewrite f(x)f(x) in terms of u=sin⁡x+cos⁡xu=\sin x+\cos x.

I.

f(x)=sin⁡x+cos⁡xsin⁡xcos⁡xf(x)=\frac{\sin x+\cos x}{\sin x\cos x}.

[1]
II.

Let u=sin⁡x+cos⁡xu=\sin x+\cos x. f(x)=2uu2−1f(x)=\frac{2u}{u^2-1}.

[3]
B

the range of possible values of uu on 0<x<π20<x<\frac{\pi}{2}.

[2]
C

the minimum value of f(x)f(x) and the value of xx where it occurs.

[4]
D

sec⁡x+csc⁡x=4\sec x+\csc x=4 for 0<x<π20<x<\frac{\pi}{2}.

[3]

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Geometry & Shapes

Vectors