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Topic 3.10 · HL only

Unit circle, pythagorean identity, tan(x), ambiguous case: notes and practice questions

Summary
  • Unit Circle: radius 1, centered at (0,0)(0,0).
  • Angle Measurement: positive angles are anti-clockwise from positive x-axis.
  • Point (x,y)(x, y) on unit circle: x=cos⁡θx = \cos \theta, y=sin⁡θy = \sin \theta.
  • Tangent: gradient from origin to (x,y)(x,y) is tan⁡θ\tan \theta.
  • CAST Rule (Quadrants):
  • 1st Q (0∘−90∘0^\circ - 90^\circ): All ratios positive.
  • 2nd Q (90∘−180∘90^\circ - 180^\circ): Sine positive.
  • 3rd Q (180∘−270∘180^\circ - 270^\circ): Tangent positive.
  • 4th Q (270∘−360∘270^\circ - 360^\circ): Cosine positive.
  • Tangent Identity: tan⁡θ=sin⁡θcos⁡θ \tan \theta = \frac{\sin \theta}{\cos \theta}
  • Pythagorean Identity: sin⁡2θ+cos⁡2θ=1 \sin^2 \theta + \cos^2 \theta = 1 (useful for converting mixed sin/cos equations).
  • Sine/Cosine graphs: periodic (360∘360^\circ or 2π2\pi), range [−1,1][-1, 1].
  • cos⁡(−x)=cos⁡(x)\cos(-x) = \cos(x)
  • sin⁡(−x)=−sin⁡(x)\sin(-x) = -\sin(x)
  • Tangent graph: periodic (180∘180^\circ or π\pi), vertical asymptotes at ±90∘,±270∘\pm 90^\circ, \pm 270^\circ.
  • Ambiguous Case of Sine Rule: occurs when given two sides and a non-included angle.
  • Two possible angles: acute (calculator's answer) and obtuse.
  • Obtuse Angle: 180∘−Acute Angle 180^\circ - \text{Acute Angle} (or π−Acute Angle\pi - \text{Acute Angle}).
  • GDC Tip: Always check angle mode (degrees/radians) based on question domain.
  • GDC Tip: Solve equations by graphing y1y_1 and y2y_2 and finding intersections.
  • GDC Tip: Verify identities by graphing original and simplified expressions; they should overlap.

How it is examined

Never ask for exact values, that is explicitly not assessed. Trigonometric equations are solved graphically over a stated interval, so the question must give the interval and the answer count depends on it. The ambiguous case turns up as a two-answer sine rule problem where the student has to justify which triangle fits the context, and that justification is usually the extra mark.

Given in the booklet

The Pythagorean identity and the definition of tan⁡θ\tan\theta.

Key ideas
  • The definitions of cos⁡θ\cos\theta and sin⁡θ\sin\theta in terms of the unit circle.
  • The Pythagorean identity, cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1.
  • The definition of tan⁡θ\tan\theta as sin⁡θcos⁡θ\dfrac{\sin\theta}{\cos\theta}.
  • The extension of the sine rule to the ambiguous case.
Not assessed

**Knowledge of exact values of cos⁡θ\cos\theta, sin⁡θ\sin\theta and tan⁡θ\tan\theta will not be assessed on examinations**, but may aid understanding of trigonometric functions. This is a hard difference from AA, where exact values are examinable.

Linking questions

  • Other contexts: generation of sinusoidal voltage in electrical engineering.
  • International-mindedness: the origin of the word "sine"; trigonometry was developed by successive civilizations and cultures. How is mathematical knowledge considered from a sociocultural perspective?
  • TOK: to what extent is mathematical knowledge embedded in particular traditions or bound to particular cultures? How have events in the history of mathematics shaped its current form and methods?

Practice questions

13 questions · 9 medium · 4 hard
Showing 13 of 13

Question 1

MediumPaper 1 · calculator7 marks
(a)

A large observation wheel, 'The SkyGazer', rotates at a constant speed. The height, hh metres, of a passenger capsule above the ground can be modelled by the function h(t)=asin⁡(bt)+ch(t) = a \sin(bt) + c, where tt is the time in minutes since the capsule started its ascent from the wheel's horizontal midline. The ride starts at 10:00 AM.

At 10:00 AM, the capsule is at its midline height and rising. It reaches its maximum height of 65 m at 10:04 AM. Its lowest point is 5 m above the ground.

(a) Find the value of aa.

[1]
(b)

(b) Find the value of bb.

[2]
(c)

(c) Find the first time after the ride starts when the capsule reaches a height of 60 m. Give your answer to the nearest minute.

[4]

Question 2

HardPaper 2 · calculator14 marks
(a)

(a) A drone launches a rescue package from an initial position of (50)\begin{pmatrix} 5 \\ 0 \end{pmatrix} metres, relative to an origin on the ground. The package is launched with an initial speed of 15ms−115\text{ms}^{-1} at an angle θ\theta to the horizontal ground, where 0<θ<π20 < \theta < \frac{\pi}{2}.

The velocity components of the package, tt seconds after it is launched, are given by vx(t)=15cos⁡θv_x(t) = 15\cos\theta and vy(t)=15sin⁡θ−9.8tv_y(t) = 15\sin\theta - 9.8t.

Find an expression for xx, the horizontal displacement from the origin, in terms of θ\theta and tt.

[3]
(b)

(b) It is given that the vertical displacement of the package from the ground is y=(15sin⁡θ)t−4.9t2y = (15\sin\theta)t - 4.9t^2. When the package hits the ground, show that t=150sin⁡θ49t = \frac{150\sin\theta}{49}.

[3]
(c)

(c) Let xgx_g be the value of xx when the package hits the ground.

Find an expression for xgx_g in terms of θ\theta only.

[2]
(d)

(d) Hence, find the value of θ\theta which maximizes the value of xgx_g.

[2]
(e)(i)

(e) The model is adapted to account for a horizontal wind with speed 2ms−12\text{ms}^{-1} acting in the opposite direction to the initial horizontal motion.

(i) In this new model, the horizontal velocity component is vx(t)=15cos⁡θ−2v_x(t) = 15\cos\theta - 2. The time taken for the package to hit the ground remains t=150sin⁡θ49t = \frac{150\sin\theta}{49}.

Find an expression for xgx_g, the value of xx when the package hits the ground, in terms of θ\theta only.

[2]
(e)(ii)

(ii) Hence, find the value of θ\theta which maximizes the value of xgx_g.

[2]

Question 3

MediumPaper 1 · calculator8 marks
(a)

A drone's camera gimbal can rotate to adjust its view. The transformation matrix R(θ)R(\theta) that rotates a point (x,y)(x, y) on the camera's image plane counter-clockwise about the origin through an angle θ\theta is given by:

R(θ)=(cos⁡θ−sin⁡θsin⁡θcos⁡θ)R(\theta) = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}

The drone needs to perform a total rotation of 2θ2\theta for a panoramic shot. Write down the matrix R(2θ)R(2\theta) that represents this single combined rotation.

[2]
(b)

Alternatively, the drone's gimbal controller performs two consecutive rotations, each through an angle of θ\theta. Calculate the resulting transformation matrix when R(θ)R(\theta) is applied twice, i.e., R(θ)×R(θ)R(\theta) \times R(\theta).

[2]
(c)(i)

By comparing your results from part (a) and part (b), explain how the identity sin⁡(2θ)=2sin⁡(θ)cos⁡(θ)\sin(2\theta) = 2\sin(\theta)\cos(\theta) can be derived.

[2]
(c)(ii)

Using the same comparison as in part (c.i), and the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1, show that cos⁡(2θ)=1−2sin⁡2(θ)\cos(2\theta) = 1 - 2\sin^2(\theta).

[2]

Question 4

HardPaper 1 · calculator22 marks
(a)

A complex number is defined by z=sin⁡(2θ)−i(1+cos⁡(2θ))z = \sin(2\theta) - \text{i}(1 + \cos(2\theta)) for −π2≤θ≤π2-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}.

(a) Solve 3cos⁡(x−45∘)=sin⁡(x+45∘)3 \cos(x - 45^{\circ}) = \sin(x + 45^{\circ}) for 0∘≤x≤360∘0^{\circ} \le x \le 360^{\circ}.

[5]
(b)

(b) Show that cos⁡15∘−sin⁡15∘=12\cos 15^{\circ} - \sin 15^{\circ} = \frac{1}{\sqrt{2}}.

[3]
(c)

(c) Find the modulus and argument of zz in terms of θ\theta. Express each answer in its simplest form.

[9]
(d)

(d) Hence find the fourth roots of zz in modulus-argument form.

[5]

Question 5

MediumPaper 1 · calculator8 marks

(a) Two oscillating electrical signals, S1(t)S_1(t) and S2(t)S_2(t), are given by the functions:

S1(t)=3sin⁡(3t+π6)S_1(t) = 3 \sin \left(3t + \frac{\pi}{6}\right)

S2(t)=4sin⁡(3t+π3)S_2(t) = 4 \sin \left(3t + \frac{\pi}{3}\right)

The combined signal is given by S(t)=S1(t)+S2(t)S(t) = S_1(t) + S_2(t), which can be expressed in the form Rsin⁡(3t+α)R\sin(3t + \alpha), where R>0R > 0 and 0<α<π20 < \alpha < \frac{\pi}{2}.

Determine the values of RR and α\alpha, giving your answers to 33 significant figures.

Question 6

HardPaper 1 · calculator13 marks
(a)

A design studio is manufacturing a decorative wall sconce. The cross-sectional profile of the sconce is defined by two mathematical curves. The sconce is formed by rotating the region between these curves through π\pi radians about the xx-axis, creating a flat rear surface that mounts flush against a wall. All linear dimensions are in centimetres.

The curve of the outer profile is given by f(x)=6cos⁡(π12x)f(x) = 6 \cos\left(\frac{\pi}{12}x\right), for 0≤x≤60 \le x \le 6.

The curve of the inner profile, g(x)g(x), is formed by translating the graph of ff by 0.60.6 units to the left and 1.21.2 units down.

(a) Write down an expression for g(x)g(x).

[2]
(b)

(b) The curve g(x)g(x) intersects the xx-axis at (a,0)(a, 0), where 0≤x≤a0 \le x \le a defines the inner boundary of the sconce.

Find the value of aa. Give your answer to three significant figures.

[3]
(c)(i)

(c.i) Write down an expression for the volume of the solid formed.

[5]
(c)(ii)

(c.ii) Hence find the volume of material used in the sconce. Give your answer to three significant figures.

[3]

Question 7

MediumPaper 1 · calculator5 marks

Two sound waves, originating from different sources, combine to form a resultant wave. The displacement of the first wave at a point is given by y1(t)=5cos⁡(2t+45∘)y_1(t) = 5\cos(2t + 45^\circ) and the displacement of the second wave is y2(t)=5sin⁡(2t+135∘)y_2(t) = 5\sin(2t + 135^\circ), where tt is time in seconds and displacements are in metres.

The resultant wave's displacement is given by Y(t)=y1(t)+y2(t)Y(t) = y_1(t) + y_2(t).

Determine the amplitude of the resultant wave Y(t)Y(t).

Question 8

HardPaper 1 · calculator13 marks
(a)

A design studio is developing a wall-mounted decorative bracket. The bracket is designed to mount flush against a flat vertical wall. The cross-sectional profile of the bracket in the xyx y-plane is defined by two mathematical curves, where all linear dimensions are measured in centimetres.

The curve of the outer profile is given by f(x)=8cos⁡(π16x)f(x) = 8 \cos\left(\frac{\pi}{16}x\right), for 0≤x≤80 \le x \le 8.

The curve of the inner profile, g(x)g(x), is formed by translating the graph of ff by 0.80.8 units to the left and 1.21.2 units downwards.

(a) Write down an expression for g(x)g(x).

[2]
(b)

The inner profile curve g(x)g(x) intersects the xx-axis at the point (a,0)(a, 0). The relevant portion of the inner profile is restricted to 0≤x≤a0 \le x \le a.

(b) Find the value of aa. Give your answer to three significant figures.

[3]
(c)(i)

The bracket is modelled by the solid formed when the region between the profiles is rotated through π\pi radians about the xx-axis. This region is bounded by f(x)f(x) from x=0x = 0 to x=8x = 8, the line x=0x = 0, the xx-axis, and g(x)g(x) from x=0x = 0 to x=ax = a.

(c.i) Write down an expression for the volume of the solid formed.

[5]
(c)(ii)

(c.ii) Hence find the volume of material used in the bracket. Give your answer to three significant figures.

[3]

Question 9

MediumPaper 1 · calculator6 marks

(a) The height, HH metres, of a buoy above sea level at time tt hours is modelled by the function H(t)=1.5cos⁡(0.5t)+3H(t) = 1.5 \cos(0.5t) + 3, where the angle 0.5t0.5t is measured in radians.

Calculate the times, tt, when the buoy is at a height of 2.252.25 metres above sea level, for 0≤t≤200 \le t \le 20 hours. Give your answers to three significant figures.

Question 10

MediumPaper 1 · calculator5 marks
(a)

The height of the tide in a harbour, HH metres, tt hours after midnight, can be modelled by the function H(t)=3.5+1.8cos⁡(πt6)H(t) = 3.5 + 1.8 \cos \left( \frac{\pi t}{6} \right).

(a) Find the minimum and maximum heights of the tide.

[2]
(b)

(b) Determine the first two positive values of tt for which the tide height is 2.62.6 metres.

[3]

Question 11

MediumPaper 2 · calculator8 marks
(a)

(a) The temperature TT (in degrees Celsius) inside a specialized plant incubator varies with time tt (in minutes) according to the model T=25.0−1.5cos⁡(π30t)T = 25.0 - 1.5 \cos\left(\frac{\pi}{30}t\right).

Determine the maximum and minimum temperatures inside the incubator.

[3]
(b)

(b) Find the first time after 1010 hours at which the temperature is 24.0 ∘C24.0 \, ^{\circ}\text{C}. Give your answer to three significant figures.

[5]

Question 12

MediumPaper 1 · calculator8 marks
(a)

A team of architects is designing a new domed stadium. The cross-section of the dome's roof can be modelled by the function f(x)=0.8e−0.5x2f(x) = 0.8e^{-0.5x^2}, where −3<x<3-3 < x < 3. The xx-axis represents the ground level and the yy-axis represents the height above the ground, with units in metres.

A support beam is to be anchored to the dome's roof at a point, P, where the roof's height is 0.60.6 m above the ground. The coordinates of P are (a,0.6)(a, 0.6), where a>0a > 0.

Calculate the value of aa.

[2]
(b)

Find an expression for f′(x)f'(x).

[2]
(c)

The support beam is designed to be perpendicular to the dome's roof at point P. Find the angle, θ\theta, that this support beam makes with the horizontal ground.

[4]

Question 13

MediumPaper 1 · calculator13 marks
(a)

A company is designing a decorative stand for a new line of luxury smart speakers. The profile of the stand's base is defined by two mathematical curves. The stand itself is formed by rotating the region between these curves through π\pi radians about the xx-axis. All dimensions are in centimetres.

The curve of the outer profile is given by f(x)=5cos⁡(π10x)f(x) = 5 \cos\left(\frac{\pi}{10}x\right), for 0≤x≤50 \le x \le 5.

The curve of the inner profile, g(x)g(x), is formed by translating the graph of ff by 0.40.4 units to the left and 0.60.6 units down.

(a) Write down an expression for g(x)g(x).

[2]
(b)

The inner profile curve g(x)g(x) intersects the xx-axis at (a,0)(a, 0). The relevant portion of g(x)g(x) for the stand is restricted to 0≤x≤a0 \le x \le a.

(b) Find the value of aa. Give your answer to three significant figures.

[3]
(c)(i)

The decorative stand is modelled by the solid formed when the region RR is rotated through π\pi radians about the xx-axis. The region RR is defined by the area under f(x)f(x) from x=0x=0 to x=5x=5, excluding the area under g(x)g(x) from x=0x=0 to x=ax=a.

(c.i) Write down an expression for the volume of the solid formed.

[5]
(c)(ii)

(c.ii) Hence find the volume of material used in the stand. Give your answer to three significant figures.

[3]

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What does Unit circle, pythagorean identity, tan(x), ambiguous case cover in IB Maths AI?

Unit Circle: radius 1, centered at (0,0). Angle Measurement: positive angles are anti-clockwise from positive x-axis. Point (x, y) on unit circle: x = cos θ, y = sin θ.

Is Unit circle, pythagorean identity, tan(x), ambiguous case SL or HL?

Unit circle, pythagorean identity, tan(x), ambiguous case is HL only. SL students are not examined on it.

How do I revise Unit circle, pythagorean identity, tan(x), ambiguous case for IB Maths AI?

Start from the core idea: unit Circle: radius 1, centered at (0,0). In the exam: never ask for exact values, that is explicitly not assessed. Trigonometric equations are solved graphically over a stated interval, so the question must give the interval and the answer count depends on it. Then practise exam-style questions, easiest first, writing out every step of your working before you check it.

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