Skip to content
  1. IB Question Bank
  2. Maths AA
  3. Geometry & Trigonometry
Topic 3.08 · SL and HL

Trig identities (pythagorean identity, double angle) & relationships: notes and practice questions

Summary
  • Pythagorean identity:

sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1
Derived forms:
tan⁡2θ+1=sec⁡2θ\tan^2\theta + 1 = \sec^2\theta, 1+cot⁡2θ=csc⁡2θ1 + \cot^2\theta = \csc^2\theta

  • Double angle formulas:

sin⁡(2θ)=2sin⁡θcos⁡θ\sin(2\theta) = 2\sin\theta\cos\theta
cos⁡(2θ)=cos⁡2θ−sin⁡2θ\cos(2\theta) = \cos^2\theta - \sin^2\theta or 1−2sin⁡2θ1 - 2\sin^2\theta or 2cos⁡2θ−12\cos^2\theta - 1.
tan⁡(2θ)=2tan⁡θ1−tan⁡2θ\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}

How it is examined

"Without finding θ\theta" is the giveaway phrase and it means an exact-value chain through the identities, on Paper 1. The sign ambiguity when taking a square root of the Pythagorean identity is where marks go, and the quadrant information in the question is what resolves it. 4 to 6 marks.

Given in the booklet

The Pythagorean identity and the double angle identities for sine and cosine are given, with all three forms of cos⁡2θ\cos 2\theta.

Key ideas
  • The Pythagorean identity cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1.
  • Double angle identities for sine and cosine.
  • The relationship between trigonometric ratios.
At HL

Extended at AHL 3.9 (the other two Pythagorean identities) and AHL 3.10 (compound angles, and the double angle identity for tan).

Linking questions

  • Feeds SL 3.8, where identities are the route into solving equations.

Practice questions

75 questions · 1 easy · 45 medium · 29 hard
Showing 20 of 20

Question 1

EasyPaper 1 · no calculator5 marks

Solve csc⁡2(x)−2cot⁡(x)=0\csc^{2}{(x)} - 2\cot(x) = 0 in the interval 0≤x≤2π0 \leq x \leq 2\pi

Question 2

MediumPaper 1 · no calculator5 marks
(a)

aa Find ∫cos⁡2(x)dx\int_{}^{}{\cos^{2}(x)}dx.

[3]
(b)

bb Hence, evaluate ∫0π2cos⁡2(x)dx\int_{0}^{\frac{\pi}{2}}{\cos^{2}(x)}dx.

[2]

Question 3

HardPaper 1 · no calculator7 marks

Solve the equation log2(2cos⁡(x)+1)+log2(2cos⁡(x)−1)=−1log_2(\sqrt{2} \cos(x) + 1) + log_2(\sqrt{2} \cos(x) - 1) = -1, for −π<x<π-π < x < π.

Question 4

MediumPaper 1 · no calculator5 marks
(a)

aa Find ∫cos⁡2(x)dx\int_{}^{}{\cos^{2}(x)}dx.

[3]
(b)

bb Hence, evaluate ∫0π2cos⁡2(x)dx\int_{0}^{\frac{\pi}{2}}{\cos^{2}(x)}dx.

[2]

Question 5

HardPaper 3 · calculator16 marks
(a)

In a study of wave propagation, a mathematical model uses the function g(x)=ex−e−x2g(x) = \frac{e^x - e^{-x}}{2}, where x∈Rx \in \mathbb{R}, to describe a certain physical quantity. This function is also known as the hyperbolic sine function, sinh⁡x\sinh x.

Verify that y=g(x)y = g(x) satisfies the differential equation d2ydx2=y\frac{d^2y}{dx^2} = y.

[2]
(b)

Another related function, the hyperbolic cosine, is defined as f(x)=ex+e−x2f(x) = \frac{e^x + e^{-x}}{2}, also known as cosh⁡x\cosh x. Show that (cosh⁡x)2−(sinh⁡x)2=1(\cosh x)^2 - (\sinh x)^2 = 1.

[3]
(c)(i)

The functions cosh⁡x\cosh x and sinh⁡x\sinh x can be extended to complex numbers. Using Euler's formula eiθ=cos⁡θ+isin⁡θe^{i\theta} = \cos \theta + i \sin \theta, where θ∈R\theta \in \mathbb{R}, express cosh⁡(iθ)\cosh(i\theta) in terms of cos⁡θ\cos \theta and sin⁡θ\sin \theta.

[3]
(c)(ii)

Similarly, express sinh⁡(iθ)\sinh(i\theta) in terms of cos⁡θ\cos \theta and sin⁡θ\sin \theta.

[2]
(d)

Hence, show that (cosh⁡(iθ))2+(sinh⁡(iθ))2=cos⁡(2θ)(\cosh(i\theta) )^2 + (\sinh(i\theta) )^2 = \cos(2\theta).

[2]
(e)

In a design project, a component's profile is described by a hyperbola with parametric equations x=Acosh⁡tx = A \cosh t and y=Bsinh⁡ty = B \sinh t, where A,BA, B are positive constants and t∈Rt \in \mathbb{R}.

Given that the component's profile passes through the point (6,0)(6, 0) and has asymptotes y=±43xy = \pm \frac{4}{3}x, find the values of AA and BB.

[4]

Question 6

MediumPaper 1 · no calculator7 marks
(a)

(a) Show that cos⁡(2x)−sin⁡(x)=(1−2sin⁡(x))(1+sin⁡(x))\cos(2x) - \sin(x) = (1 - 2\sin(x) )(1 + \sin(x) ).

[2]
(b)

(b) Hence, solve the equation cos⁡(2x)−sin⁡(x)=sin⁡(x)+1\cos(2x) - \sin(x) = \sin(x) + 1 for 0≤x<2π0 \le x < 2\pi.

[5]

Question 7

HardPaper 1 · no calculator19 marks
(a)

A ladder must be placed against a tall vertical building, clearing a monument that is 8 m high and stands on horizontal ground 1 m away from the building's base. The ladder touches the ground, the top corner of the monument, and the wall of the building.

A diagram showing a vertical building and the horizontal ground. A monument of height 8m stands 1m away from the base of the building. A ladder is shown leaning against the building, just touching the top of the monument. The angle the ladder makes with the ground is labelled as theta.

Let LL be the length of the ladder in metres.

Let θ\theta be the angle that the ladder makes with the ground, where 0<θ<π20 < \theta < \frac{\pi}{2}.

(a) Show that L=sec⁡θ+8csc⁡θL = \sec \theta + 8\csc \theta.

[2]
(b)(i)

(b) (i) Find dLdθ\frac{dL}{d\theta}.

[2]
(b)(ii)

(b) (ii) When dLdθ=0\frac{dL}{d\theta} = 0, show that tan⁡θ=2\tan\theta = 2.

[3]
(c)(i)

(c) (i) Find d2Ldθ2\frac{d^2L}{d\theta^2}.

[3]
(c)(ii)

(c) (ii) When tan⁡θ=2\tan\theta = 2, find the value of d2Ldθ2\frac{d^2L}{d\theta^2}.

[4]
(d)(i)

(d) (i) Hence, justify that LL is a minimum when tan⁡θ=2\tan\theta = 2.

[1]
(d)(ii)

(d) (ii) Determine this minimum value of LL.

[2]
(e)

(e) A construction company only has ladders with a maximum length of 11 m. Determine whether it is possible to position a ladder against the building over the monument, giving a reason for your answer.

[2]

Question 8

MediumPaper 1 · no calculator4 marks

It is given that sec⁡θ=−3\sec \theta = -3, where π<θ<2π\pi < \theta < 2\pi. Find the exact value of tan⁡θ\tan \theta.

Question 9

HardPaper 1 · no calculator16 marks
(a)

(a) Find the binomial expansion of (cos⁡θ+isin⁡θ)4(\cos \theta + i \sin \theta)^4. Give your answer in the form a+bia + bi where aa and bb are expressed in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta.

[4]
(b)

(b) By using De Moivre's theorem and your answer to part (a), show that cos⁡4θ=8cos⁡4θ−8cos⁡2θ+1\cos 4\theta = 8 \cos^4\theta - 8 \cos^2\theta + 1.

[5]
(c)

(c) Hence, find the four distinct roots of the equation 8x4−8x2+1=08x^4 - 8x^2 + 1 = 0, expressing them in the form cos⁡(α)\cos(\alpha) where 0<α<π0 < \alpha < \pi.

[4]
(d)

(d) By considering the roots of the equation in part (c), or otherwise, find the exact value of cos⁡(π8)cos⁡(3π8)\cos(\frac{\pi}{8})\cos(\frac{3\pi}{8}).

[3]

Question 10

MediumPaper 1 · no calculator7 marks
(a)

Consider the functions f(x)=sin⁡x−cos⁡xf(x) = \sin x - \cos x and h(x)=x+π4h(x) = x + \frac{\pi}{4} for x∈Rx \in \mathbb{R}.

(a) Find an expression for (f∘h)(x)(f \circ h)(x).

[2]
(b)

(b) Hence, solve the equation (f∘h)(x)=1(f \circ h)(x) = 1 for 0≤x≤2π0 \le x \le 2\pi.

[5]

Question 11

HardPaper 1 · no calculator8 marks
(a)

Consider the function g(x)=1−cos⁡(ax)x2g(x)=\frac{1-\cos(ax)}{x^2}, where x≠0x \neq 0 and a∈R+a \in \mathbb{R}^+.

(a) Show that gg is an even function.

[2]
(b)

(b) Given that lim⁡x→0g(x)=8\lim_{x\to0} g(x) = 8, find the value of aa.

[6]

Question 12

MediumPaper 1 · no calculator7 marks
(a)

Consider the functions f(x)=−3cos⁡x+5f(x) = -3\cos x + 5 and g(x)=−3cos⁡(x+π2)+5−kg(x) = -3\cos\left(x+\frac{\pi}{2}\right) + 5 - k, where x∈Rx \in \mathbb{R} and k>0k > 0.

The graph of gg is obtained by two transformations of the graph of ff.

Describe these two transformations.

[2]
(b)

The yy-intercept of the graph of gg is at (0,p)(0, p).

Given that the maximum value of g(x)g(x) is less than or equal to 1, find the largest possible value of pp.

[5]

Question 13

HardPaper 1 · no calculator17 marks
(a)

Find the binomial expansion of (cos⁡θ+isin⁡θ)4(\cos \theta + i \sin \theta)^4. Give your answer in the form a+bia + bi where aa and bb are expressed in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta.

[4]
(b)

By using De Moivre's theorem and your answer to part (a), show that cos⁡4θ=8cos⁡4θ−8cos⁡2θ+1\cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1.

[6]
(c)(i)

Hence, show that θ=π8\theta = \frac{\pi}{8} and θ=3π8\theta = \frac{3\pi}{8} are solutions of the equation 8cos⁡4θ−8cos⁡2θ+1=08\cos^4\theta - 8\cos^2\theta + 1 = 0.

[3]
(c)(ii)

Hence, find the exact value of cos⁡(π8)cos⁡(3π8)\cos(\frac{\pi}{8})\cos(\frac{3\pi}{8}).

[4]

Question 14

MediumPaper 1 · no calculator6 marks

Solve the equation cos⁡(2x)+3cos⁡(x)+2=0\cos(2x) + 3\cos(x) + 2 = 0 for 0≤x≤2π0 \leq x \leq 2\pi.

Question 15

HardPaper 1 · no calculator8 marks
(a)

Show that 1−cos⁡(2x)−sin⁡(2x)=2sin⁡x(sin⁡x−cos⁡x)1 - \cos(2x) - \sin(2x) = 2\sin x(\sin x - \cos x).

[2]
(b)

Hence, solve the equation 1−cos⁡(2x)−sin⁡(2x)+sin⁡x−cos⁡x=01 - \cos(2x) - \sin(2x) + \sin x - \cos x = 0 for 0<x<2π0 < x < 2\pi.

[6]

Question 16

MediumPaper 1 · no calculator6 marks

The diagram shows a parallelogram PQRS where PQ = 626\sqrt{2} cm, PS = 7 cm and cos⁡(SP^Q)=35\cos(\text{S}\hat{\text{P}}\text{Q}) = \frac{3}{5}.

Diagram of parallelogram PQRS. Side PQ is the base. Angle SPQ is indicated.

Find the exact area of the parallelogram PQRS.

Question 17

HardPaper 1 · no calculator7 marks
(a)

Consider the functions f(x)=sin⁡xf(x) = \sin x and g(x)=cos⁡2xg(x) = \cos 2x, where 0≤x≤π0 \le x \le \pi.

The graphs of ff and gg are shown in the following diagram.

Graph of sin(x) and cos(2x) intersecting, with shaded region R

The graphs intersect at points P and Q. The region enclosed by the two graphs is shaded and labelled R.

(a) Find the xx-coordinates of P and Q.

[3]
(b)

(b) Find the area of R.

[4]

Question 18

MediumPaper 1 · no calculator7 marks

Solve the equation 2sin⁡2x−cos⁡x=12\sin^2x - \cos x = 1, for 0≤x≤2π0 \le x \le 2\pi.

Question 19

HardPaper 1 · no calculator14 marks
(a)

A curve is given by the equation ey=cos⁡(x)e^y = \cos(x) for x∈(−π2,π2)x \in (-\frac{\pi}{2}, \frac{\pi}{2}).

(a) Use implicit differentiation to show that dydx=−tan⁡(x)\frac{dy}{dx} = -\tan(x).

[3]
(b)

(b) Show that d2ydx2+(dydx)2+1=0\frac{d^2y}{dx^2} + (\frac{dy}{dx})^2 + 1 = 0.

[3]
(c)

(c) Find an expression for d3ydx3\frac{d^3y}{dx^3} in terms of dydx\frac{dy}{dx} and d2ydx2\frac{d^2y}{dx^2}.

[3]
(d)

(d) Hence, find the Maclaurin series for y=ln⁡(cos⁡(x))y = \ln(\cos(x) ) up to and including the term in x4x^4.

[5]

Question 20

MediumPaper 1 · no calculator7 marks

The following diagram shows triangle LMN, with LM = 2152\sqrt{15}, MN = xx and LN = 3x3x.

Triangle LMN with sides 2*sqrt(15), x, and 3x. Angle LNM is at vertex N.

Given that cos⁡(∠LNM)=23\cos(\angle LNM) = \frac{2}{3}, find the area of the triangle.

Give your answer in the form aba\sqrt{b} where a,b∈Z+a, b \in \mathbb{Z}^+.

55 more Trig identities (pythagorean identity, double angle) & relationships questions in the app

Every answer is marked mark by mark, IB-style, and the AI tutor helps when you are stuck.

Where marks are lost

  • Using your own wrong value after failing a "show that".
  • Using an alternative method after "Hence".
Free. Every IB subject.
No card, no trial that runs out. Just a free account.
  • 50 marked answers a month
    Marked mark by mark, IB-style
  • Hints and mark schemes
    On every part of every question
  • 3,000+ questions
    All 6 subjects, SL and HL, mapped to the syllabus
  • Progress that adapts
    Your Study Profile picks what to practise next

Practise this topic as a session

Pick a difficulty and paper, and FourtyFive tracks your progress on this topic as you go.

or with email
FAQ

Questions,
answered.

Can't find what you're looking for? Email our student team.

What does Trig identities (pythagorean identity, double angle) & relationships cover in IB Maths AA?

Pythagorean identity:. sin^2θ + cos^2θ = 1. Derived forms:.

Is Trig identities (pythagorean identity, double angle) & relationships SL or HL?

Both. SL and HL students study Trig identities (pythagorean identity, double angle) & relationships, and HL goes further: Extended at AHL 3.9 (the other two Pythagorean identities) and AHL 3.10 (compound angles, and the double angle identity for tan).

How do I revise Trig identities (pythagorean identity, double angle) & relationships for IB Maths AA?

Start from the core idea: pythagorean identity:. In the exam: "Without finding θ" is the giveaway phrase and it means an exact-value chain through the identities, on Paper 1. The sign ambiguity when taking a square root of the Pythagorean identity is where marks go, and the quadrant information in the question is what resolves it. Then practise exam-style questions, easiest first, writing out every step of your working before you check it.

How does FourtyFive help me practise Trig identities (pythagorean identity, double angle) & relationships?

FourtyFive has 75 Trig identities (pythagorean identity, double angle) & relationships questions. Every answer you write is marked mark by mark, IB-style, and you see where each mark was won or lost. Every part has a hint, the AI tutor helps you through the step you are stuck on, and your Study Profile picks what to practise next.

Is FourtyFive free for Trig identities (pythagorean identity, double angle) & relationships practice?

Yes. A free account gives you 50 marked answers a month, and you do not need a card to sign up.

Can I handwrite Trig identities (pythagorean identity, double angle) & relationships answers on an iPad?

Yes. In the FourtyFive iPad app you write your working by hand with Apple Pencil, the way you would on paper, and it is marked the same way.

Start with the IB question
bank built for you.

Free to start, no card needed. Thousands of syllabus-mapped questions, AI Examiner marking, your weakest topics first.