Module 26 of 46Lesson 1 of 10 · HL · AA 3.9 — Reciprocal trigonometric ratios and inverse trigonometric functions

Course map · Topic 3: Geometry and Trigonometry

3.9 Reciprocal trigonometric ratios and inverse trigonometric functions

HL only12 worked questions

Three new ratios — sec, cosec and cot — are just “1 over” the ratios you know, and they bring two new Pythagorean identities. The inverse functions arcsin, arccos and arctan undo sin, cos and tan, but only on a restricted domain, so their ranges matter a lot.

✏️ Start with the questions. Try each one on paper, then open the solution and compare.

Stuck? Open the notes ↓

✏️ Practice questions

Try each question on paper first. Then open the solution and compare every step.

Question 1●○○ Warm-up✍️ By hand · Paper 1 · no calculator[3 marks]

Given that

\cos\theta=\dfrac{3}{5}

and

\dfrac{3\pi}{2}\lt \theta\lt 2\pi

, find the exact values of

\sec\theta

\csc\theta

and

\cot\theta

Show the full step-by-step solution

Step 1🧠 ThinkFind sin θ first

sec, cosec and cot are all built from sin and cos. We know cos, so we need sin.

\theta

is in the 4th quadrant (between

\dfrac{3\pi}{2}

and

2\pi

), where cos is positive and sin is negative. A 3–4–5 triangle gives the size of sin.

Step 2✍️ By handWork it out

\sin^2\theta=1-\dfrac{9}{25}=\dfrac{16}{25}\Rightarrow\sin\theta=-\dfrac{4}{5}

(4th quadrant)

\sec\theta=\dfrac{1}{\cos\theta}=\dfrac{5}{3}

\csc\theta=\dfrac{1}{\sin\theta}=-\dfrac{5}{4}

\cot\theta=\dfrac{\cos\theta}{\sin\theta}=\dfrac{\frac{3}{5}}{-\frac{4}{5}}=-\dfrac{3}{4}

✅ Check it Signs: sec has the sign of cos (+), cosec and cot have the sign of sin and tan (−) in the 4th quadrant. ✔

Final answer

\sec\theta=\dfrac{5}{3},\ \csc\theta=-\dfrac{5}{4},\ \cot\theta=-\dfrac{3}{4}

Question 2●○○ Warm-up✍️ By hand · Paper 1 · no calculator[4 marks]

Find the exact value of each of the following.

(a) $\sec\dfrac{2\pi}{3}$ [1]
(b) $\csc\dfrac{7\pi}{6}$ [1]
(c) $\arccos\left(-\dfrac{1}{2}\right)$ [1]
(d) $\arctan(-1)$ [1]

Show the full step-by-step solution

Step 1🧠 ThinkTwo different jobs

In (a) and (b) we know the angle and want a ratio: find cos or sin, then flip it. In (c) and (d) we know the ratio and want the angle — and the angle must be inside the range of arccos

[0,\pi]

or arctan

\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)

Part (a)

Step 1✍️ By hand

\cos\dfrac{2\pi}{3}=-\dfrac{1}{2}

\sec\dfrac{2\pi}{3}=-2

Part (b)

Step 1✍️ By hand

\sin\dfrac{7\pi}{6}=-\dfrac{1}{2}

(3rd quadrant)

\csc\dfrac{7\pi}{6}=-2

Part (c)

Step 1✍️ By hand

Which angle between 0 and

\pi

has cosine

-\dfrac{1}{2}

? cos is negative in the 2nd quadrant:

\pi-\dfrac{\pi}{3}

\arccos\left(-\dfrac{1}{2}\right)=\dfrac{2\pi}{3}

Part (d)

Step 1✍️ By hand

Which angle between

-\dfrac{\pi}{2}

and

\dfrac{\pi}{2}

has tangent

-1

\arctan(-1)=-\dfrac{\pi}{4}

Final answer (a)

-2

(b)

-2

(c)

\dfrac{2\pi}{3}

(d)

-\dfrac{\pi}{4}

⚠️ Common mistake Answering (d) with

\dfrac{3\pi}{4}

: it has tangent −1 too, but it is not in the range of arctan.

Question 3●○○ Warm-up🖩 GDC · Paper 2[3 marks]

Use your calculator to find, correct to 3 significant figures, (a)

\arccos(-0.3)

, (b)

\cot 2.5

, (c)

\sec 1.2

. (d) Explain why

\arcsin 1.2

cannot be found.

Show the full step-by-step solution

Step 1🧠 ThinkRadians, and no sec key needed

The angles here have no degree sign, so they are in radians — check that the screen says RAD. The TI has keys for sin, cos, tan and their inverses (press trig). For cot and sec just type 1 over tan or cos.

Step 2🖩 On the GDCType the three values

trig💻 type sin( cos( tan(›cos⁻¹(-0.3)›enter💻 Enter›1/tan(2.5)›enter💻 Enter›1/cos(1.2)›enter💻 Enter

📍 Where to find it on the TI-Nspire

⌨️ sin, cos, tan: click trig on the keypad, or type sin(

arccos(−0.3), cot 2.5, sec 1.2 in RAD

Step 3✍️ By handRound and explain (d)

(a)

1.88

(b)

-1.34

(c)

2.76

(d)

\sin x

is always between

-1

and

1

, so no angle has

\sin x=1.2

: 1.2 is not in the domain of arcsin.R1

💡 Remember If you type

\sin^{-1}(1.2)

the calculator gives an error (non-real result). The error is the calculator telling you the same thing as part (d).

Final answer (a)

1.88

(b)

-1.34

(c)

2.76

(d) the domain of arcsin is

-1\le x\le 1

Question 4●●○ Standard✍️ By hand · Paper 1 · no calculator[4 marks]

Given that

\tan\theta=-2

and

\dfrac{\pi}{2}\lt \theta\lt \pi

, find the exact value of (a)

\sec\theta

; (b)

\sin\theta

Show the full step-by-step solution

Step 1🧠 ThinkUse the identity that links tan and sec

1+\tan^2\theta=\sec^2\theta

gives

\sec^2\theta

straight away. Taking the square root gives

\pm

; the 2nd quadrant (cos negative) tells us which sign.

Part (a)

Step 1✍️ By hand

\sec^2\theta=1+(-2)^2=5

\sec\theta=-\sqrt5

since

\cos\theta\lt 0

in the 2nd quadrantA1

Part (b)

Step 1✍️ By hand

\cos\theta=\dfrac{1}{\sec\theta}=-\dfrac{1}{\sqrt5}

, and

\sin\theta=\tan\theta\cos\theta

\sin\theta=(-2)\times\left(-\dfrac{1}{\sqrt5}\right)=\dfrac{2}{\sqrt5}

=\dfrac{2\sqrt5}{5}

✅ Check it

\sin^2\theta+\cos^2\theta=\dfrac{4}{5}+\dfrac{1}{5}=1

✔ and sin is positive in the 2nd quadrant ✔.

Final answer (a)

\sec\theta=-\sqrt5

(b)

\sin\theta=\dfrac{2\sqrt5}{5}

Question 5●●○ Standard✍️ By hand · Paper 1 · no calculator[5 marks]

Solve the equation

2\sec^2x-3\tan x-1=0

for

0\le x\lt \pi

Show the full step-by-step solution

Step 1🧠 ThinkMake everything tan

The equation mixes sec and tan. Replace

\sec^2x

1+\tan^2x

. Then it is a quadratic in

\tan x

— like

2t^2-3t+1=0

Step 2✍️ By handSubstitute and solve the quadratic

2(1+\tan^2x)-3\tan x-1=0

2\tan^2x-3\tan x+1=0

(2\tan x-1)(\tan x-1)=0\Rightarrow\tan x=\dfrac{1}{2}\ \text{or}\ \tan x=1

Step 3✍️ By handFind the angles in $0\le x\lt \pi$

Both values of tan are positive, so each gives one angle in the first quadrant (tan repeats every

\pi

, so there is no second angle before

\pi

\tan x=1\Rightarrow x=\dfrac{\pi}{4}

\tan x=\dfrac{1}{2}\Rightarrow x=\arctan\dfrac{1}{2}

✅ Check it At home you can check on the GDC:

\tan^{-1}(0.5)=0.463648

, and

2\sec^2(0.463648)-3\tan(0.463648)-1=0

Final answer

x=\dfrac{\pi}{4}

x=\arctan\dfrac{1}{2}\ (\approx0.464)

Question 6●●○ Standard✍️ By hand · Paper 1 · no calculator[4 marks]

Show that

\sec^2\theta+\csc^2\theta=\sec^2\theta\,\csc^2\theta

Show the full step-by-step solution

Step 1🧠 ThinkStart from one side and change it into the other

Write everything in sin and cos. Start with the left-hand side, add the two fractions, and use

\sin^2\theta+\cos^2\theta=1

Step 2✍️ By hand

\text{LHS}=\dfrac{1}{\cos^2\theta}+\dfrac{1}{\sin^2\theta}

=\dfrac{\sin^2\theta+\cos^2\theta}{\cos^2\theta\sin^2\theta}

=\dfrac{1}{\cos^2\theta\sin^2\theta}

=\dfrac{1}{\cos^2\theta}\times\dfrac{1}{\sin^2\theta}=\sec^2\theta\csc^2\theta=\text{RHS}

💡 Remember In a “show that”, never work on both sides at once and never start from the answer. One side → steps → the other side.

Final answer Shown:

\dfrac{1}{\cos^2\theta}+\dfrac{1}{\sin^2\theta}=\dfrac{1}{\sin^2\theta\cos^2\theta}=\sec^2\theta\csc^2\theta

Question 7●●○ Standard✍️ By hand · Paper 1 · no calculator[5 marks]

Find the exact value of each of the following.

(a) $\sin\left(\arccos\dfrac{1}{3}\right)$ [2]
(b) $\sec\left(\arctan\dfrac{3}{4}\right)$ [2]
(c) $\arcsin\left(\sin\dfrac{5\pi}{6}\right)$ [1]

Show the full step-by-step solution

Step 1🧠 ThinkName the angle

Call the inside angle

\alpha

. For (a),

\alpha=\arccos\dfrac{1}{3}

means

\cos\alpha=\dfrac{1}{3}

and

0\le\alpha\le\pi

. Draw a right-angled triangle or use the Pythagorean identity to find the other ratio.

Part (a)

Step 1✍️ By hand

\cos\alpha=\dfrac{1}{3}\gt 0

, so

\alpha

is in the first quadrant and

\sin\alpha\gt 0

\sin\alpha=\sqrt{1-\dfrac{1}{9}}=\sqrt{\dfrac{8}{9}}

=\dfrac{2\sqrt2}{3}

Part (b)

Step 1✍️ By hand

\beta=\arctan\dfrac{3}{4}

: a 3–4–5 triangle, first quadrant.

\sec^2\beta=1+\dfrac{9}{16}=\dfrac{25}{16}

\sec\beta=\dfrac{5}{4}

Part (c)

Step 1✍️ By hand

\sin\dfrac{5\pi}{6}=\dfrac{1}{2}

. Now find the angle in

\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]

with sine

\dfrac{1}{2}

\arcsin\dfrac{1}{2}=\dfrac{\pi}{6}

Final answer (a)

\dfrac{2\sqrt2}{3}

(b)

\dfrac{5}{4}

(c)

\dfrac{\pi}{6}

⚠️ Common mistake (c): “arcsin and sin cancel” is only true inside the range of arcsin.

\dfrac{5\pi}{6}

is outside it.

Question 8●●○ Standard🖩 GDC · Paper 2[4 marks]

Solve the equation

\csc x=3-x

for

0\lt x\lt \pi

. Give your answers to 3 significant figures.

Show the full step-by-step solution

Step 1🧠 ThinkNo algebra will do this — use a graph

The equation mixes a trig function with

x

on its own, so we cannot rearrange it. Draw

y=\csc x

and

y=3-x

and find where they cross. Remember

\csc x=\dfrac{1}{\sin x}

Step 2🖩 On the GDCGraph both sides (RAD) and use Intersection

1/sin(x)›enter💻 Enter›3-x›enter💻 Enter›menu💻 click menu on keypad›6: Analyze Graph›4: Intersection

📍 Where to find it on the TI-Nspire

⌨️ sin, cos, tan: click trig on the keypad, or type sin(

Enter f1 and f2 on a Graphs page, set the window to

0\le x\le 3.2

0\le y\le 5

. Choose Intersection and give a lower and upper bound around each crossing.

two intersections in (0, π)

Step 3✍️ By handWrite the solutions

x=0.394

x=1.93

A1A1

✅ Check it

\csc x\ge1

(0,\pi)

and its lowest point is

(\dfrac{\pi}{2},1)

. The line

y=3-x

crosses the U-shaped curve on both sides of the bottom. ✔

Final answer

x=0.394

x=1.93

Question 9●●● Exam-level✍️ + 🖩 · Paper 2[7 marks]

Let

f(x)=2\arcsin x

and

g(x)=\arccos(x^2)

(a) Write down the largest possible domain of $f$ and the range of $f$ on that domain.[2]
(b) Use your calculator to find the coordinates of the point where the graphs of $f$ and $g$ intersect.[2]
(c) By considering $\cos(2\arcsin x)$ , show that the $x$ -coordinate is exactly $\dfrac{1}{\sqrt3}$ .[3]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkDomain of arcsin, range doubled

arcsin only accepts inputs from −1 to 1, and its outputs go from

-\dfrac{\pi}{2}

\dfrac{\pi}{2}

. Multiplying by 2 doubles the outputs.

Step 2✍️ By hand

Domain

-1\le x\le1

Range

-\pi\le f(x)\le\pi

Part (b)

Step 1🖩 On the GDCGraph both and intersect (RAD)

trig💻 type sin( cos( tan(›2·sin⁻¹(x)›enter💻 Enter›trig💻 type sin( cos( tan(›cos⁻¹(x^2)›enter💻 Enter›menu💻 click menu on keypad›6: Analyze Graph›4: Intersection

one intersection

Step 2✍️ By hand

(0.577,\ 1.23)

A1A1

Part (c)

Step 1🧠 ThinkTake cos of both sides

At the intersection

2\arcsin x=\arccos(x^2)

. Taking cos of the right side just gives

x^2

. For the left side let

\alpha=\arcsin x

, so

\sin\alpha=x

, and use

\cos2\alpha=1-2\sin^2\alpha

Step 2✍️ By hand

\cos(2\arcsin x)=1-2\sin^2(\arcsin x)=1-2x^2

1-2x^2=x^2\Rightarrow x^2=\dfrac{1}{3}

From the graph

x\gt 0

, so

x=\dfrac{1}{\sqrt3}

✅ Check it

\dfrac{1}{\sqrt3}=0.577350\ldots

— the same as the calculator. ✔

Final answer (a)

-1\le x\le1

;

-\pi\le f(x)\le\pi

(b)

(0.577,\ 1.23)

Question 10●●● Exam-level✍️ By hand · Paper 1 · no calculator[8 marks]

The function

f

is defined by

f(x)=3\arccos(2x-1)

(a) Find the largest possible domain of $f$ .[2]
(b) Write down the range of $f$ .[1]
(c) Find the exact value of $f\left(\dfrac{1}{4}\right)$ .[2]
(d) Find an expression for $f^{-1}(x)$ , stating its domain.[3]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkThe inside of arccos must be between −1 and 1

arccos can only take inputs from −1 to 1. So we need

-1\le 2x-1\le1

. Solve this double inequality.

Step 2✍️ By hand

-1\le2x-1\le1\Rightarrow0\le2x\le2

0\le x\le1

Part (b)

Step 1✍️ By hand

arccos gives

0

\pi

; times 3.

0\le f(x)\le3\pi

Part (c)

Step 1✍️ By hand

f\left(\dfrac{1}{4}\right)=3\arccos\left(-\dfrac{1}{2}\right)=3\times\dfrac{2\pi}{3}

=2\pi

Part (d)

Step 1🧠 ThinkSwap x and y, then undo each step

Write

x=3\arccos(2y-1)

. Divide by 3, take cos of both sides (cos undoes arccos), then solve for

y

. The domain of

f^{-1}

is the range of

f

Step 2✍️ By hand

\dfrac{x}{3}=\arccos(2y-1)\Rightarrow\cos\dfrac{x}{3}=2y-1

f^{-1}(x)=\dfrac{1+\cos\dfrac{x}{3}}{2}

domain

0\le x\le3\pi

✅ Check it

f^{-1}(2\pi)=\dfrac{1+\cos\dfrac{2\pi}{3}}{2}=\dfrac{1-\frac12}{2}=\dfrac14

— matches part (c). ✔

Final answer (a)

0\le x\le1

(b)

0\le f(x)\le3\pi

(c)

2\pi

(d)

f^{-1}(x)=\dfrac{1+\cos(\frac{x}{3})}{2}

0\le x\le3\pi

Question 11●●● Exam-level✍️ By hand · Paper 1 · no calculator[6 marks]

Solve

\cot^2x-\csc x-1=0

for

0\le x\le2\pi

Show the full step-by-step solution

Step 1🧠 ThinkTurn cot² into cosec²

Use

\cot^2x=\csc^2x-1

. Then the equation only has

\csc x

in it — a quadratic. Once we know

\csc x

, flip it to get

\sin x

Step 2✍️ By hand

\csc^2x-1-\csc x-1=0

\csc^2x-\csc x-2=0\Rightarrow(\csc x-2)(\csc x+1)=0

\csc x=2\Rightarrow\sin x=\dfrac{1}{2}

;\quad

\csc x=-1\Rightarrow\sin x=-1

\sin x=\dfrac12

x=\dfrac{\pi}{6},\ \dfrac{5\pi}{6}

A1A1

\sin x=-1

x=\dfrac{3\pi}{2}

✅ Check it None of the answers makes

\sin x=0

, so cot and cosec exist at all of them. Always check this with reciprocal ratios —

x=0,\pi,2\pi

can never be solutions here.

Final answer

x=\dfrac{\pi}{6},\ \dfrac{5\pi}{6},\ \dfrac{3\pi}{2}

Question 12●●● Exam-level✍️ + 🖩 · Paper 3[12 marks]

Investigation. This question explores the sum

S(x)=\arctan x+\arctan\dfrac{1}{x}

x\ne0

(a) Use your calculator to find $S(1)$ , $S(2)$ , $S(5)$ and $S(0.3)$ .[2]
(b) Make a conjecture about $S(x)$ for $x\gt 0$ , giving the exact value.[1]
(c) By drawing a right-angled triangle with shorter sides $1$ and $x$ , prove your conjecture for $x\gt 0$ .[4]
(d) Find $S(-2)$ and $S(-0.5)$ . Using the fact that $\arctan$ is an odd function, state and prove the value of $S(x)$ for $x\lt 0$ .[3]
(e) Hence find the exact value of $\arctan 3+\arctan\dfrac{1}{3}-\arctan(-4)-\arctan\left(-\dfrac{1}{4}\right)$ .[2]

Show the full step-by-step solution

Part (a)

Step 1🖩 On the GDCType the sums (RAD)

trig💻 type sin( cos( tan(›tan⁻¹(2)+tan⁻¹(1/2)›enter💻 Enter

Do the same for 1, 5 and 0.3. Press ctrlenter if the screen shows an exact form.

always 1.5708

for x < 0: −1.5708

Step 2✍️ By hand

S(1)=S(2)=S(5)=S(0.3)=1.5708

(to 5 s.f.)A2

Part (b)

Step 1✍️ By hand

Conjecture:

S(x)=\dfrac{\pi}{2}

for all

x\gt 0

(since

\dfrac{\pi}{2}=1.5708

)A1

Part (c)

Step 1🧠 ThinkThe two acute angles of a right triangle

Draw a right-angled triangle with legs

x

and

1

. Call the angle opposite

x

\alpha

and the angle opposite

1

\beta

. Then

\tan\alpha=\dfrac{x}{1}

and

\tan\beta=\dfrac{1}{x}

. The angles of a triangle add to

\pi

, and one angle is

\dfrac{\pi}{2}

Step 2✍️ By hand

\tan\alpha=x\Rightarrow\alpha=\arctan x

(

\alpha

acute)M1

\tan\beta=\dfrac{1}{x}\Rightarrow\beta=\arctan\dfrac1x

(

\beta

acute)A1

\alpha+\beta+\dfrac{\pi}{2}=\pi

\arctan x+\arctan\dfrac1x=\dfrac{\pi}{2}

for

x\gt 0

Part (d)

Step 1🧠 ThinkOdd function: arctan(−u) = −arctan(u)

x\lt 0

, write

x=-u

with

u\gt 0

. Then pull the minus signs out using the odd property and use part (c) on

u

Step 2✍️ By hand

S(-2)=S(-0.5)=-1.5708

S(-u)=\arctan(-u)+\arctan\left(-\dfrac1u\right)=-\left(\arctan u+\arctan\dfrac1u\right)

=-\dfrac{\pi}{2}

, so

S(x)=-\dfrac{\pi}{2}

for

x\lt 0

Part (e)

Step 1✍️ By hand

\left(\arctan3+\arctan\dfrac13\right)-\left(\arctan(-4)+\arctan\left(-\dfrac14\right)\right)=\dfrac{\pi}{2}-\left(-\dfrac{\pi}{2}\right)

=\pi

💡 Remember This is how Paper 3 works: calculator results suggest a pattern, then you prove it by hand, then you test where it changes (here at

x\lt 0

Final answer (a) all

1.5708

(b)

S(x)=\dfrac{\pi}{2}

-1.5708

;

S(x)=-\dfrac{\pi}{2}

for

x\lt 0

(e)

\pi

🆘 Stuck on a question? Open the notes, GDC toolkit and common mistakes

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

🎯 What you must know

$\sec\theta=\dfrac{1}{\cos\theta}$ , $\csc\theta=\dfrac{1}{\sin\theta}$ , $\cot\theta=\dfrac{1}{\tan\theta}=\dfrac{\cos\theta}{\sin\theta}$ . Each one has the same sign as the ratio it comes from.
Divide $\sin^2\theta+\cos^2\theta=1$ by $\cos^2\theta$ : $\tan^2\theta+1=\sec^2\theta$ . Divide by $\sin^2\theta$ : $1+\cot^2\theta=\csc^2\theta$ .
$y=\arcsin x$ : domain $-1\le x\le1$ , range $-\dfrac{\pi}{2}\le y\le\dfrac{\pi}{2}$ .
$y=\arccos x$ : domain $-1\le x\le1$ , range $0\le y\le\pi$ .
$y=\arctan x$ : domain all real $x$ , range $-\dfrac{\pi}{2}\lt y\lt \dfrac{\pi}{2}$ (horizontal asymptotes $y=\pm\dfrac{\pi}{2}$ ).
$\sin(\arcsin x)=x$ always (for $-1\le x\le 1$ ), but $\arcsin(\sin x)=x$ only when $x$ is in the range of arcsin.
$\sec x$ and $\csc x$ are never between $-1$ and $1$ : their graphs have vertical asymptotes where $\cos x=0$ (for sec) or $\sin x=0$ (for csc and cot).

Reciprocal ratios

\sec\theta=\dfrac{1}{\cos\theta},\ \ \csc\theta=\dfrac{1}{\sin\theta},\ \ \cot\theta=\dfrac{\cos\theta}{\sin\theta}

📘 In the formula booklet

Pythagorean identities

1+\tan^2\theta=\sec^2\theta,\qquad 1+\cot^2\theta=\csc^2\theta

📘 In the formula booklet

Ranges of the inverse functions

\arcsin x\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right],\ \arccos x\in[0,\pi],\ \arctan x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)

🧠 Learn this one

🖩 Your GDC toolkit for this subtopic

Follow the steps on your own TI-Nspire CX II. The screens show what you should see.

Inverse trig (sin⁻¹, cos⁻¹, tan⁻¹) from the trig key — check RAD at the top

trig💻 type sin( cos( tan(›cos⁻¹(-0.3)›enter💻 Enter

Press trig.
Choose cos⁻¹(-0.3).
Press enter.

arccos(−0.3), cot 2.5, sec 1.2 in RAD

Worked example: Question 3 — Type the three values

sec, cosec, cot: type them as 1 over the usual ratio

1/cos(1.2)›enter💻 Enter

📍 Where to find it on the TI-Nspire

⌨️ sin, cos, tan: click trig on the keypad, or type sin(

Type 1/cos(1.2).
Press enter.

arccos(−0.3), cot 2.5, sec 1.2 in RAD

Worked example: Question 3 — Type the three values

Solve an equation with reciprocal ratios: graph both sides and intersect

menu💻 click menu on keypad›6: Analyze Graph›4: Intersection

Start on a Graphs page (ctrl doc adds one).
Press menu.
Choose 6: Analyze Graph.
Choose 4: Intersection.
Move the pointer to the LEFT of the crossing point and press click, then to the RIGHT and press click. The intersection point and its coordinates appear.

The menu path

two intersections in (0, π)

Worked example: Question 8 — Graph both sides (RAD) and use Intersection

Enter f1 and f2 on a Graphs page, set the window to

0\le x\le 3.2

0\le y\le 5

. Choose Intersection and give a lower and upper bound around each crossing.

⚠️ Where students lose marks

Thinking $\sec\theta=\cos^{-1}\theta$ . The $^{-1}$ means the inverse function (arccos), not 1 over. $\sec\theta=(\cos\theta)^{-1}$ .
Giving $\arccos\left(-\dfrac{1}{2}\right)=-\dfrac{\pi}{3}$ . arccos never gives a negative answer: it is $\dfrac{2\pi}{3}$ .
Writing $\arcsin(\sin\dfrac{5\pi}{6})=\dfrac{5\pi}{6}$ . $\dfrac{5\pi}{6}$ is outside the range of arcsin, so the answer is $\dfrac{\pi}{6}$ .
Losing the sign when using $\sec^2\theta=1+\tan^2\theta$ : the square root gives $\pm$ , and the quadrant decides.
Calculator in DEG mode when the question is in radians (or the other way round).

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL HL · AA 3.9 — Reciprocal trigonometric ratios and inverse trigonometric functions

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL

3.9 Reciprocal trigonometric ratios and inverse trigonometric functions

✏️ Practice questions

🎯 What you must know

🖩 Your GDC toolkit for this subtopic

⚠️ Where students lose marks