HL · Further trig 1 · Reciprocal ratios & the new Pythagorean identities

Module 23 of 46Lesson 1 of 7 · HL · Further trig 1 · Reciprocal ratios & the new Pythagorean identities

Reciprocal & inverse trig functions, compound angle identities · Essential understanding 1

Reciprocal ratios & the new Pythagorean identities

IB AA AHL 3.9with Prof. Kitso

💻 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.

The big idea: There are three extra trig ratios, made by flipping the first three upside down:

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

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

and

\cot\theta = \dfrac{1}{\tan\theta}

. Divide

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

\cos^2\theta

or by

\sin^2\theta

and you get two more identities:

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

and

1 + \cot^2\theta = \csc^2\theta

. They turn messy trig into plain quadratics.

By the end of this lesson you can:

write down $\sec$ , $\csc$ and $\cot$ and find their exact values at the standard angles
say when each reciprocal ratio is undefined and what values it can take
use $1 + \tan^2\theta = \sec^2\theta$ and $1 + \cot^2\theta = \csc^2\theta$ to find exact values (choosing the sign from the quadrant)
prove identities that contain sec, csc and cot
solve equations that become quadratics in sec x, tan x or csc x, and check them on the TI-Nspire CX II

🧑🏾 Imagine this…

A ramp is “1 in 12” if it rises 1 m for every 12 m along the ground. Builders can also describe it the other way up: “12 across for every 1 up”. Same slope, flipped fraction. Trig does the same.

\tan\theta

is $\dfrac{\text{up}}{\text{across}}$ ; cot

\,\theta

is $\dfrac{\text{across}}{\text{up}}$ . Nothing new is being measured. We simply flip the fraction because some formulas, especially in calculus, look tidier that way.

\sec\theta

(secant)

\dfrac{1}{\cos\theta}

. Undefined when

\cos\theta = 0

, that is

\theta = \dfrac{\pi}{2} + k\pi

\csc\theta

(cosecant)

\dfrac{1}{\sin\theta}

. The IB writes it cosec. Undefined when

\sin\theta = 0

, that is

\theta = k\pi

\cot\theta

(cotangent)

\dfrac{1}{\tan\theta} = \dfrac{\cos\theta}{\sin\theta}

. Undefined when

\sin\theta = 0

, that is

\theta = k\pi

Reciprocal

the number you multiply by to get 1. The reciprocal of

\dfrac{2}{3}

\dfrac{3}{2}

. Zero has no reciprocal.

★ Reciprocal ratios and Pythagorean identities (in the formula booklet)

$\sec\theta = \dfrac{1}{\cos\theta}$ , $\csc\theta = \dfrac{1}{\sin\theta}$ , $\cot\theta = \dfrac{1}{\tan\theta} = \dfrac{\cos\theta}{\sin\theta}$ .
Where they come from. Divide $\sin^2\theta + \cos^2\theta = 1$ by $\cos^2\theta$ : $\tan^2\theta + 1 = \sec^2\theta$ . Divide it by $\sin^2\theta$ : $1 + \cot^2\theta = \csc^2\theta$ .
Rearranged: $\tan^2\theta = \sec^2\theta - 1$ , $\cot^2\theta = \csc^2\theta - 1$ , $\sec^2\theta - \tan^2\theta = 1$ .
Values: $|\sin\theta| \le 1$ , so $|\csc\theta| \ge 1$ . Also $|\sec\theta| \ge 1$ . $\sec\theta$ and $\csc\theta$ are never strictly between $-1$ and $1$ . $\cot\theta$ can be any real number.
Signs follow the original ratio: $\sec$ has the sign of $\cos$ , $\csc$ the sign of $\sin$ , $\cot$ the sign of $\tan$ (CAST still works).

⚠️ Do not mix up the “inverse” and the “reciprocal”

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

is the reciprocal. It is not

\cos^{-1}\theta

(arccos), which is a different idea (Lesson 3). The pairs are: cos with sec, sin with csc (the “co” goes with the sine!), tan with cot. A good check:

\sec 0 = 1

but

\csc 0

does not exist.

⚠️ No calculator key for sec, csc, cot

On the TI-Nspire you type them as fractions:

\sec x

is 1/cos(x),

\csc x

is 1/sin(x) and

\cot x

is 1/tan(x). Do not type cos(x)^-1 inside a larger expression without brackets, and never press the sin⁻¹ key, which is arcsin.

💡 Trading squares

When an equation has both

\tan^2 x

and

\sec x

, replace

\tan^2 x

\sec^2 x - 1

. Now everything is in sec x and you have an ordinary quadratic. Likewise swap

\cot^2 x

for

\csc^2 x - 1

when you see

\csc x

🧪 Try it yourself1 + tan²x is exactly sec²x

The blue and red curves are drawn from two different formulas,

\dfrac{1}{\cos^2} x

and

1 + \tan^2 x

, yet you see only one curve because they are the same function. Both are always 1 or more (so

|\sec x| \ge 1

), and both shoot up to infinity at

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

(about 1.57), where cos is 0.

Worked examples

Example 1.1Exact values of sec, csc and cot

Find the exact values of (a)

\sec\dfrac{\pi}{3}

(b)

\csc\dfrac{\pi}{4}

(c)

\cot\dfrac{2\pi}{3}

(d)

\sec\dfrac{7\pi}{6}

🧠 Think it through

Find cos, sin or tan first (a quadrant sketch and the 30-45-60 values), then flip.
Rationalise any denominator that has a surd: multiply top and bottom by the surd.

✍️ By hand

(a)

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

, so

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

(A1)

(b)

\sin\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}

, so

\csc\dfrac{\pi}{4} = \dfrac{2}{\sqrt{2}} = \sqrt{2}

(A1)

(c)

\tan\dfrac{2\pi}{3} = -\sqrt{3}

(Q2, reference angle

\dfrac{\pi}{3}

), so

\cot\dfrac{2\pi}{3} = -\dfrac{1}{\sqrt{3}} = -\dfrac{\sqrt{3}}{3}

(A1)

(d)

\dfrac{7\pi}{6}

is in Q3 with reference angle

\dfrac{\pi}{6}

, so

\cos\dfrac{7\pi}{6} = -\dfrac{\sqrt{3}}{2}

and

\sec\dfrac{7\pi}{6} = -\dfrac{2}{\sqrt{3}} = -\dfrac{2\sqrt{3}}{3}

(A1)

🔢 On your TI-Nspire CX II

Check in RAD mode. Type each reciprocal as a fraction. Compare with the exact answers as decimals: √2 = 1.41421, − $\dfrac{\sqrt{3}}{3}$ = −0.57735, − $\dfrac{2\sqrt{3}}{3}$ = −1.1547.
All four match the exact values ✓

✅ Answer (a)

2

(b)

\sqrt{2}

(c)

-\dfrac{\sqrt{3}}{3}

(d)

-\dfrac{2\sqrt{3}}{3}

On Paper 1 you must give the exact form. A decimal such as −1.15 loses the accuracy mark.

Example 1.2All six ratios from one

Given that

\sec\theta = 3

and

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

, find the exact values of

\sin\theta

\tan\theta

\csc\theta

and

\cot\theta

🧠 Think it through

Flip sec to get cos. Then use $\sin^2\theta = 1 - \cos^2\theta$ .
The angle is in Q4, where cos is positive but sin and tan are negative.
Flip sin and tan to get csc and cot.

✍️ By hand

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

, so

\sin^2\theta = 1 - \dfrac{1}{9} = \dfrac{8}{9}

(M1)

Q4:

\sin\theta = -\dfrac{\sqrt{8}}{3} = -\dfrac{2\sqrt{2}}{3}

(A1)

\tan\theta = \dfrac{\sin\theta}{\cos\theta} = -\dfrac{2\sqrt{2}}{3} \times 3 = -2\sqrt{2}

(A1)

\csc\theta = -\dfrac{3}{2\sqrt{2}} = -\dfrac{3\sqrt{2}}{4}

\cot\theta = -\dfrac{1}{2\sqrt{2}} = -\dfrac{\sqrt{2}}{4}

(A1)

🔢 On your TI-Nspire CX II

The calculator’s cos⁻¹( $\dfrac{1}{3}$ ) = 1.23096 is the Q1 angle. The Q4 angle is 2π − 1.23096 = 5.05223. Check the values at that angle.
csc θ = −1.06066 = − $\dfrac{3\sqrt{2}}{4}$ ✓

✅ Answer

\sin\theta = -\dfrac{2\sqrt{2}}{3}

\tan\theta = -2\sqrt{2}

\csc\theta = -\dfrac{3\sqrt{2}}{4}

\cot\theta = -\dfrac{\sqrt{2}}{4}

You could also use 1 + tan²θ = sec²θ: 1 + tan²θ = 9, so tan θ = ±2√2, and the negative root is chosen because the angle is in Q4.

Example 1.3Proving identities with sec, csc and cot

Prove that (a)

\sec\theta - \cos\theta \equiv \sin\theta\tan\theta

(b)

\tan\theta + \cot\theta \equiv \sec\theta\csc\theta

(c)

(\csc\theta - \cot\theta)(\csc\theta + \cot\theta) \equiv 1

🧠 Think it through

Change everything to sin and cos. This is almost always a safe first move.
Combine fractions over a common denominator and use $\sin^2\theta + \cos^2\theta = 1$ .
For (c), expand the brackets as a difference of squares first.

✍️ By hand
(a) LHS    (M1)
 = RHS   (AG)
(b) LHS 
   (M1)
 = RHS   (AG)
(c) LHS    (M1)
 = RHS   (AG)

🔢 On your TI-Nspire CX II

A numerical spot-check at θ = 0.8 supports (a) and (b). It never replaces the algebra.
Each pair of lines agrees ✓

✅ Answer (a), (b) and (c) are all proved.

Always start from one side only and finish on exactly what you were asked to show. In (c) it is also fine to start from the right-hand side if you prefer.

Example 1.4A quadratic in sec x (exact answers)

Solve

\tan^2 x + \sec x = 1

for

0 \le x \le 2\pi

, giving exact answers.

🧠 Think it through

There is a tan² and a sec. Replace tan²x by sec²x − 1.
Rearrange to a quadratic in sec x and factorise.
Turn each value of sec x back into cos x and solve. Use symmetry for the second angle.

✍️ By hand

(\sec^2 x - 1) + \sec x = 1

, so

\sec^2 x + \sec x - 2 = 0

(M1)

(\sec x + 2)(\sec x - 1) = 0

, so

\sec x = -2

\sec x = 1

(A1)

\sec x = 1

\cos x = 1

, so

x = 0

x = 2\pi

(A1)

\sec x = -2

\cos x = -\dfrac{1}{2}

, so

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

x = \dfrac{4\pi}{3}

(A1)(A1)

🔢 On your TI-Nspire CX II

Check one answer: type tan(2π/3)^2+1/cos(2π/3). It should equal 1. Then check x = 0.
📍 Where to find it on the TI-Nspire
⌨️ sin, cos, tan: click trig on the keypad, or type sin(
All give 1 ✓

✅ Answer

x = 0,\ \dfrac{2\pi}{3},\ \dfrac{4\pi}{3},\ 2\pi

Do not forget the end points 0 and 2π, which are included because the interval has ≤ signs.

Example 1.5An unfriendly answer: use the GDCPaper 2 style

Solve

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

for

0 \le x \le 2\pi

. Give non-exact answers to 3 significant figures.

🧠 Think it through

Same trick: replace tan²x with sec²x − 1.
Factorise. One factor gives cos x = 1 (exact), the other gives a cos x that is not a standard value.
A value of sec x with absolute value less than 1 would be impossible. Check each one.

✍️ By hand
, so    (M1)
, so  or    (A1)
:  or 
:  (valid, since )
and    (A1)(A1)

🔢 On your TI-Nspire CX II

In a Graphs page enter f1(x) = 2*tan(x)^2+1/cos(x) and f2(x) = 1. Use a window about 0 ≤ x ≤ 6.5, −2 ≤ y ≤ 6, then menu💻 click menu on keypad▸6: Analyze Graph▸4: Intersection for each crossing point.
📍 Where to find it on the TI-Nspire
⌨️ sin, cos, tan: click trig on the keypad, or type sin(
Crossings at x = 2.30 and 3.98; the end points 0 and 2π are also solutions

✅ Answer

x = 0,\ 2.30,\ 3.98,\ 2\pi

(to 3 s.f. where not exact)

The GDC graph also shows the end-point solutions only if your window starts exactly at 0. Always list the exact ones from your algebra too.

Challenge questions

⭐ Challenge 1.6A negative sinePaper 1 style

The angle

\theta

satisfies

\csc\theta = -\dfrac{5}{3}

and

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

(a) Find the exact values of $\sec\theta$ and $\cot\theta$ .
(b) Show that $\sec^2\theta\csc^2\theta \equiv \sec^2\theta + \csc^2\theta$ , and check it for the value of $\theta$ above.

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

Flip csc to get sin = − $\dfrac{3}{5}$ . Use the identity, and Q3 means both cos and sin are negative.
For (b) write the left side with sin and cos and use sin² + cos² = 1 on the top.

✍️ By hand
(a) , so , and in Q3 
,   , so 
(b) LHS 
 = RHS
Check: 
and  ✓

✅ Answer (a)

\sec\theta = -\dfrac{5}{4}

\cot\theta = \dfrac{4}{3}

. (b) proved; both sides equal

\dfrac{625}{144}

⭐ Challenge 1.7A ladder over a fencePaper 2 style

A garden fence is

1.5

m high and stands

2

m from a tall wall. A ladder rests on the ground, touches the top of the fence and leans on the wall. It makes an angle

\theta

with the ground.

(a) Show that the length of the ladder is $L = 1.5\csc\theta + 2\sec\theta$ .
(b) Find $L$ when $\theta = 50^\circ$ .
(c) Use your GDC to find the smallest possible length of the ladder, and the angle at which it happens.
(d) The derivative of $L$ is $1.5\left(-\csc\theta\cot\theta\right) + 2\sec\theta\tan\theta$ . Show that $L' = 0$ leads to $\tan^3\theta = \dfrac{3}{4}$ , and verify your angle.

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

Split the ladder into two pieces at the fence top. Each piece is the hypotenuse of a right-angled triangle.
Use RAD or DEG carefully: 50° is 0.8727 radians.
Graph L(θ) for 0 < θ < $\dfrac{\pi}{2}$ and use Analyze Graph ▸ Minimum.
For (d) write csc, cot, sec, tan as sin and cos and cross-multiply.

✍️ By hand
(a) Lower piece: height , so length .   Upper piece: horizontal distance , so length . Total    (AG)
(b) 
 m   (A1)
(d) : , so , giving    (AG)
, so  rad 

🔢 On your TI-Nspire CX II

Part (c): graph f1(x) = 1.5/sin(x)+2/cos(x) in RAD with window 0.2 ≤ x ≤ 1.4, 3 ≤ y ≤ 9, then menu💻 click menu on keypad▸6: Analyze Graph▸2: Minimum.
📍 Where to find it on the TI-Nspire
⌨️ sin, cos, tan: click trig on the keypad, or type sin(
Minimum length 4.93 m at θ = 0.738 rad

✅ Answer (b)

5.07

m (c) shortest ladder

4.93

m, at

\theta = 0.738

rad (

42.3^\circ

) (d) shown,

\tan\theta = \sqrt[3]{0.75}

Any ladder shorter than 4.93 m cannot reach the wall over the fence. The shortest one that can is the minimum of L.

Next: Lesson 2 · Graphs of sec, csc & cot

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL HL · Further trig 1 · Reciprocal ratios & the new Pythagorean identities

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

Reciprocal ratios & the new Pythagorean identities

Worked examples

Challenge questions