Trig functions 1 · The Pythagorean identity & exact values

Module 21 of 46Lesson 1 of 7 · Trig functions 1 · The Pythagorean identity & exact values

Trigonometric identities, circular functions & equations · Essential understanding 1

The Pythagorean identity & exact values

IB AA SL 3.6with 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: For every angle,

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

and

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

. These two facts let you find the other trig ratios from just one of them, exactly, with no calculator.

By the end of this lesson you can:

use $\sin^2\theta + \cos^2\theta = 1$ and $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ to find exact values
use the quadrant (the sign of sin, cos, tan) to choose the correct sign
write one ratio in terms of another or in terms of a letter such as p
simplify expressions and prove identities, using “show that” setting-out
check your exact answers on the TI-Nspire CX II

🧑🏾 Imagine this…

Draw a circle of radius 1 and put a dot on it. How far right is the dot? That is

\cos\theta

. How far up? That is

\sin\theta

. The dot, the point below it on the x-axis, and the centre make a right-angled triangle with hypotenuse 1. Pythagoras says: (across)² + (up)² = 1², so

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

. Move the dot anywhere and it is still true.

Identity

an equation that is true for every value of the variable. We write it with ≡ or just =. Example:

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

. An equation is only true for some values.

\sin^2\theta

means

(\sin\theta)^2

: find sin first, then square. It is not

\sin(\theta^2)

Quadrant

which quarter of the circle the angle ends in. Q1:

0

\dfrac{\pi}{2}

; Q2:

\dfrac{\pi}{2}

\pi

; Q3:

\pi

\dfrac{3\pi}{2}

; Q4:

\dfrac{3\pi}{2}

2\pi

Exact value

an answer left as a fraction or surd, such as

-\dfrac{\sqrt{15}}{4}

, not a rounded decimal like

-0.968

★ The two identities (both are in the formula booklet)

$\sin^2\theta + \cos^2\theta = 1$ . Rearranged: $\sin^2\theta = 1 - \cos^2\theta$ and $\cos^2\theta = 1 - \sin^2\theta$ .
$\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ (not defined when $\cos\theta = 0$ ).
Signs (CAST): Q1 all positive; Q2 only sin positive; Q3 only tan positive; Q4 only cos positive.
To find an exact value: (1) put the known value into the identity, (2) solve for the square, (3) take $\pm\sqrt{\ }$ , (4) pick the sign from the quadrant.

⚠️ Do not forget the ±

\sin^2\theta = \dfrac{9}{25}

then

\sin\theta = \pm\dfrac{3}{5}

. Only the quadrant (or the information in the question) tells you which one. Writing just the positive root loses the mark.

⚠️ Illegal cancelling

You cannot “cancel” the squares:

\sin^2\theta + \cos^2\theta

is not

\sin\theta + \cos\theta

, and

\dfrac{\sin\theta}{\cos\theta}

does not cancel to a number. Cancel only common factors of the whole top and bottom.

💡 How to answer “show that”

Start from one side only (usually the more complicated one) and work to the other side line by line. Never write the answer on both sides and work towards the middle. The last line must be exactly what you were asked to show. Marks: M1 for a correct substitution, A1 for correct simplification, AG for the given result.

🧪 Try it yourselfsin² and cos² always add up to 1

The blue curve is

\sin^2 x

, the red curve is

\cos^2 x

and the green curve is their sum. Trace along: wherever sin² is high, cos² is low, and the sum is the flat line at 1. Check x = 0.8: sin² = 0.515, cos² = 0.485.

Worked examples

Example 1.1Both ratios from one, acute angle

Given that

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

and

\theta

is acute, find the exact values of

\cos\theta

and

\tan\theta

🧠 Think it through

Use $\sin^2\theta + \cos^2\theta = 1$ and put in the value of sin.
Solve for $\cos^2\theta$ , then take the square root. Acute means Q1, so cos is positive.
Then $\tan = \dfrac{\sin}{\cos}$ .

✍️ By hand

\cos^2\theta = 1 - \sin^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}

(M1)

\cos\theta = \dfrac{4}{5}

because

\theta

is acute so

\cos\theta > 0

(A1)

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

(A1)

🔢 On your TI-Nspire CX II

Check in RAD mode: type cos(sin⁻¹(3/5)) and tan(sin⁻¹(3/5)). (Press trig💻 type sin( cos( tan( and choose sin⁻¹ from the list.)
📍 Where to find it on the TI-Nspire
⌨️ sin, cos, tan: click trig on the keypad, or type sin(
0.8 = $\dfrac{4}{5}$ and 0.75 = $\dfrac{3}{4}$ ✓

✅ Answer

\cos\theta = \dfrac{4}{5}

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

This is the 3-4-5 triangle. You can also draw the triangle with opposite 3 and hypotenuse 5, but the identity method works in every quadrant, so learn it.

Example 1.2Second quadrant: the sign matters

Given that

\sin\theta = \dfrac{5}{13}

and

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

, find the exact values of

\cos\theta

and

\tan\theta

🧠 Think it through

Same identity. This time the angle is obtuse (Q2).
In Q2 only sin is positive, so cos is negative and tan is negative.

✍️ By hand

\cos^2\theta = 1 - \dfrac{25}{169} = \dfrac{144}{169}

(M1)

\cos\theta = \pm\dfrac{12}{13}

. In Q2,

\cos\theta < 0

, so

\cos\theta = -\dfrac{12}{13}

(A1)

\tan\theta = \dfrac{\frac{5}{13}}{-\frac{12}{13}} = -\dfrac{5}{12}

(A1)

🔢 On your TI-Nspire CX II

θ is not the calculator’s own answer: sin⁻¹( $\dfrac{5}{13}$ ) gives the Q1 angle 0.395. The Q2 angle is π − 0.395 = 2.747. Type these.
cos θ = −0.923077 = − $\dfrac{12}{13}$ ✓

✅ Answer

\cos\theta = -\dfrac{12}{13}

\tan\theta = -\dfrac{5}{12}

The calculator’s inverse functions give only one angle. Always use the quadrant given in the question.

Example 1.3Third quadrant with a surd

Given that

\cos\theta = -\dfrac{1}{4}

and

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

, find the exact values of

\sin\theta

and

\tan\theta

🧠 Think it through

Use $\sin^2\theta = 1 - \cos^2\theta$ .
Q3: sin is negative and tan is positive.
The square root will not be a whole number, so leave it as a surd.

✍️ By hand

\sin^2\theta = 1 - \left(-\dfrac{1}{4}\right)^2 = 1 - \dfrac{1}{16} = \dfrac{15}{16}

(M1)

\sin\theta = -\dfrac{\sqrt{15}}{4}

since

\sin\theta < 0

in Q3 (A1)

\tan\theta = \dfrac{-\frac{\sqrt{15}}{4}}{-\frac{1}{4}} = \sqrt{15}

(A1)

🔢 On your TI-Nspire CX II

The calculator’s cos⁻¹(− $\dfrac{1}{4}$ ) = 1.823 is the Q2 angle. The Q3 angle with the same cosine is 2π − 1.823 = 4.4597 (use the exact calculator value, not the rounded one). Check:
− $\dfrac{\sqrt{15}}{4}$ = −0.968, √15 = 3.87 ✓

✅ Answer

\sin\theta = -\dfrac{\sqrt{15}}{4}

\tan\theta = \sqrt{15}

Surd answers are the “exact” answers Paper 1 wants. If you give −0.968 you lose the accuracy mark.

Example 1.4Writing in terms of p

Given that

\sin x = p

, where

\dfrac{\pi}{2} < x < \pi

, write down expressions in terms of

p

for (a)

\cos x

(b)

\tan x

(c)

\cos^2 x - \sin^2 x

🧠 Think it through

Treat p exactly like a number. Use cos² = 1 − sin² = 1 − p².
Q2 again, so cos is negative: choose the negative root.
For (c) you do not need roots at all, because cos² is already 1 − p².

✍️ By hand

(a)

\cos^2 x = 1 - p^2

, so

\cos x = -\sqrt{1 - p^2}

(negative in Q2) (M1)(A1)

(b)

\tan x = \dfrac{\sin x}{\cos x} = -\dfrac{p}{\sqrt{1 - p^2}}

(A1)

(c)

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

(A1)

🔢 On your TI-Nspire CX II

Test your answers with a value, say p = 0.6 (so x = π − sin⁻¹ 0.6 = 2.4981). The formulas give cos x = −0.8, tan x = −0.75, and 1 − 2p² = 0.28.
Matches −√(1−0.36) = −0.8 and 1 − 2(0.36) = 0.28 ✓

✅ Answer (a)

-\sqrt{1 - p^2}

(b)

-\dfrac{p}{\sqrt{1 - p^2}}

(c)

1 - 2p^2

Substituting a test number is a quick way to catch sign errors. It is not a proof, but it tells you whether to look again.

Example 1.5Show that…

(a) Show that

\tan x + \dfrac{1}{\tan x} \equiv \dfrac{1}{\sin x \cos x}

. (b) Simplify

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

. (c) Given that

3\sin\theta = 2\cos\theta

and

\theta

is acute, find the exact value of

\sin\theta

🧠 Think it through

(a) Replace tan by $\dfrac{\sin}{\cos}$ on the left, add the two fractions over a common denominator, then use the identity on the top.
(b) Use the rearranged identities on the top and the bottom.
(c) Divide both sides by cos θ (it is not zero for acute θ) to get tan θ, then use a triangle or the identity.

✍️ By hand
(a) LHS    (M1)(A1)
 = RHS   (AG)
(b) 
(c) . Then , so , 
 (positive, acute)   (A1)

🔢 On your TI-Nspire CX II

Check (a) numerically at x = 0.8 and (c) with tan⁻¹.
Both sides agree ✓

✅ Answer (a) shown (b)

\dfrac{1}{\tan^2\theta}

(c)

\sin\theta = \dfrac{2}{\sqrt{13}} = \dfrac{2\sqrt{13}}{13}

A numerical check on the GDC supports your algebra but never replaces it. In a “show that” question the algebra is what earns the marks.

Challenge questions

⭐ Challenge 1.6Tangent givenPaper 1 style

The angle

\theta

satisfies

\tan\theta = -\dfrac{5}{12}

and

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

(a) Find the exact values of $\sin\theta$ and $\cos\theta$ .
(b) Show that $1 + \tan^2\theta = \dfrac{1}{\cos^2\theta}$ , and verify this for the value of $\tan\theta$ above.

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

Show the full solution

🧠 Think it through

Write sin = tan × cos, put it in sin² + cos² = 1 and solve for cos².
For (b) divide sin² + cos² = 1 by cos² on both sides.

✍️ By hand

(a)

\sin\theta = -\dfrac{5}{12}\cos\theta

, so

\dfrac{25}{144}\cos^2\theta + \cos^2\theta = 1

\dfrac{169}{144}\cos^2\theta = 1

\cos^2\theta = \dfrac{144}{169}

\cos\theta = -\dfrac{12}{13}

(Q2),

\sin\theta = -\dfrac{5}{12}\times\left(-\dfrac{12}{13}\right) = \dfrac{5}{13}

(b)

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

, so

\tan^2\theta + 1 = \dfrac{1}{\cos^2\theta}

Check:

1 + \dfrac{25}{144} = \dfrac{169}{144}

and

\dfrac{1}{\cos^2\theta} = \dfrac{169}{144}

✓

✅ Answer (a)

\sin\theta = \dfrac{5}{13}

\cos\theta = -\dfrac{12}{13}

. (b) shown; both sides equal

\dfrac{169}{144}

⭐ Challenge 1.7A difference of fourth powersPaper 2 style

Neo claims that

\sin^4 x - \cos^4 x \equiv 2\sin^2 x - 1

(a) Prove her claim. (Hint: factorise as a difference of two squares.)
(b) Use your GDC to draw the graphs of both sides for 0 ≤ x ≤ π and describe what you see.
(c) Hence solve $\sin^4 x - \cos^4 x = \dfrac{1}{2}$ for $0 \le x \le \pi$ , giving exact answers, and confirm one of them on your GDC.

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

Show the full solution

🧠 Think it through

a² − b² = (a − b)(a + b) with a = sin²x and b = cos²x.
Plot y1 = sin(x)^4 − cos(x)^4 and y2 = 2 sin(x)^2 − 1.
Replace the left side by 2 sin²x − 1, then solve for sin x. Remember ±.

✍️ By hand

(a)

\sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x)

(M1)

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

(AG)

(c)

2\sin^2 x - 1 = \dfrac{1}{2}

, so

\sin^2 x = \dfrac{3}{4}

\sin x = \pm\dfrac{\sqrt{3}}{2}

. On

0 \le x \le \pi

sin is never negative, so

\sin x = \dfrac{\sqrt{3}}{2}

x = \dfrac{\pi}{3}

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

🔢 On your TI-Nspire CX II

Open a Graphs page ctrl💻 Ctrl doc ▾💻 File menu ▸ Add Graphs. Enter f1(x) = sin(x)^4-cos(x)^4 and f2(x) = 2*sin(x)^2-1. Set the window to 0 ≤ x ≤ π, −1.2 ≤ y ≤ 1.2. The two curves lie exactly on top of each other.
📍 Where to find it on the TI-Nspire
⌨️ sin, cos, tan: click trig on the keypad, or type sin(
One curve drawn over the other
For (c) add f3(x) = 0.5. Use menu💻 click menu on keypad▸6: Analyze Graph▸4: Intersection to find both crossing points.
1.0472 = $\dfrac{\pi}{3}$ and 2.0944 = $\dfrac{2\pi}{3}$ , matching the exact answers.

✅ Answer (a) proved (b) the graphs coincide (c)

x = \dfrac{\pi}{3}

\dfrac{2\pi}{3}

Next: Lesson 2 · Double angle identities

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL Trig functions 1 · The Pythagorean identity & exact values

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

The Pythagorean identity & exact values

Worked examples

Challenge questions