Mathematics Advanced • Year 11 • Module 2 • Lesson 7
Pythagorean Identities
Build fluency in stating the three Pythagorean identities, deriving the second and third by division, and recovering missing trig values via quadrant analysis.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete the three Pythagorean identities:
________ + ________ = 1
1 + ________ = ________
________ + cot²θ = ________
Q1.2 Which identity do you divide by $\cos^2 \theta$ to obtain the tangent-secant identity?
Q1.3 Write the rearrangement of $\sin^2 \theta + \cos^2 \theta = 1$ that isolates $\sin^2 \theta$.
sin²θ = ____________________
2. Worked example, finding tanθ when secθ is known
Follow each line. The reasoning column tells you why each algebraic move is legal.
Problem. If $\sec \theta = \frac{13}{5}$ and $\theta$ is acute, find $\tan \theta$.
Step 1, Choose the identity that contains both sec and tan.
1 + tan²θ = sec²θ
Reason: Identity 2 directly relates the two functions in question.
Step 2, Substitute the known value.
1 + tan²θ = (13/5)² = 169/25
Reason: square the fraction (top and bottom).
Step 3, Isolate tan²θ with a common denominator.
tan²θ = 169/25 − 25/25 = 144/25
Reason: write 1 as 25/25 to subtract.
Step 4, Take the square root with the correct sign.
tanθ = ±12/5. Acute θ ⇒ tanθ > 0, so tanθ = +12/5.
Reason: ASTC, in Quadrant I, tangent is positive.
Conclusion. $\tan \theta = \mathbf{\frac{12}{5}}$.
3. Faded example, fill in the missing steps
If $\csc \theta = 3$ and $\frac{\pi}{2} < \theta < \pi$, find $\cot \theta$. Fill in each blank line. 4 marks
Step 1, Pick the identity linking csc and cot.
1 + cot²θ = ________
Step 2, Substitute cscθ = 3 and square.
1 + cot²θ = ________ ² = ________
Step 3, Solve for cot²θ.
cot²θ = ________ − 1 = ________
Step 4, Take square root, choose sign by quadrant.
$\theta$ lies in Quadrant ________. In this quadrant cot is ________ (positive / negative).
cotθ = ____________________ (simplify the surd)
Conclusion. $\cot \theta$ = ____________.
4. Graduated practice
Show one identity citation and one line of algebra at minimum. Sign of the answer must reflect the given quadrant.
Foundation, recognise the identity (4 questions)
| Q | Expression | Identity used | Simplification |
|---|---|---|---|
| 4.1 1 | $1 - \sin^2 \theta$ | ||
| 4.2 1 | $\sec^2 \theta - \tan^2 \theta$ | ||
| 4.3 1 | $\csc^2 \theta - \cot^2 \theta$ | ||
| 4.4 1 | $\sin^2 \theta(1 + \cot^2 \theta)$ |
Standard, find the missing value (6 questions)
Show identity, substitution, and quadrant analysis.
4.5 If $\sin \theta = \frac{3}{5}$ and $\theta$ is acute, find $\cos \theta$ exactly. 2 marks
4.6 If $\tan \theta = 2$ and $\theta$ is acute, find $\sec \theta$ exactly. 2 marks
4.7 If $\cos \theta = -\frac{2}{3}$ and $\pi < \theta < \frac{3\pi}{2}$, find $\sin \theta$ exactly. 2 marks
4.8 If $\sec \theta = -\frac{5}{3}$ and $\frac{\pi}{2} < \theta < \pi$, find $\tan \theta$ exactly. 2 marks
4.9 Simplify $\frac{\sec^2 \theta - 1}{\tan \theta}$ to a single trigonometric function. 2 marks
4.10 Simplify $\cos^2 \theta(1 + \tan^2 \theta)$. 2 marks
Extension, prove an identity (2 questions)
4.11 Prove $\frac{1 - \cos^2 \theta}{\sin \theta} = \sin \theta$, stating any restrictions. 3 marks
4.12 Prove $\sec^2 \theta + \csc^2 \theta = \sec^2 \theta \cdot \csc^2 \theta$. Hint: rewrite the LHS over a common denominator. 3 marks
5. Self-check the easy 3
Tick the first three once you've verified your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1, The three identities
$\sin^2 \theta + \cos^2 \theta = 1$ · $1 + \tan^2 \theta = \sec^2 \theta$ · $1 + \cot^2 \theta = \csc^2 \theta$.
Q1.2, Identity used
Divide the fundamental identity $\sin^2 \theta + \cos^2 \theta = 1$ by $\cos^2 \theta$ (provided $\cos \theta \neq 0$) to obtain $\tan^2 \theta + 1 = \sec^2 \theta$.
Q1.3, Rearrangement
$\sin^2 \theta = \mathbf{1 - \cos^2 \theta}$ (subtract $\cos^2 \theta$ from both sides).
Q3, Faded example: $\csc \theta = 3$, QII
Step 1: $1 + \cot^2 \theta = \mathbf{\csc^2 \theta}$.
Step 2: $1 + \cot^2 \theta = 3^2 = 9$.
Step 3: $\cot^2 \theta = 9 - 1 = 8$.
Step 4: $\theta$ in QII; cot is negative in QII. $\cot \theta = -\sqrt{8} = -2\sqrt{2}$.
Conclusion: $\cot \theta = \mathbf{-2\sqrt{2}}$.
Q4.1, $1 - \sin^2 \theta$
Identity 1 rearranged: $1 - \sin^2 \theta = \mathbf{\cos^2 \theta}$.
Q4.2, $\sec^2 \theta - \tan^2 \theta$
Identity 2 rearranged: $\sec^2 \theta - \tan^2 \theta = \mathbf{1}$.
Q4.3, $\csc^2 \theta - \cot^2 \theta$
Identity 3 rearranged: $\csc^2 \theta - \cot^2 \theta = \mathbf{1}$.
Q4.4, $\sin^2 \theta(1 + \cot^2 \theta)$
$1 + \cot^2 \theta = \csc^2 \theta = \frac{1}{\sin^2 \theta}$. Multiply: $\sin^2 \theta \cdot \frac{1}{\sin^2 \theta} = \mathbf{1}$.
Q4.5, $\sin \theta = \frac{3}{5}$, acute
$\cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25}$, so $|\cos \theta| = \frac{4}{5}$. Acute (QI) ⇒ cos positive. $\cos \theta = \mathbf{\frac{4}{5}}$.
Q4.6, $\tan \theta = 2$, acute
$\sec^2 \theta = 1 + 4 = 5$, so $|\sec \theta| = \sqrt{5}$. Acute ⇒ sec positive. $\sec \theta = \mathbf{\sqrt{5}}$.
Q4.7, $\cos \theta = -\frac{2}{3}$, QIII
$\sin^2 \theta = 1 - \frac{4}{9} = \frac{5}{9}$, so $|\sin \theta| = \frac{\sqrt{5}}{3}$. In QIII sin negative. $\sin \theta = \mathbf{-\frac{\sqrt{5}}{3}}$.
Q4.8, $\sec \theta = -\frac{5}{3}$, QII
$\tan^2 \theta = \sec^2 \theta - 1 = \frac{25}{9} - 1 = \frac{16}{9}$, so $|\tan \theta| = \frac{4}{3}$. In QII tan negative. $\tan \theta = \mathbf{-\frac{4}{3}}$.
Q4.9, $\frac{\sec^2 \theta - 1}{\tan \theta}$
$\sec^2 \theta - 1 = \tan^2 \theta$. So $\frac{\tan^2 \theta}{\tan \theta} = \mathbf{\tan \theta}$ (where defined, needs $\tan \theta$ defined and non-zero).
Q4.10, $\cos^2 \theta(1 + \tan^2 \theta)$
$1 + \tan^2 \theta = \sec^2 \theta = \frac{1}{\cos^2 \theta}$. Multiply: $\cos^2 \theta \cdot \frac{1}{\cos^2 \theta} = \mathbf{1}$.
Q4.11, Prove $\frac{1 - \cos^2 \theta}{\sin \theta} = \sin \theta$
LHS = $\frac{1 - \cos^2 \theta}{\sin \theta} = \frac{\sin^2 \theta}{\sin \theta} = \sin \theta = $ RHS (using Identity 1 rearranged). Restriction: $\sin \theta \neq 0$, i.e. $\theta \neq n\pi$.
Q4.12, Prove $\sec^2 \theta + \csc^2 \theta = \sec^2 \theta \csc^2 \theta$
LHS = $\frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} = \frac{1}{\sin^2 \theta \cos^2 \theta}$ (using Identity 1 in the numerator).
RHS = $\sec^2 \theta \csc^2 \theta = \frac{1}{\cos^2 \theta} \cdot \frac{1}{\sin^2 \theta} = \frac{1}{\sin^2 \theta \cos^2 \theta}$. LHS = RHS. ✓ Restriction: $\sin \theta \neq 0$ and $\cos \theta \neq 0$, i.e. $\theta \neq \frac{n\pi}{2}$.