Mathematics Advanced • Year 11 • Module 2 • Lesson 7
Pythagorean Identities
Apply the three Pythagorean identities to physical contexts: simple harmonic motion energy, unit-circle reconstruction, ladder geometry, identity-driven simplification, and signal power calculations.
Problem 1 — Energy conservation in a pendulum (physical)
For a small-angle pendulum, the kinetic energy fraction is modelled as $K(\theta) = \sin^2 \theta$ and the potential energy fraction is $U(\theta) = \cos^2 \theta$, with $\theta$ the phase angle in radians. (See the lesson's "Why physicists love these identities" callout.)
Set up: What are we solving for?
(i) Show that the total energy fraction $K(\theta) + U(\theta)$ is constant for every $\theta$. Cite the identity. 2 marks
(ii) At $\theta = \frac{\pi}{6}$, find the exact value of $K(\theta)$ and confirm that $U(\theta) = 1 - K(\theta)$ produces the expected exact value of $\cos^2 \frac{\pi}{6}$. 2 marks
(iii) A second pendulum has $K(\theta) = \frac{1}{4}$. Without solving for $\theta$, find $U(\theta)$ exactly and explain why this single piece of information is enough. 2 marks
Stuck? Revisit lesson § Deriving and using the identities — the SHM callout.Problem 2 — Recovering all six trig values (geometric)
A particle sits on the unit circle at the terminal side of an angle $\theta$ with $\sin \theta = -\frac{\sqrt{5}}{3}$ and $\theta$ in Quadrant IV.
Set up: What are we solving for?
(i) Use Identity 1 to find $\cos \theta$ exactly. State the sign justification. 2 marks
(ii) Find $\tan \theta$, $\sec \theta$, $\csc \theta$ and $\cot \theta$ exactly, rationalising where required. 4 marks
(iii) Verify your $\tan \theta$ and $\sec \theta$ satisfy Identity 2 ($1 + \tan^2 \theta = \sec^2 \theta$) by direct substitution. 2 marks
Problem 3 — Ladder against a wall (real-world geometry)
A 13 m ladder rests against a wall, making angle $\theta$ with the ground. The horizontal distance from the wall to the base is $13 \cos \theta$ metres; the height reached is $13 \sin \theta$ metres. Engineers worry about $13^2 \cos^2 \theta + 13^2 \sin^2 \theta$ — the squared total reach.
Set up: What are we solving for?
(i) Simplify $169 \cos^2 \theta + 169 \sin^2 \theta$ using Identity 1, citing the identity explicitly. 2 marks
(ii) The ladder is set so that $\cos \theta = \frac{5}{13}$. Find $\sin \theta$ exactly and state the exact height reached, $13 \sin \theta$. 2 marks
(iii) Explain in one sentence why the result $169 \cos^2 \theta + 169 \sin^2 \theta = 169$ is "just Pythagoras' theorem applied to the right triangle the ladder makes". 2 marks
Stuck on (iii)? The horizontal and vertical legs are $13 \cos \theta$ and $13 \sin \theta$; the hypotenuse is the ladder.Problem 4 — Simplifying compound expressions
Three lines of a textbook reduce to single trig functions via the Pythagorean identities. Complete each.
Set up: What are we solving for?
(i) Simplify $\frac{\sin^2 \theta}{1 - \cos \theta}$ for $\cos \theta \neq 1$. (Hint: factor the numerator as a difference of squares.) 3 marks
(ii) Simplify $(\sec \theta + \tan \theta)(\sec \theta - \tan \theta)$. 2 marks
(iii) Simplify $\frac{1}{1 + \sin \theta} + \frac{1}{1 - \sin \theta}$ to a single expression involving $\sec^2 \theta$. State restrictions. 3 marks
Problem 5 — AC signal average power (engineering)
For an AC voltage $v(t) = V_0 \sin(\omega t)$, instantaneous power into a 1-Ω resistor at any phase angle $\theta = \omega t$ is $P = V_0^2 \sin^2 \theta$. The cosine-form companion is $V_0^2 \cos^2 \theta$.
Set up: What are we solving for?
(i) Using Identity 1, show that $V_0^2 \sin^2 \theta + V_0^2 \cos^2 \theta$ equals $V_0^2$ for any $\theta$. 2 marks
(ii) At $\theta = \frac{\pi}{3}$, find $\sin^2 \theta$ and $\cos^2 \theta$ exactly, then verify their sum equals 1. 2 marks
(iii) A student writes "$\tan^2 \theta = \csc^2 \theta$" by analogy with the Pythagorean identities. Evaluate the claim — state which trap from the lesson it falls into, then give the correct identity that links $\tan$ to its Pythagorean partner. 2 marks
Stuck on (iii)? Trap 1 in the lesson is exactly "tangent always pairs with secant".How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — SHM energy
Set up. We're applying Identity 1 to confirm energy conservation and to recover one component from the other.
(i) $K(\theta) + U(\theta) = \sin^2 \theta + \cos^2 \theta = 1$ (Identity 1). The total is the constant 1 for all $\theta$.
(ii) $K\left(\frac{\pi}{6}\right) = \sin^2 \frac{\pi}{6} = \left(\frac{1}{2}\right)^2 = \mathbf{\frac{1}{4}}$. By the identity, $U = 1 - K = \frac{3}{4}$. Check: $\cos^2 \frac{\pi}{6} = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$. ✓
(iii) $U(\theta) = 1 - K(\theta) = 1 - \frac{1}{4} = \mathbf{\frac{3}{4}}$. One piece of information suffices because Identity 1 ties the two together as $K + U = 1$.
Problem 2 — Recovering trig values
Set up. We're using Identity 1 to recover $\cos \theta$, then forming the other four trig values and confirming Identity 2.
(i) $\cos^2 \theta = 1 - \frac{5}{9} = \frac{4}{9}$, so $|\cos \theta| = \frac{2}{3}$. In QIV cos positive, so $\cos \theta = \mathbf{\frac{2}{3}}$.
(ii) $\tan \theta = \frac{-\sqrt{5}/3}{2/3} = \mathbf{-\frac{\sqrt{5}}{2}}$. $\sec \theta = \frac{1}{\cos \theta} = \mathbf{\frac{3}{2}}$. $\csc \theta = \frac{1}{\sin \theta} = -\frac{3}{\sqrt{5}} = \mathbf{-\frac{3\sqrt{5}}{5}}$. $\cot \theta = \frac{1}{\tan \theta} = -\frac{2}{\sqrt{5}} = \mathbf{-\frac{2\sqrt{5}}{5}}$.
(iii) $1 + \tan^2 \theta = 1 + \frac{5}{4} = \frac{9}{4}$. $\sec^2 \theta = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$. ✓ Identity 2 holds.
Problem 3 — Ladder
Set up. We're applying Identity 1 to recognise the squared-reach quantity is constant, then evaluating at a specific angle, and explaining the geometric origin.
(i) $169(\cos^2 \theta + \sin^2 \theta) = 169 \cdot 1 = \mathbf{169}$ (by Identity 1). The squared reach is constant: a 13 m ladder always has horizontal+vertical squared-leg sum = 169.
(ii) $\sin^2 \theta = 1 - \frac{25}{169} = \frac{144}{169}$, so $|\sin \theta| = \frac{12}{13}$. For a ladder $\theta$ is acute (positive), so $\sin \theta = \frac{12}{13}$. Height reached: $13 \cdot \frac{12}{13} = \mathbf{12}$ m.
(iii) The ladder is the hypotenuse (length 13). Horizontal leg $13 \cos \theta$, vertical leg $13 \sin \theta$. Pythagoras gives $(13 \cos \theta)^2 + (13 \sin \theta)^2 = 13^2 = 169$ — exactly Identity 1 multiplied through by $13^2$.
Problem 4 — Simplifying
Set up. We're applying Identity 1 (and its rearrangements) plus difference-of-squares factorisations to collapse compound expressions.
(i) $\frac{\sin^2 \theta}{1 - \cos \theta} = \frac{1 - \cos^2 \theta}{1 - \cos \theta} = \frac{(1 - \cos \theta)(1 + \cos \theta)}{1 - \cos \theta} = \mathbf{1 + \cos \theta}$ (provided $\cos \theta \neq 1$).
(ii) $(\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = \sec^2 \theta - \tan^2 \theta = \mathbf{1}$ (Identity 2 rearranged).
(iii) Common denominator: $\frac{(1 - \sin \theta) + (1 + \sin \theta)}{(1 + \sin \theta)(1 - \sin \theta)} = \frac{2}{1 - \sin^2 \theta} = \frac{2}{\cos^2 \theta} = \mathbf{2 \sec^2 \theta}$. Restrictions: $\sin \theta \neq \pm 1$ (i.e. $\cos \theta \neq 0$, so $\theta \neq \frac{\pi}{2} + n\pi$).
Problem 5 — AC signal
Set up. We're confirming Identity 1 at a chosen $\theta$, then critiquing a common misapplication (Trap 1).
(i) $V_0^2 \sin^2 \theta + V_0^2 \cos^2 \theta = V_0^2(\sin^2 \theta + \cos^2 \theta) = V_0^2 \cdot 1 = \mathbf{V_0^2}$. Total power capacity is constant.
(ii) $\sin^2 \frac{\pi}{3} = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$. $\cos^2 \frac{\pi}{3} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$. Sum: $\frac{3}{4} + \frac{1}{4} = 1$. ✓
(iii) Claim is incorrect. It commits Trap 1 from the lesson: "tangent always pairs with secant" (not cosecant). The correct identity is $\mathbf{1 + \tan^2 \theta = \sec^2 \theta}$. The cosecant analogue uses cotangent: $1 + \cot^2 \theta = \csc^2 \theta$.