Mathematics Advanced • Year 11 • Module 2 • Lesson 15
Review and Connections
Rebuild fluency across the whole module — exact values, Pythagorean identities, quadrant signs, transformed graph features, and the equation-graph link.
1. Quick recall — the three pillars
Three warm-ups. 1 mark each
Q1.1 State the three Pythagorean identities:
$\sin^2 x + $ ________ $ = 1$. $1 + \tan^2 x = $ ________. $1 + \cot^2 x = $ ________.
Q1.2 ASTC mnemonic — which trig functions are positive in each quadrant?
Q1: ________. Q2: ________. Q3: ________. Q4: ________.
Q1.3 General form for sinusoidal modelling:
$y = $ ________ $\cdot \sin($ ________ $(x - $ ________ $)) + $ ________. Amplitude $= $ ________, period $= $ ________, range $= $ ________.
2. Worked example — combined identity + exact value
Problem. If $\sin \theta = 1/3$ and $\pi/2 < \theta < \pi$ (i.e. $\theta$ is in Q2), find the exact value of $\tan \theta$.
Step 1 — Use the Pythagorean identity to get $\cos^2\theta$.
$\cos^2\theta = 1 - \sin^2\theta = 1 - 1/9 = 8/9$.
Step 2 — Determine the sign using the quadrant.
In Q2, cosine is negative: $\cos\theta = -\sqrt{8}/3 = -2\sqrt{2}/3$.
Reason: ASTC says only sine is positive in Q2 (Trap 02 — forgetting quadrant signs).
Step 3 — Compute $\tan\theta$.
$\tan\theta = \sin\theta / \cos\theta = (1/3) / (-2\sqrt{2}/3) = -1/(2\sqrt{2}) = -\sqrt{2}/4$.
Conclusion. $\tan\theta = $ $-\sqrt{2}/4$.
3. Faded example — fill in the missing steps
If $\cos\theta = -3/5$ and $\pi < \theta < 3\pi/2$ (Q3), find $\sin\theta$ and $\tan\theta$. 4 marks
Step 1 — Pythagorean identity.
$\sin^2\theta = 1 - \cos^2\theta = 1 - $ ________ $ = $ ________.
Step 2 — Sign from quadrant.
In Q3, sine is ________ (positive / negative): $\sin\theta = $ ________.
Step 3 — Compute $\tan\theta$.
$\tan\theta = \sin\theta / \cos\theta = $ ________.
Conclusion. $\sin\theta = $ ________, $\tan\theta = $ ________.
4. Graduated practice
Foundation — exact values and quadrants (4 questions)
| Q | Find | Answer |
|---|---|---|
| 4.1 1 | $\sin(\pi/3)$ exact value | |
| 4.2 1 | $\cos(2\pi/3)$ exact value | |
| 4.3 1 | $\tan(5\pi/4)$ exact value | |
| 4.4 1 | Quadrant of $11\pi/6$ |
Standard — identities, graphs, equations (6 questions)
4.5 Simplify $\dfrac{1 - \cos^2 x}{\sin x \cos x}$. 2 marks
4.6 State the amplitude, period, and range of $y = 4\cos(3x) + 2$. 3 marks
4.7 How many solutions does $\sin x = -0.5$ have in $[0, 4\pi]$? Justify with a period count. 2 marks
4.8 Rewrite $y = 3\sin(2x - \pi/3)$ in the form $y = a\sin(b(x - c))$ and state the phase shift. 2 marks
4.9 A daylight model is $D(t) = 12 + 2\sin(2\pi t /365)$ where $t$ is days from March 21. Find the maximum daylight and the day it occurs. 2 marks
4.10 Find the period and the asymptotes (in $[0, 2\pi)$) of $y = \tan(2x)$. 2 marks
Extension — combine the whole module (2 questions)
4.11 Sketch $y = 3\sin\!\left(2x - \dfrac{\pi}{3}\right)$ for one complete period starting from the first positive $x$-intercept. State amplitude, period, phase shift, and the $(x, y)$ coordinates of the maximum, minimum, and the intercepts within that period. 4 marks
4.12 Prove the identity $\dfrac{\sin x}{1 + \cos x} + \dfrac{1 + \cos x}{\sin x} = \dfrac{2}{\sin x}$. (Combine the two fractions, then use $\sin^2 x + \cos^2 x = 1$.) 4 marks
5. Self-check the easy 3
Tick the first three once you've verified each one.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Pythagorean identities
$\sin^2 x + $ $\cos^2 x$ $ = 1$. $1 + \tan^2 x = $ $\sec^2 x$. $1 + \cot^2 x = $ $\csc^2 x$.
Q1.2 — ASTC quadrant signs
Q1: All (sin, cos, tan all positive). Q2: Sine only. Q3: Tangent only. Q4: Cosine only.
Q1.3 — General form
$y = $ $a$ $\cdot \sin($$b$$(x - $ $c$$)) + $ $d$. Amplitude $= |a|$, period $= 2\pi/|b|$, range $= [d - |a|, d + |a|]$.
Q3 — Faded example: $\cos\theta = -3/5$, $\theta \in $ Q3
$\sin^2\theta = 1 - 9/25 = 16/25$. In Q3, sine is negative: $\sin\theta = -4/5$. $\tan\theta = (-4/5)/(-3/5) = 4/3$. $\sin\theta = -4/5$, $\tan\theta = 4/3$.
Q4.1 — $\sin(\pi/3)$
$\sqrt{3}/2$.
Q4.2 — $\cos(2\pi/3)$
Q2 (between $\pi/2$ and $\pi$). Reference angle $\pi/3$. Cosine negative in Q2: $\cos(2\pi/3) = $ $-1/2$.
Q4.3 — $\tan(5\pi/4)$
Q3 (between $\pi$ and $3\pi/2$). Reference angle $\pi/4$. Tangent positive in Q3: $\tan(5\pi/4) = $ $1$.
Q4.4 — Quadrant of $11\pi/6$
$11\pi/6$ lies between $3\pi/2 = 9\pi/6$ and $2\pi = 12\pi/6$, so it is in Q4.
Q4.5 — Simplify $(1 - \cos^2 x)/(\sin x \cos x)$
$1 - \cos^2 x = \sin^2 x$ (Pythagorean). So expression $= \sin^2 x / (\sin x \cos x) = \sin x / \cos x = $ $\tan x$.
Q4.6 — $y = 4\cos(3x) + 2$
Amplitude $4$, period $= 2\pi/3$, range $= [2 - 4, 2 + 4] = $ $[-2, 6]$.
Q4.7 — $\sin x = -0.5$ on $[0, 4\pi]$
2 per period × 2 periods = $4$ solutions.
Q4.8 — Factor $3\sin(2x - \pi/3)$
$2x - \pi/3 = 2(x - \pi/6)$. So $y = $ $3\sin(2(x - \pi/6))$; phase shift $= $ $\pi/6$ right (Trap 02: not $\pi/3$).
Q4.9 — Daylight max
Max $= 12 + 2 = $ $14$ h when $\sin = 1 \Rightarrow 2\pi t/365 = \pi/2 \Rightarrow t = 365/4 \approx 91$ days after March 21, i.e. around June 21.
Q4.10 — $y = \tan(2x)$
Period $= \pi/2$. Asymptotes where $\cos(2x) = 0 \Rightarrow 2x = \pi/2 + n\pi \Rightarrow x = \pi/4 + n\pi/2$. In $[0, 2\pi)$: $x = \pi/4, 3\pi/4, 5\pi/4, 7\pi/4$.
Q4.11 — Sketch $y = 3\sin(2x - \pi/3)$
Factor: $y = 3\sin(2(x - \pi/6))$. Amplitude $3$, period $\pi$, phase shift $\pi/6$ right. First positive $x$-intercept at $x = \pi/6$ (going up). Max at quarter-period later: $x = \pi/6 + \pi/4 = 5\pi/12$, $y = 3$. Next zero at $x = \pi/6 + \pi/2 = 2\pi/3$. Min at $x = \pi/6 + 3\pi/4 = 11\pi/12$, $y = -3$. Returns to zero at $x = \pi/6 + \pi = 7\pi/6$. Key points: $(\pi/6, 0), (5\pi/12, 3), (2\pi/3, 0), (11\pi/12, -3), (7\pi/6, 0)$.
Q4.12 — Prove the identity
LHS $= \dfrac{\sin x}{1 + \cos x} + \dfrac{1 + \cos x}{\sin x}$. Common denominator $\sin x(1 + \cos x)$:
$= \dfrac{\sin^2 x + (1 + \cos x)^2}{\sin x(1 + \cos x)} = \dfrac{\sin^2 x + 1 + 2\cos x + \cos^2 x}{\sin x(1 + \cos x)}$.
Use $\sin^2 x + \cos^2 x = 1$:
$= \dfrac{1 + 1 + 2\cos x}{\sin x(1 + \cos x)} = \dfrac{2 + 2\cos x}{\sin x(1 + \cos x)} = \dfrac{2(1 + \cos x)}{\sin x(1 + \cos x)} = \dfrac{2}{\sin x}$. ✓ (QED)