Advanced Trigonometry

Question 1

(a) (4 marks) From $\cos(A+B)=\cos A\cos B - \sin A\sin B$, set $B=A$: $\cos 2A = \cos A\cos A - \sin A\sin A = \cos^2 A - \sin^2 A. \quad(1)$ Using $\sin^2 A = 1-\cos^2 A$: $\cos 2A = \cos^2 A - (1-\cos^2 A) = 2\cos^2 A - 1. \quad(1)$ Using $\cos^2 A = 1-\sin^2 A$: $\cos 2A = (1-\sin^2 A) - \sin^2 A = 1 - 2\sin^2 A. \quad(1)$ Correct citation of Pythagorean identity $\sin^2+\cos^2=1$ at each step (1).

(b) (5 marks) From $\cos 2A = 1-2\sin^2 A$, put $A=\theta/2$: $\cos\theta = 1 - 2\sin^2(\theta/2)$, so $\sin^2\!\left(\tfrac{\theta}{2}\right)=\frac{1-\cos\theta}{2}. \quad(2)$ Set $\theta = 30^\circ$ so $\theta/2 = 15^\circ$, with $\cos30^\circ=\tfrac{\sqrt3}{2}$: $\sin^2 15^\circ = \frac{1-\tfrac{\sqrt3}{2}}{2} = \frac{2-\sqrt3}{4}. \quad(1)$ Since $15^\circ$ is in Q1, $\sin15^\circ>0$: $\sin15^\circ = \frac{\sqrt{2-\sqrt3}}{2}=\frac{\sqrt6-\sqrt2}{4}. \quad(2)$ (Equivalence $\sqrt{2-\sqrt3}=\tfrac{\sqrt6-\sqrt2}{2}$ shown by squaring: $\tfrac{6+2-2\sqrt{12}}{4}=\tfrac{8-4\sqrt3}{4}=2-\sqrt3$.)

(c) (6 marks) Use $\cos 2x = 1-2\sin^2 x$: $2(1-2\sin^2 x)+4\sin x = 1 \Rightarrow -4\sin^2 x + 4\sin x + 1 = 1. \quad(2)$ So $-4\sin^2 x + 4\sin x = 0 \Rightarrow 4\sin x(1-\sin x)=0$. (1) Thus $\sin x = 0$ or $\sin x = 1$. (1)

• $\sin x = 0$: $x = 0,\ \pi$.
• $\sin x = 1$: $x = \tfrac{\pi}{2}$.

Solutions in $[0,2\pi)$: $\boxed{x = 0,\ \tfrac{\pi}{2},\ \pi}$. (2)

(d) (5 marks) Combine over common denominator $\sin\theta\cos\theta$: $\frac{\sin3\theta\cos\theta - \cos3\theta\sin\theta}{\sin\theta\cos\theta}. \quad(1)$ Numerator is $\sin(3\theta-\theta)=\sin2\theta$ (difference formula). (2) $=\frac{\sin2\theta}{\sin\theta\cos\theta}=\frac{2\sin\theta\cos\theta}{\sin\theta\cos\theta}=2. \quad(1)$ Domain: $\sin\theta\neq0$ and $\cos\theta\neq0$, i.e. $\theta\neq \tfrac{n\pi}{2},\ n\in\mathbb Z$. (1)

Question 2

(a) (4 marks) $R = \sqrt{3^2+4^2}=5$ (1). Match $R\sin(2t+\phi)=R\cos\phi\sin2t + R\sin\phi\cos2t$ with $3\sin2t+4\cos2t$: $R\cos\phi = 3,\ R\sin\phi = 4$ (1). So $\tan\phi = 4/3$, both positive ⇒ $\phi$ in Q1 (1): $\phi = \arctan(4/3)\approx 0.927 \text{ rad}. \quad(1)$ Thus $x(t)=5\sin(2t+0.927)$.

(b) (4 marks) Amplitude $=5$ cm (1). Period $=\dfrac{2\pi}{2}=\pi$ s (1). Max when $2t+\phi=\tfrac{\pi}{2}$ ⇒ $t=\dfrac{\pi/2-0.927}{2}\approx\dfrac{1.5708-0.9273}{2}\approx 0.322$ s (2).

(c) (4 marks) $v(t)=\dfrac{d}{dt}[5\sin(2t+\phi)]=10\cos(2t+\phi)$ (2), which is $10\sin(2t+\phi+\tfrac{\pi}{2})$. Max speed $=10$ cm/s (2).

(d) (4 marks) $\sin(2t+\phi)\cdot 5\sin(2t+\tfrac{\pi}{3})$. Product-to-sum: $\sin P\sin Q=\tfrac12[\cos(P-Q)-\cos(P+Q)]$ (1). With $P=2t+\phi,\ Q=2t+\tfrac{\pi}{3}$: $=\frac52\Big[\cos(\phi-\tfrac{\pi}{3})-\cos(4t+\phi+\tfrac{\pi}{3})\Big]. \quad(2)$ First term constant (frequency 0); second term oscillates at angular frequency $4$ rad/s (i.e. double the original). (1)

(e) (6 marks) Reduce $x(t)=y(t)$: $5\sin(2t+\phi)=5\sin(2t+\tfrac{\pi}{3})$ ⇒ $\sin(2t+\phi)=\sin(2t+\tfrac{\pi}{3})$ (1). Two cases: (i) $2t+\phi = 2t+\tfrac{\pi}{3}+2k\pi$ ⇒ $\phi=\tfrac{\pi}{3}$, false (no solutions this way); (ii) $2t+\phi = \pi-(2t+\tfrac{\pi}{3})+2k\pi$ ⇒ $4t = \pi - \tfrac{\pi}{3}-\phi + 2k\pi = \tfrac{2\pi}{3}-\phi+2k\pi$ (2). $t = \dfrac{2\pi/3-\phi+2k\pi}{4}$. With $\phi\approx0.927$: $t\approx\dfrac{2.094-0.927}{4}\approx0.292$ s ($k=0$); next $k=1$ gives $t\approx1.862$ s (outside $[0,\pi]$ if $\pi\approx3.14$? actually $1.862<\pi$, included). So on $[0,\pi]$: $t\approx0.292,\ 1.862$ s. First is $0.292$ s (1).

Pseudocode (bisection sign-change) (2):

def f(t): return x(t) - y(t)   # continuous
t, dt, prev = 0, 0.001, f(0)
while t <= pi:
    cur = f(t+dt)
    if prev*cur < 0:           # sign change ⇒ root bracketed
        # bisection between t and t+dt
        a,b = t, t+dt
        for _ in range(50):
            m=(a+b)/2
            if f(a)*f(m)<=0: b=m
            else: a=m
        print("root ~", (a+b)/2); break
    prev=cur; t+=dt

Question 3

(a) (5 marks) SSA data (two sides + non-included angle) ⇒ ambiguous case (1). Law of Sines: $\dfrac{\sin B}{b}=\dfrac{\sin A}{a}$ ⇒ $\sin B = \dfrac{b\sin A}{a}=\dfrac{5\sin40^\circ}{8}=\dfrac{5(0.64279)}{8}=0.40174$ (2). Since $a>b$, angle $A$ opposite larger side ⇒ only one triangle (the obtuse alternative $B'$ would give $A+B'>180^\circ$? check: $B_1=23.69^\circ$, $B_2=156.31^\circ$; $A+B_2=196^\circ>180^\circ$ invalid) (1). Unique: $B \approx 23.69^\circ$ (1).

(b) (4 marks) $C = 180^\circ - 40^\circ - 23.69^\circ = 116.31^\circ$ (1). $c = \dfrac{a\sin C}{\sin A}=\dfrac{8\sin116.31^\circ}{\sin40^\circ}=\dfrac{8(0.89649)}{0.64279}\approx 11.16$ m (2). Area $=\tfrac12 bc\sin A=\tfrac12(5)(11.16)\sin40^\circ\approx \tfrac12(5)(11.16)(0.64279)\approx 17.93$ m² (1).

(c) (5 marks) Drop perpendicular from $B$ to line $CA$, foot $H$, length $h$. Then $BH=h=c\sin A$ and $AH=c\cos A$, so $CH=b-c\cos A$ (2). In right triangle $BHC$: $a^2 = h^2 + CH^2 = (c\sin A)^2 + (b-c\cos A)^2. \quad(1)$ Expand: $c^2\sin^2A + b^2 - 2bc\cos A + c^2\cos^2A = b^2 + c^2(\sin^2A+\cos^2A) - 2bc\cos A$ (1) $= b^2 + c^2 - 2bc\cos A. \quad(1)$

(d) (4 marks) $\text{Area}^2 = \tfrac14 b^2c^2\sin^2A=\tfrac14 b^2c^2(1-\cos^2A)$ (1). Substitute $\cos A=\tfrac{b^2+c^2-a^2}{2bc}$: $=\tfrac14 b^2c^2\left[1-\left(\tfrac{b^2+c^2-a^2}{2bc}\right)^2\right]=\tfrac{1}{16}\big[(2bc)^2-(b^2+c^2-a^2)^2\big]. \quad(1)$ Difference of squares: $=\tfrac{1}{16}(2bc+b^2+c^2-a^2)(2bc-b^2-c^2+a^2)=\tfrac{1}{16}[(b+c)^2-a^2][a^2-(b-c)^2].$ Each bracket factors, giving $(b+c+a)(b+c-a)(a+b-c)(a-b+c)$. With $s=\tfrac12(a+b+c)$ these equal $2s\cdot2(s-a)\cdot2(s-c)\cdot2(s-b)$ (1): $\text{Area}^2 = \tfrac{1}{16}\cdot16\,s(s-a)(s-b)(s-c)=s(s-a)(s-b)(s-c).$ $\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}. \quad(1)$

[
  {"claim":"sin15 = (sqrt6-sqrt2)/4 equals sqrt((2-sqrt3)/... ) numeric",
   "code":"lhs=(sqrt(6)-sqrt(2))/4; rhs=sqrt((1-sqrt(3)/2)/2); result=simplify(lhs-rhs)==0"},
  {"claim":"Q1c solutions satisfy 2cos2x+4sinx=1",
   "code":"xs=[0,pi/2,pi]; result=all(simplify(2*cos(2*x)+4*sin(x)-1)==0 for x in x