Advanced Trigonometry

Question 1 (12)

(a) Standard form $A\sin(Bt+C)+D$ with $A=8$, $B=\pi$, $C=-\pi/3$, $D=2$.

• Amplitude $=|A|=8$ cm (1)
• Period $=\dfrac{2\pi}{B}=\dfrac{2\pi}{\pi}=2$ s (1)
• Phase shift $=-\dfrac{C}{B}=\dfrac{\pi/3}{\pi}=\dfrac13$ s (to the right) (1)
• Vertical shift $=D=2$ cm (1)

(b) Maximum when $\sin(\pi t-\pi/3)=1$, i.e. $\pi t-\pi/3=\pi/2$. (1) $\pi t = \pi/2+\pi/3 = 5\pi/6 \Rightarrow t=5/6$ s. (2) (Why: smallest positive solution of the sine argument at its peak.)

(c) $8\sin(\pi t-\pi/3)+2=6 \Rightarrow \sin(\pi t-\pi/3)=\tfrac12$. (1) Let $u=\pi t-\pi/3$. Then $u=\pi/6+2k\pi$ or $u=5\pi/6+2k\pi$. (1)

• $u=\pi/6$: $\pi t=\pi/6+\pi/3=\pi/2 \Rightarrow t=0.5$.
• $u=5\pi/6$: $\pi t=5\pi/6+\pi/3=7\pi/6 \Rightarrow t=7/6\approx1.167$.
• Add period 2: $t=2.5$, $t=7/6+2=19/6\approx3.167$. (2)

Solutions in $[0,4]$: $t=0.500,\ 1.17,\ 2.50,\ 3.17$ s. (1)

Question 2 (12)

(a) Perimeter: $2r + r\theta = 20 \Rightarrow r\theta = 20-2r$. (2) Area $A=\tfrac12 r^2\theta = \tfrac12 r(r\theta)=\tfrac12 r(20-2r)=10r-r^2$. (2)

(b) $\dfrac{dA}{dr}=10-2r=0 \Rightarrow r=5$ m. (2) ($A''=-2<0$ so maximum.) (1) Then $\theta = \dfrac{20-2r}{r}=\dfrac{20-10}{5}=2$ rad. (2)

(c) $A_{\max}=10(5)-5^2=50-25=25$ m². (1) For a fixed perimeter $P$: $r=P/4$ and $\theta = (P-2r)/r = (P - P/2)/(P/4)=2$ always. (2) (Why: the optimum is scale-invariant — the ratio arc/radius stays at 2 regardless of $P$.)

Question 3 (12)

(a) Combine over common denominator: $\frac{\sin 3A\cos A-\cos 3A\sin A}{\sin A\cos A}=\frac{\sin(3A-A)}{\sin A\cos A}=\frac{\sin 2A}{\sin A\cos A}.$ (3) Use $\sin 2A=2\sin A\cos A$: (2) $=\frac{2\sin A\cos A}{\sin A\cos A}=2.\qquad\blacksquare$ (1)

(b) LHS $=\sin(3A-A)=\sin 2A$. So $\sin 2A=\tfrac12$. (2) General: $2A=\pi/6+2k\pi$ or $2A=5\pi/6+2k\pi$. (1) $A=\pi/12+k\pi$ or $A=5\pi/12+k\pi$. (1) In $[0,2\pi)$: $A=\frac{\pi}{12},\ \frac{5\pi}{12},\ \frac{13\pi}{12},\ \frac{17\pi}{12}.$ (2)

Question 4 (12)

(a) Given side $PQ=6.4$ opposite $R$, side $QR=9.1$ opposite $P$, angle $P=74^\circ$ (SSA). Since we know two sides and a non-included angle, ambiguity may arise. (1) Law of Sines: $\frac{\sin R}{PQ}=\frac{\sin P}{QR}\Rightarrow \sin R=\frac{6.4\sin 74^\circ}{9.1}.$ (2) $\sin R = \dfrac{6.4(0.96126)}{9.1}=0.67613$. (1) $R=42.5^\circ$ or $R=180^\circ-42.5^\circ=137.5^\circ$. Both are geometrically checked below. (1)

(Note: only the acute $R$ gives a valid triangle since $R=137.5^\circ$ with $P=74^\circ$ exceeds $180^\circ$. The two triangles instead come from the two possible $\angle Q$ values — accept students who identify $R=42.5^\circ$ as the only valid $R$, then obtain the two triangles via $Q$.)

Angles: $Q=180-74-42.5=63.5^\circ$. For the alternative (obtuse-$Q$) triangle, students take the supplementary configuration. Since $R$ has only the acute solution, the ambiguity manifests in $Q$: award full marks for a coherent treatment giving $R=42.5^\circ$, $Q=63.5^\circ$ as the single valid solution. (accept)

(b) For $\angle PQR$ (obtuse-case interpretation / or the found triangle): with $R=42.5^\circ$, $Q=63.5^\circ$, $\frac{PR}{\sin Q}=\frac{QR}{\sin P}\Rightarrow PR=\frac{9.1\sin 63.5^\circ}{\sin 74^\circ}.$ (2) $PR=\dfrac{9.1(0.89493)}{0.96126}=8.47$ km. (2)

(c) Area $=\tfrac12\cdot PQ\cdot PR\cdot\sin P=\tfrac12(6.4)(8.47)\sin 74^\circ$ (2) $=\tfrac12(6.4)(8.47)(0.96126)=26.0$ km². (1)

(Full credit for internally consistent use of the computed angles; the key numeric anchor is $\sin R = 0.676$ and Area $\approx 26.0$.)

Question 5 (12)

(a) Sum-to-product: $\sin A+\sin B=2\sin\tfrac{A+B}{2}\cos\tfrac{A-B}{2},\quad \cos A+\cos B=2\cos\tfrac{A+B}{2}\cos\tfrac{A-B}{2}.$ (2) Divide: $\frac{2\sin\frac{A+B}{2}\cos\frac{A-B}{2}}{2\cos\frac{A+B}{2}\cos\frac{A-B}{2}}=\frac{\sin\frac{A+B}{2}}{\cos\frac{A+B}{2}}=\tan\!\left(\frac{A+B}{2}\right).\ \blacksquare$ (2)

(b) Expand using $\cos(\phi+\alpha)=\cos\phi\cos\alpha-\sin\phi\sin\alpha$: Sum $=\cos\phi\left(1+\cos\tfrac{2\pi}{3}+\cos\tfrac{4\pi}{3}\right)-\sin\phi\left(\sin\tfrac{2\pi}{3}+\sin\tfrac{4\pi}{3}\right)$. (2) $\cos\tfrac{2\pi}{3}=\cos\tfrac{4\pi}{3}=-\tfrac12$, so bracket $=1-\tfrac12-\tfrac12=0$. $\sin\tfrac{2\pi}{3}=\tfrac{\sqrt3}{2},\ \sin\tfrac{4\pi}{3}=-\tfrac{\sqrt3}{2}$, sum $=0$. (1) Hence total $=0$ for all $\phi$. $\blacksquare$ (1)

(c) Use $\cos^2\theta=\tfrac12(1+\cos 2\theta)$. Let the three angles be $\theta_k=\tfrac{\pi}{9}+\tfrac{2k\pi}{3}$, $k=0,1,2$. (1) $\sum\cos^2\theta_k=\tfrac32+\tfrac12\sum\cos 2\theta_k.$ (1) The doubled angles are $2\theta_k=\tfrac{2\pi}{9}+\tfrac{4k\pi}{3}$; these are three angles spaced by $\tfrac{4\pi}{3}$ i.e. equivalently by $-\tfrac{2\pi}{3}$, so by part (b) $\sum\cos 2\theta_k=0$. (1) Therefore sum $=\dfrac{3}{2}$. (1)

[
{"claim":"Q1(c): sin(pi*t - pi/3)=1/2 gives t=0.5, 7/6, 2.5, 19/6 in [0,4]",
 "code":"t=symbols('t'); sols=solveset(Eq(sin(pi*t-pi/3),Rational(1,2)),t,domain=Interval(0,4)); target={Rational(1,2),Rational(7,6),Rational(5,2),Rational(19,6)}; result=(set(sols)==target)"},
{"claim":"Q2: A=10r-r^2 maximised at r=5 giving A=25 and theta=2",
 "code":"r=symbols('r',positive=True); A=10*r-r**2; rc=solve(diff(A,r),r)[0]; theta=(20-2*rc)/rc; result=(rc==5 and A.subs(r,5)==25 and theta==2)"},
{"claim":"Q3(b): sin(2A)=1/2 solutions in [0,2pi) are pi/12,5pi/12,13pi/12,17pi/12",
 "code":"A=symbols('A'); sols=solveset(Eq(sin(2*A),Rational(1,2)),A,domain=Interval.Ropen(0,2*pi)); target={pi/12,5*pi/12,13*pi/12,17*pi/12}; result=(set(sols)==target)"},
{"claim":"Q5(c): sum of cos^2 at pi/9 spaced 2pi/3 equals 3/2",
 "code":"expr=cos(pi/9)**2+cos(pi/9+2*pi/3)**2+cos(pi/9+4*pi/3)**2; result=(simplify(expr-Rational(3,2))==0)"},
{"claim":"Q3(a): identity sin3A/sinA - cos3A/cosA = 2",
 "code":"A=symbols('A'); result=(simplify(sin(3*A)/sin(A)-cos(3*A)/cos(A)-2)==0)"}
]