Advanced Trigonometry

Question 1

(a) Place $P=(\cos A,\sin A)$ and $Q=(\cos B,\sin B)$ on the unit circle. The angle between $OP$ and $OQ$ is $A-B$. Chord distance: $PQ^2 = (\cos A-\cos B)^2 + (\sin A-\sin B)^2 = 2 - 2(\cos A\cos B+\sin A\sin B)$. (1) By the law of cosines on the unit vectors: $PQ^2 = 1+1-2\cos(A-B) = 2-2\cos(A-B)$. (1) Equate: $\cos(A-B) = \cos A\cos B + \sin A\sin B$. (1) Replace $B\to -B$, using $\cos(-B)=\cos B$, $\sin(-B)=-\sin B$: (1) $\cos(A+B) = \cos A\cos B - \sin A\sin B.\quad\blacksquare$ (1)

(b) Set $B=A$: $\cos 2A = \cos^2A - \sin^2A$. (1) Use $\sin^2A = 1-\cos^2A$: $\cos 2A = 2\cos^2A - 1$. (1) Use $\cos^2A = 1-\sin^2A$: $\cos 2A = 1 - 2\sin^2A$. (1)

(c) Use the form $\cos 2A = 1-2\sin^2 A$ because it isolates $\sin^2$. Let $A\to A/2$: $\cos A = 1 - 2\sin^2(A/2)$. (1) Rearrange: $\sin^2(A/2) = \dfrac{1-\cos A}{2}$. $\blacksquare$ (1)

Question 2

(a) By definition, a point at angle $\theta$ on the unit circle is $(\cos\theta,\sin\theta)$. (1) It lies on the circle $x^2+y^2=1$. (1) Substituting gives $\cos^2\theta+\sin^2\theta=1$. (1) This holds for any $\theta$ (positive, negative, beyond $2\pi$) because every angle corresponds to a point on the unit circle regardless of quadrant/rotation. (1)

(b) Divide $\sin^2\theta+\cos^2\theta=1$ by $\cos^2\theta$: $\tan^2\theta+1 = \sec^2\theta$, valid when $\cos\theta\neq0$ i.e. $\theta\neq\tfrac\pi2+n\pi$. (2) Divide by $\sin^2\theta$: $1+\cot^2\theta=\csc^2\theta$, valid when $\sin\theta\neq0$ i.e. $\theta\neq n\pi$. (2)

(c) At $\theta=\tfrac\pi2$, $\cos\theta=0$ so $\tan\theta$ and $\sec\theta$ are undefined (division by zero); the identity's derivation required dividing by $\cos^2\theta\neq0$. (2)

Question 3

(a) Place $C$ at origin, $B=(a,0)$, and $A=(b\cos C, b\sin C)$. (1) Then $c^2 = |AB|^2 = (a-b\cos C)^2 + (b\sin C)^2$. (1) $= a^2 - 2ab\cos C + b^2\cos^2C + b^2\sin^2C$. (1) $= a^2 - 2ab\cos C + b^2(\cos^2C+\sin^2C)$. (1) $= a^2+b^2 - 2ab\cos C.\quad\blacksquare$ (1)

(b) $c^2 = 49+64 - 2(7)(8)\cos60^\circ = 113 - 112\cdot\tfrac12 = 113-56 = 57$. (2) $c = \sqrt{57}$. (1)

(c) Area $= \tfrac12\cdot7\cdot8\cdot\sin60^\circ = 28\cdot\tfrac{\sqrt3}{2} = 14\sqrt3$. (2)

Question 4

(a) $\sin(A+B) = \sin A\cos B + \cos A\sin B$ $\sin(A-B) = \sin A\cos B - \cos A\sin B$. (1) Add: $\sin(A+B)+\sin(A-B) = 2\sin A\cos B$. (1) So $\sin A\cos B = \tfrac12[\sin(A+B)+\sin(A-B)].$ (1)

(b) Let $P=A+B$, $Q=A-B$, so $A=\tfrac{P+Q}{2}$, $B=\tfrac{P-Q}{2}$. (1) From (a), $\sin(A+B)+\sin(A-B) = 2\sin A\cos B$ becomes (1) $\sin P + \sin Q = 2\sin\!\left(\tfrac{P+Q}{2}\right)\cos\!\left(\tfrac{P-Q}{2}\right).\quad\blacksquare$ (2)

(c) $\sin75^\circ+\sin15^\circ = 2\sin45^\circ\cos30^\circ = 2\cdot\tfrac{\sqrt2}{2}\cdot\tfrac{\sqrt3}{2} = \tfrac{\sqrt6}{2}.$ (3)

Question 5

(a) Factor: $\cos x(2\cos x - \sqrt3)=0$. (1) Case $\cos x=0$: $x=\tfrac\pi2, \tfrac{3\pi}{2}$. (2) Case $\cos x=\tfrac{\sqrt3}{2}$: $x=\tfrac\pi6, \tfrac{11\pi}{6}$. (2) Solutions: $\left\{\tfrac\pi6, \tfrac\pi2, \tfrac{3\pi}{2}, \tfrac{11\pi}{6}\right\}$.

(b) $\tan\theta=1 \Rightarrow \theta = \tfrac\pi4 + n\pi$. So $2x-\tfrac\pi3 = \tfrac\pi4+n\pi$. (1) $2x = \tfrac\pi4+\tfrac\pi3+n\pi = \tfrac{7\pi}{12}+n\pi$. (1) $x = \tfrac{7\pi}{24} + \tfrac{n\pi}{2},\ n\in\mathbb Z.$ (1)

(c) Period of solutions is $\tfrac\pi2$; over an interval of length $2\pi$ there are $2\pi \div \tfrac\pi2 = 4$ solutions. (2) ($n=0,1,2,3$: $x=\tfrac{7\pi}{24}, \tfrac{19\pi}{24}, \tfrac{31\pi}{24}, \tfrac{43\pi}{24}$, all in $[0,2\pi)$.)

Question 6

(a) Write as $3\sin\!\big(2(x-\tfrac\pi4)\big)+1$. (1) Amplitude $=3$; (1) Period $=\tfrac{2\pi}{2}=\pi$; (1) Phase shift $=\tfrac\pi4$ right; Vertical shift $=+1$ up. (1) Order: horizontal stretch by $\tfrac12$ (period halved), shift right $\tfrac\pi4$, vertical stretch $\times3$, shift up $1$. (1)

(b) $\arcsin$: domain $[-1,1]$, range $[-\tfrac\pi2,\tfrac\pi2]$. $\arccos$: domain $[-1,1]$, range $[0,\pi]$. (2) $\sin$ is not one-to-one over $\mathbb R$; we restrict its domain to $[-\tfrac\pi2,\tfrac\pi2]$ where it is monotonic increasing and covers all values $[-1,1]$ exactly once, giving a well-defined inverse. (1)

(c) $\cos\tfrac{5\pi}{4} = -\tfrac{\sqrt2}{2}$. (1) $\arccos$ must return a value in $[0,\pi]$; $\tfrac{5\pi}{4}\notin[0,\pi]$, so the answer is not $\tfrac{5\pi}{4}$ but $\arccos(-\tfrac{\sqrt2}{2}) = \tfrac{3\pi}{4}$. (1)

[
  {"claim":"Q3b: c=sqrt(57) when a=7,b=8,C=60deg","code":"c2=7**2+8**2-2*7*8*cos(rad(60)); result = simplify(c2-57)==0"},
  {"claim":"Q3c: area = 14*sqrt(3)","code":"A=Rational(1,2)*7*8*sin(rad(60)); result = simplify(A-14*sqrt(3))==0"},
  {"claim":"Q4c: sin75+sin15 = sqrt(6)/2","code":"result = simplify(sin(rad(75))+sin(rad(15)) - sqrt(6)/2)==0"},
  {"claim":"Q5a: solutions of 2cos^2 x - sqrt3 cos x=0 in [0,2pi)","code":"sols=solveset(Eq(2*cos(x)**2-sqrt(3)*cos(x),0),x,Interval.Ropen(0,2*pi)); result = sols==FiniteSet(pi/6, pi/2, 3*pi/2, 11*pi/6)"},
  {"claim":"Q6c: arccos(cos(5pi/4)) = 3pi/4","code":"result = simplify(acos(cos(5*pi/4)) - 3*pi/4)==0"}
]