TMUA Trigonometry questions

Answer: C · Method: Use the cosine rule and ½ab sin C

1. Call the angle $\theta$. Cosine rule: $\cos\theta = \dfrac{6^2 + 10^2 - 14^2}{2 \times 6 \times 10} = \dfrac{-60}{120} = -\dfrac{1}{2}$.
2. So $\theta = 120^\circ$.

Why the tempting wrong answers are wrong

• B: $\cos$ is negative here, so the angle is obtuse.

Answer: C · Method: Use the cosine rule and ½ab sin C

1. Area $= \tfrac{1}{2} \times 7 \times 15 \times \sin 60^\circ$, and $\sin 60^\circ = \tfrac{\sqrt{3}}{2}$.
2. That's $\dfrac{105\sqrt{3}}{4}$.

Why the tempting wrong answers are wrong

• B: $\sin 60^\circ = \tfrac{\sqrt{3}}{2}$, not $\tfrac{1}{2}$.
• E: The formula is $\tfrac{1}{2}ab\sin C$: don't forget the $\tfrac{1}{2}$.

Answer: D · Method: Use the cosine rule and ½ab sin C

1. Cosine rule: $AB^2 = 6^2 + 10^2 - 2 \times 6 \times 10 \cos 120^\circ$.
2. $\cos 120^\circ = -\tfrac{1}{2}$, so $AB^2 = 136 + 60 = 196$ and $AB = 14$.

Why the tempting wrong answers are wrong

• A: $\cos 120^\circ = -\tfrac{1}{2}$. Check the sign in the cosine rule.
• B: That's Pythagoras, which only works for a right angle.

Answer: C · Method: Arc length, sector and segment in radians

1. Arc length $= r\theta = 4 \times \dfrac{\pi}{6} = \dfrac{2\pi}{3}$.
2. Perimeter $=$ arc $+ 2r = \dfrac{2\pi}{3} + 8$.

Why the tempting wrong answers are wrong

• A: That's just the arc. The perimeter also includes the two radii.
• B: There are two straight edges, each a radius.

Answer: C · Method: Reduce to sin or cos = constant, then count per period

1. Write $\sin 2x = 2\sin x\cos x$, so $\cos x\,(2\sin x - 1) = 0$.
2. $\cos x = 0$ gives $x = \tfrac{\pi}{2}, \tfrac{3\pi}{2}$; $\sin x = \tfrac{1}{2}$ gives $x = \tfrac{\pi}{6}, \tfrac{5\pi}{6}$. That is four solutions.

Why the tempting wrong answers are wrong

• A: Dividing both sides by $\cos x$ throws away the solutions where $\cos x = 0$. Factorise instead.

Answer: D · Method: Reduce to sin or cos = constant, then count per period

1. Use $\cos^2 x = 1 - \sin^2 x$: $2 - 2\sin^2 x + 3\sin x = 3$, so $2\sin^2 x - 3\sin x + 1 = 0$, giving $\sin x = 1$ or $\sin x = \tfrac{1}{2}$.
2. In each interval of length $2\pi$, $\sin x = 1$ has one solution and $\sin x = \tfrac{1}{2}$ has two: three in total.
3. $0 \le x \le 4\pi$ covers two full periods (the endpoints aren't solutions), so there are $6$ solutions.

Why the tempting wrong answers are wrong

• A: The interval $0 \le x \le 4\pi$ covers two full periods, not one.
• F: $\sin x = 1$ has only one solution per period, at the top of the curve.

Answer: C · Method: Reduce to sin or cos = constant, then count per period

1. Factorise: $\cos x = \dfrac{1}{2}$ or $\cos x = -\dfrac{1}{2}$.
2. $\cos x = \dfrac{1}{2}$ has 3 solutions in the interval. $\cos x = -\dfrac{1}{2}$ has 3 solutions in the interval.
3. That's $6$ solutions in total.

Why the tempting wrong answers are wrong

• A: The interval is $3\pi$ long, not one period of $2\pi$.

Answer: D · Method: Substitute to get a quadratic, then use the substitute's range

1. Let $s = \cos x$, so the expression is $2s^2 - 2s - 1$.
2. $s$ takes every value in $-1 \le s \le 1$. The turning point is at $s = \dfrac{1}{2}$, which is inside this range.
3. Check the turning point and the ends: the values are $-1$, $3$, $-\dfrac{3}{2}$. The greatest is $3$ and the least is $-\dfrac{3}{2}$, a difference of $\dfrac{9}{2}$.

Why the tempting wrong answers are wrong

• C: Check the turning point as well as the ends: it's inside the range.

Answer: B · Method: Arc length, sector and segment in radians

1. Sector area $= \tfrac{1}{2}r^2\theta = \tfrac{1}{2} \times 144 \times \dfrac{\pi}{6} = 12\pi$.
2. Triangle area $= \tfrac{1}{2}r^2\sin\theta = \tfrac{1}{2} \times 144 \times \tfrac{1}{2} = 36$.
3. Segment $=$ sector $-$ triangle $= 12\pi - 36$.

Why the tempting wrong answers are wrong

• C: The sector is $\tfrac{1}{2}r^2\theta$ and the triangle is $\tfrac{1}{2}r^2\sin\theta$: don't forget the halves.
• E: That's the whole sector. Subtract the triangle formed by the two radii and the chord.

Answer: C · Method: Reduce to sin or cos = constant, then count per period

1. Use $\cos^2 x = 1 - \sin^2 x$ to write everything in terms of $\sin x$: $\sin^2 x + \sin x - 2 = 0$.
2. Factorise: $\sin x = 1$ or $\sin x = -2$.
3. $\sin x = 1$ has 2 solutions in the interval. $\sin x = -2$ is impossible.
4. That's $2$ solutions in total.

Why the tempting wrong answers are wrong

• B: The interval is $4\pi$ long, not one period of $2\pi$.
• D: $\sin x = \pm 1$ has only one solution per period, at the peak or trough.

Answer: C · Method: Substitute to get a quadratic, then use the substitute's range

1. Use $\cos^2 x = 1 - \sin^2 x$: the expression is $-2s^2 - 3s + 4$ with $s = \sin x$.
2. $s$ takes every value in $-1 \le s \le 1$. The turning point is at $s = -\dfrac{3}{4}$, which is inside this range.
3. Check the turning point and the ends: the values are $-1$, $5$, $\dfrac{41}{8}$. The greatest is $\dfrac{41}{8}$ and the least is $-1$, a difference of $\dfrac{49}{8}$.

Why the tempting wrong answers are wrong

• B: Check the turning point as well as the ends: it's inside the range.

Answer: C · Method: Substitute to get a quadratic, then use the substitute's range

1. Let $s = \sin x$, so the expression is $2s^2 + 5s - 3$.
2. $s$ takes every value in $-1 \le s \le 1$. The turning point is at $s = -\dfrac{5}{4}$, which is outside this range.
3. So the extremes are at the ends: $s = 1$ gives $4$ and $s = -1$ gives $-6$. The greatest is $4$.