Mathbench

Chapter 6 — Trigonometry

ACT counts trigonometry inside Geometry, but it deserves its own chapter because this is where the ACT goes furthest beyond the SAT. The SAT tests right triangles and almost nothing else. The ACT adds the law of sines, the law of cosines, radians, identities, and the graphs of sine and cosine.

There are usually three or four trigonometry questions on a paper. They are completely learnable — the whole topic rests on one mnemonic and three formulas, and the ACT prints none of them.

Topics covered: SOHCAHTOA · finding a side · finding an angle · the special right triangles · trigonometric identities · the law of sines · the law of cosines · degrees and radians · the graphs of sine and cosine


Section 1 — Right-Triangle Trigonometry

Label the sides relative to the angle you are using:

Then:

sinθ=oppositehypotenusecosθ=adjacenthypotenusetanθ=oppositeadjacent\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} \qquad \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} \qquad \tan \theta = \frac{\text{opposite}}{\text{adjacent}}

SOH-CAH-TOA. Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent.

The word "opposite" is the trap: opposite the angle you are working with, which changes when you switch angles. Always mark your angle on the sketch first.


Q1Basic

Topic: Finding a trigonometric ratio in a right triangle

A ramp is modelled by right triangle ABCABC with the right angle at CC. The sloping edge ABAB measures 13 units and the side BCBC, which lies opposite angle AA, measures 5 units. What is the value of sinA\sin A?

A) 1213\frac{12}{13}

B) 512\frac{5}{12}

C) 135\frac{13}{5}

D) 513\frac{5}{13}

Show the worked solution

Answer: D

Explanation

The right angle is at CC, so the hypotenuse is AB=13AB = 13 — the side opposite the right angle.

For angle AA, the opposite side is BC=5BC = 5 (the question says so).

sinA=oppositehypotenuse=513\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{13}

The third side is AC=13252=144=12AC = \sqrt{13^2 - 5^2} = \sqrt{144} = 12, which is where the other options come from. 5, 12, 13 is a Pythagorean triple worth memorising.

Why each wrong option is wrong:

  • A, 1213\frac{12}{13}1213\frac{12}{13} is cosA\cos A: adjacent over hypotenuse.
  • B, 512\frac{5}{12}512\frac{5}{12} is tanA\tan A: opposite over adjacent.
  • C, 135\frac{13}{5} — inverted the ratio. Sine and cosine are always at most 1, because the hypotenuse is the longest side, so any answer bigger than 1 is wrong immediately.

Takeaway: mark the angle, label opposite / adjacent / hypotenuse relative to it, then apply SOHCAHTOA. Sine and cosine can never exceed 1.


Q2Medium

Topic: Using a ratio to find a side

A builder is installing a wheelchair ramp that rises at an angle of 12°12° to the horizontal ground, and the sloping surface measures 9 metres from end to end. Building regulations require the vertical rise to be recorded. Which expression gives that height, in metres?

F) 9cos12°9\cos 12°

G) 9sin12°9\sin 12°

H) 9sin12°\dfrac{9}{\sin 12°}

J) 9tan12°9\tan 12°

Show the worked solution

Answer: G

Explanation

Sketch it: the ramp is the hypotenuse (9 m), the vertical rise is opposite the 12°12° angle, and the ground is adjacent.

Opposite and hypotenuse means sine:

sin12°=height9\sin 12° = \frac{\text{height}}{9}

Multiply both sides by 9:

height=9sin12°\text{height} = 9 \sin 12°

Sanity check without a calculator: sin12°\sin 12° is a small number well below 1, so the height is a small fraction of 9 metres. That is what a gentle ramp should be.

Why each wrong option is wrong:

  • F, 9cos12°9\cos 12° — cosine gives the horizontal distance along the ground, not the rise.
  • H, 9sin12°\dfrac{9}{\sin 12°} — dividing by sin12°\sin 12° gives about 43 metres, which is far longer than the ramp itself. Impossible: a leg cannot exceed the hypotenuse.
  • J, 9tan12°9\tan 12° — tangent uses the adjacent side, which is not given here.

Takeaway: pick the ratio that uses the side you have and the side you want. Then check the size: a leg is always shorter than the hypotenuse.


Q3Medium

Topic: A special right triangle

A support bracket is cut as a 30°30°60°60°90°90° triangle. The side opposite the 30°30° angle has length 7. What is the length of the hypotenuse?

A) 737 \sqrt{3}

B) 1414

C) 72\frac{7}{2}

D) 727 \sqrt{2}

Show the worked solution

Answer: B

Explanation

In a 30°30°60°60°90°90° triangle the sides are always in the ratio

1:3:21 : \sqrt{3} : 2

with the 1 opposite the 30°30°, the 3\sqrt{3} opposite the 60°60°, and the 2 as the hypotenuse.

Here the side opposite 30°30° is 7, so that is the "1" of the ratio and everything scales by 7:

hypotenuse=2×7=14\text{hypotenuse} = 2 \times 7 = 14

The shortcut worth remembering: the hypotenuse is twice the shortest side, and the shortest side is always the one facing the smallest angle.

Why each wrong option is wrong:

  • A, 737 \sqrt{3} — gave the side opposite 60°60°, which is 737\sqrt{3}.
  • C, 72\frac{7}{2} — halved rather than doubled. The hypotenuse is the longest side, so it cannot be smaller than a leg.
  • D, 727 \sqrt{2} — used the 45°45°45°45°90°90° ratio 1:1:21 : 1 : \sqrt{2}.

Takeaway: 303060609090 gives 1:3:21 : \sqrt{3} : 2, and 454545459090 gives 1:1:21 : 1 : \sqrt{2}. Match the side you are given to its place in the ratio, then scale.


Q4Hard

Topic: A trigonometric identity

A calculation gives tanα=512\tan \alpha = \dfrac{5}{12} and sinα=513\sin \alpha = \dfrac{5}{13} for an acute angle α\alpha, and cosα\cos \alpha is needed. What is cosα\cos \alpha?

F) 513\frac{5}{13}

G) 1312\frac{13}{12}

H) 512\frac{5}{12}

J) 1213\frac{12}{13}

Show the worked solution

Answer: J

Explanation

The identity that links all three:

tanα=sinαcosα\tan \alpha = \frac{\sin \alpha}{\cos \alpha}

Rearrange it to make cosα\cos \alpha the subject:

cosα=sinαtanα=5/135/12=513×125=1213\cos \alpha = \frac{\sin \alpha}{\tan \alpha} = \frac{5/13}{5/12} = \frac{5}{13} \times \frac{12}{5} = \frac{12}{13}

Check with the other identity, sin2+cos2=1\sin^2 + \cos^2 = 1:

(513)2+(1213)2=25+144169=169169=1  \left(\frac{5}{13}\right)^2 + \left(\frac{12}{13}\right)^2 = \frac{25 + 144}{169} = \frac{169}{169} = 1 \; \checkmark

Why each wrong option is wrong:

  • F, 513\frac{5}{13} and H, 512\frac{5}{12} — repeated a quantity already given.
  • G, 1312\frac{13}{12} — divided tangent by sine instead of sine by tangent, giving 1312\frac{13}{12}. A cosine can never exceed 1, so this is wrong on sight.

Takeaway: tan=sincos\tan = \frac{\sin}{\cos} and sin2+cos2=1\sin^2 + \cos^2 = 1. Between them they connect any two ratios. And sine and cosine are never above 1 — use that to reject an answer instantly.


Section 2 — Beyond Right Triangles

When a triangle has no right angle, SOHCAHTOA does not apply. Two formulas cover every such case, and the ACT expects both from memory.

The law of sines — use it when you have an angle and the side opposite it:

asinA=bsinB=csinC\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

The law of cosines — use it when you have two sides and the angle between them, or all three sides:

c2=a2+b22abcosCc^2 = a^2 + b^2 - 2ab\cos C

Notice that if C=90°C = 90° then cosC=0\cos C = 0 and the law of cosines collapses to c2=a2+b2c^2 = a^2 + b^2 — Pythagoras. It is the general version of a rule you already know.

Which to use: if you can pair a side with the angle opposite it, use sines. Otherwise use cosines.


Q5Hard

Topic: The law of cosines

A surveyor needs the distance across a lake. Two sides of a triangular plot measure 7 and 9 metres, and the angle between those two sides is 60°60°. What is the length of the third side?

A) 1616

B) 67\sqrt{67}

C) 130\sqrt{130}

D) 44

Show the worked solution

Answer: B

Explanation

Two sides and the angle between them: that is the law of cosines.

c2=a2+b22abcosCc^2 = a^2 + b^2 - 2ab\cos C

with a=7a = 7, b=9b = 9 and C=60°C = 60°. Since cos60°=12\cos 60° = \frac{1}{2}:

c2=49+812(7)(9)(12)=13063=67c^2 = 49 + 81 - 2(7)(9)\left(\tfrac{1}{2}\right) = 130 - 63 = 67

c=678.19c = \sqrt{67} \approx 8.19

Sanity check: the third side of a triangle must be less than the sum of the other two (1616) and more than their difference (22). 8.198.19 sits comfortably between. Correct.

Why each wrong option is wrong:

  • A, 1616 — added the two sides, which is the absolute maximum a third side could approach and never reach.
  • C, 130\sqrt{130} — used Pythagoras, which assumes a right angle. The angle here is 60°60°, so the 2abcosC-2ab\cos C term cannot be dropped.
  • D, 44 — subtracted the sides.

Takeaway: two sides and the angle between them means the law of cosines. The 2abcosC-2ab\cos C term is the correction that turns Pythagoras into a rule for any triangle.


Q6Hard

Topic: The law of sines

In triangle PQRPQR, angle PP measures 30°30°, angle QQ measures 45°45°, and the side qq opposite angle QQ measures 10 units. The side pp opposite angle PP is required. What is the length of pp?

F) 1515

G) 10210 \sqrt{2}

H) 55

J) 525 \sqrt{2}

Show the worked solution

Answer: J

Explanation

Angles paired with their opposite sides: use the law of sines.

psinP=qsinQ\frac{p}{\sin P} = \frac{q}{\sin Q}

psin30°=10sin45°\frac{p}{\sin 30°} = \frac{10}{\sin 45°}

p=10sin30°sin45°=10×1222=522=102=527.07p = \frac{10 \sin 30°}{\sin 45°} = \frac{10 \times \tfrac{1}{2}}{\tfrac{\sqrt{2}}{2}} = \frac{5}{\tfrac{\sqrt{2}}{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \approx 7.07

Sanity check: angle PP (30°30°) is smaller than angle QQ (45°45°), so the side opposite PP must be shorter than the side opposite QQ. 7.07<107.07 < 10. Correct.

Why each wrong option is wrong:

  • G, 10210 \sqrt{2} — inverted the ratio, giving about 14.1. That would make the side opposite the smaller angle the longer one, which the sanity check rejects.
  • H, 55 — halved 10 because 3030 is half of 6060. Sines are not proportional to angles.
  • F, 1515 — combined the angle numbers arithmetically.

Takeaway: the law of sines needs a side paired with its opposite angle. Then check the ordering: the bigger angle always faces the longer side.


Section 3 — Radians and Graphs

A radian is another unit for angle, like measuring in metres instead of feet. The one conversion to know:

π radians=180°\pi \text{ radians} = 180°

So to go from degrees to radians, multiply by π180\dfrac{\pi}{180}; the other way, multiply by 180π\dfrac{180}{\pi}. If the answer has a π\pi in it, it is in radians.

For graphs of y=asin(bx)y = a \sin(bx):

  • the amplitude is a|a| — the height from the middle to a peak
  • the period is 2πb\dfrac{2\pi}{b} — how far along before the wave repeats

Q7Medium

Topic: Converting degrees to radians

An angle of 40°40° has to be written in radians for a calculation that expects that unit. What is 40°40° in radians?

A) 40π40 \pi

B) 92π\frac{9}{2 \pi}

C) 2π9\frac{2 \pi}{9}

D) 29\frac{2}{9}

Show the worked solution

Answer: C

Explanation

Multiply by π180\dfrac{\pi}{180}:

40×π180=40π180=2π940 \times \frac{\pi}{180} = \frac{40\pi}{180} = \frac{2\pi}{9}

Cancel carefully: 40180\frac{40}{180} divides top and bottom by 20 to give 29\frac{2}{9}.

Sanity check: 2π9\frac{2\pi}{9} is about 0.700.70 radians. A right angle is π21.57\frac{\pi}{2} \approx 1.57 radians, and 40°40° is less than 90°90°, so an answer below 1.57 is right.

Why each wrong option is wrong:

  • A, 40π40 \pi — multiplied by π\pi without dividing by 180.
  • B, 92π\frac{9}{2 \pi} — used the conversion the other way round.
  • D, 29\frac{2}{9} — divided by 180 and lost the π\pi, leaving a number in no unit at all.

Takeaway: degrees to radians multiplies by π180\frac{\pi}{180}. The π\pi stays in the answer — if it has vanished, the conversion went the wrong way.


Q8Medium

Topic: The amplitude and period of a sine graph

A tide is modelled by y=3sin(2x)y = 3\sin(2x). What are the amplitude and the period of this graph?

F) (3, π)\left( 3, \ \pi\right)

G) (3, 2π)\left( 3, \ 2 \pi\right)

H) (2, 3)\left( 2, \ 3\right)

J) (6, π)\left( 6, \ \pi\right)

Show the worked solution

Answer: F

Explanation

For y=asin(bx)y = a\sin(bx):

amplitude=a=3\text{amplitude} = |a| = 3

period=2πb=2π2=π\text{period} = \frac{2\pi}{b} = \frac{2\pi}{2} = \pi

The amplitude is the coefficient outside, and it stretches the wave vertically. The number inside, multiplying xx, squeezes it horizontally — and it squeezes rather than stretches, so a larger bb gives a shorter period.

Why each wrong option is wrong:

  • H, (2, 3)\left( 2, \ 3\right) — swapped the roles of the two coefficients.
  • G, (3, 2π)\left( 3, \ 2 \pi\right) — gave 2π2\pi, the period of sinx\sin x with no coefficient inside.
  • J, (6, π)\left( 6, \ \pi\right) — doubled the amplitude, applying the inside coefficient outside.

Takeaway: outside sets the height, inside sets the width, and the period is 2πb\frac{2\pi}{b} — so the bigger the number inside, the faster the wave repeats.


Q9Medium

Topic: The Pythagorean identity

For an acute angle θ\theta, sinθ=35\sin \theta = \dfrac{3}{5}, and cosθ\cos \theta is required. What is cosθ\cos \theta?

A) 54\frac{5}{4}

B) 34\frac{3}{4}

C) 25\frac{2}{5}

D) 45\frac{4}{5}

Show the worked solution

Answer: D

Explanation

sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1

cos2θ=1(35)2=1925=1625\cos^2\theta = 1 - \left(\tfrac{3}{5}\right)^2 = 1 - \tfrac{9}{25} = \tfrac{16}{25}

cosθ=45\cos\theta = \tfrac{4}{5}

The angle is acute, so the cosine is positive. This is the 3-4-5 triangle in disguise.

Why each wrong option is wrong:

  • B, 34\frac{3}{4} — gave tanθ=34\tan\theta = \frac{3}{4}.
  • C, 25\frac{2}{5} — subtracted the ratios without squaring them.
  • A, 54\frac{5}{4} — inverted the answer, giving a cosine above 1, which is impossible.

Takeaway: sin2+cos2=1\sin^2 + \cos^2 = 1 links the two directly. Square, subtract, root — and expect an answer no greater than 1.


Q10Hard

Topic: A reciprocal trigonometric identity

For an acute angle β\beta, cosβ=817\cos \beta = \dfrac{8}{17} and sinβ=1517\sin \beta = \dfrac{15}{17}. What is tanβ\tan \beta?

F) 815\frac{8}{15}

G) 1517\frac{15}{17}

H) 158\frac{15}{8}

J) 2317\frac{23}{17}

Show the worked solution

Answer: H

Explanation

tanβ=sinβcosβ=15/178/17=1517×178=158\tan\beta = \frac{\sin\beta}{\cos\beta} = \frac{15/17}{8/17} = \frac{15}{17} \times \frac{17}{8} = \frac{15}{8}

The 17s cancel, which is why this is quicker than rebuilding the triangle.

Check with the other identity: (1517)2+(817)2=225+64289=1\left(\frac{15}{17}\right)^2 + \left(\frac{8}{17}\right)^2 = \frac{225 + 64}{289} = 1. ✓

Why each wrong option is wrong:

  • F, 815\frac{8}{15} — divided cosine by sine, giving cotβ\cot\beta.
  • G, 1517\frac{15}{17} — repeated the sine.
  • J, 2317\frac{23}{17} — added the two ratios.

Takeaway: tan=sincos\tan = \frac{\sin}{\cos}, in that order. Unlike sine and cosine, a tangent may exceed 1.


Q11Medium

Topic: A right triangle and the tangent ratio

A surveyor stands 40 metres from the base of a tower on level ground and measures the angle to its top as 35°35°. Which expression gives the tower's height in metres?

A) 40sin35°40\sin 35°

B) 40tan35°40\tan 35°

C) 40cos35°40\cos 35°

D) 40tan35°\dfrac{40}{\tan 35°}

Show the worked solution

Answer: B

Explanation

Opposite and adjacent means tangent:

tan35°=height40height=40tan35°\tan 35° = \frac{\text{height}}{40} \quad\Longrightarrow\quad \text{height} = 40 \tan 35°

Sanity check: 35°35° is less than 45°45°, so the tangent is below 1 and the tower is shorter than the 40 metres of ground. That rules out D immediately.

Why each wrong option is wrong:

  • A, 40sin35°40\sin 35° and C, 40cos35°40\cos 35° — sine and cosine both need the hypotenuse, which is not given here.
  • D, 40tan35°\dfrac{40}{\tan 35°} — dividing by the tangent gives about 57 metres, taller than the distance, which the 35°35° check rejects.

Takeaway: two legs and no hypotenuse means tangent. Below 45°45° the tangent is under 1; above it, over 1 — a free size check.


Q12Hard

Topic: The law of cosines to find an angle

All three sides of a triangle are known: 6, 7 and 9 units. What is the value of cosC\cos C for that angle?

F) 913\frac{9}{13}

G) 121- \frac{1}{21}

H) 121\frac{1}{21}

J) 00

Show the worked solution

Answer: H

Explanation

Start from the law of cosines with c=9c = 9 the side opposite CC:

c2=a2+b22abcosCc^2 = a^2 + b^2 - 2ab\cos C

81=36+4984cosC81 = 36 + 49 - 84\cos C

84cosC=8581=4cosC=484=12184\cos C = 85 - 81 = 4 \quad\Longrightarrow\quad \cos C = \frac{4}{84} = \frac{1}{21}

The cosine is small and positive, so CC is a little under 90°90° — which makes sense, since 92=819^2 = 81 is only just less than 62+72=856^2 + 7^2 = 85.

Why each wrong option is wrong:

  • F, 913\frac{9}{13} — invented a ratio of sides with no formula behind it.
  • G, 121- \frac{1}{21} — subtracted in the wrong order, giving a negative cosine and so an obtuse angle. The check above shows the angle is acute.
  • J, 00 — a cosine of 0 would mean exactly 90°90°, which needs 92=62+729^2 = 6^2 + 7^2. It is not: 81 against 85.

Takeaway: the law of cosines works in both directions. Rearranged it gives cosC=a2+b2c22ab\cos C = \frac{a^2 + b^2 - c^2}{2ab}, and the sign of the answer tells you whether the angle is acute or obtuse.


Q13Medium

Topic: Finding a leg with the cosine ratio

A ladder 10 metres long leans against a wall at 65°65° to the horizontal ground. Which expression gives that distance, in metres?

A) 10cos65°10\cos 65°

B) 10sin65°10\sin 65°

C) 10tan65°10\tan 65°

D) 10cos65°\dfrac{10}{\cos 65°}

Show the worked solution

Answer: A

Explanation

Sketch it. The ladder is the hypotenuse (10 m), the ground distance is adjacent to the 65°65° angle, and the wall is opposite.

Adjacent and hypotenuse means cosine:

cos65°=distance10distance=10cos65°\cos 65° = \frac{\text{distance}}{10} \quad\Longrightarrow\quad \text{distance} = 10\cos 65°

Since 65°65° is a steep angle, the cosine is small, so the foot of the ladder sits close to the wall — about 4.2 metres. That is what a steep ladder should do.

Why each wrong option is wrong:

  • B, 10sin65°10\sin 65° — sine gives the height reached up the wall, not the distance along the ground.
  • C, 10tan65°10\tan 65° — tangent needs the opposite and adjacent sides, and the hypotenuse is what is given.
  • D, 10cos65°\dfrac{10}{\cos 65°} — dividing gives about 24 metres, longer than the ladder. A leg cannot exceed the hypotenuse.

Takeaway: pick the ratio joining the side you have to the side you want. Then check the size against the hypotenuse.


Q14Hard

Topic: Finding an angle from two sides

A right triangle has a side of 7 opposite an angle θ\theta and a hypotenuse of 25. Which expression gives θ\theta?

F) sin(725)\sin\left(\frac{7}{25}\right)

G) sin1(257)\sin^{-1}\left(\frac{25}{7}\right)

H) sin1(725)\sin^{-1}\left(\frac{7}{25}\right)

J) cos1(725)\cos^{-1}\left(\frac{7}{25}\right)

Show the worked solution

Answer: H

Explanation

Opposite over hypotenuse is the sine:

sinθ=725\sin\theta = \frac{7}{25}

To get θ\theta itself, apply the inverse:

θ=sin1 ⁣(725)\theta = \sin^{-1}\!\left(\frac{7}{25}\right)

Why each wrong option is wrong:

  • F, sin(725)\sin\left(\frac{7}{25}\right) — no inverse, so it computes the sine of a number rather than the angle whose sine is that number.
  • G, sin1(257)\sin^{-1}\left(\frac{25}{7}\right)257\frac{25}{7} is greater than 1, and no angle has a sine above 1, so this expression has no value at all.
  • J, cos1(725)\cos^{-1}\left(\frac{7}{25}\right) — cosine uses the adjacent side, which here is 24, not 7.

Takeaway: to find a side use sin, cos or tan; to find an angle use their inverses. A sine or cosine above 1 is a signal you have inverted the ratio.


Q15Medium

Topic: A 45-45-90 triangle

A square gate is braced by a diagonal, forming a 45°45°45°45°90°90° triangle whose two equal sides each measure 6 units. What is the length of the hypotenuse?

A) 1212

B) 636 \sqrt{3}

C) 626 \sqrt{2}

D) 323 \sqrt{2}

Show the worked solution

Answer: C

Explanation

A 454545459090 triangle has sides in the ratio

1:1:21 : 1 : \sqrt{2}

with the 2\sqrt{2} on the hypotenuse. Both legs are 6, so everything scales by 6:

hypotenuse=628.49\text{hypotenuse} = 6\sqrt{2} \approx 8.49

Check with Pythagoras: 62+62=726^2 + 6^2 = 72, and 72=62\sqrt{72} = 6\sqrt{2}. ✓

Why each wrong option is wrong:

  • A, 1212 — doubled the leg, which is the 303060609090 rule.
  • B, 636 \sqrt{3} — used 3\sqrt{3}, also from the other special triangle.
  • D, 323 \sqrt{2} — gave a hypotenuse shorter than the legs, which is impossible.

Takeaway: 454545459090 is 1:1:21 : 1 : \sqrt{2}; 303060609090 is 1:3:21 : \sqrt{3} : 2. Keep the two apart by remembering which has two equal sides.


Q16Medium

Topic: Converting radians to degrees

An angle is given as 5π6\dfrac{5\pi}{6} radians and must be reported in degrees. What is the angle in degrees?

F) 3030

G) 150150

H) 300300

J) 216216

Show the worked solution

Answer: G

Explanation

Multiply by 180π\dfrac{180}{\pi}:

5π6×180π=5×1806=150°\frac{5\pi}{6} \times \frac{180}{\pi} = \frac{5 \times 180}{6} = 150°

The π\pi cancels, which is the sign the conversion is going the right way — an answer in degrees should have no π\pi left in it.

Sanity check: π\pi is 180°180°, so 5π6\frac{5\pi}{6} is a little less than that. 150 fits.

Why each wrong option is wrong:

  • F, 3030 — divided 180 by 6 and forgot the 5.
  • H, 300300 — used 360°360° instead of 180°180°.
  • J, 216216 — inverted the fraction.

Takeaway: radians to degrees multiplies by 180π\frac{180}{\pi}, and the π\pi must cancel. If a π\pi survives, the conversion went the wrong way.


Q17Hard

Topic: The period of a trigonometric graph

A tide is modelled by y=5cos(3x)y = 5\cos(3x). What is the period of this graph?

A) 2π3\frac{2 \pi}{3}

B) 2π2 \pi

C) 6π6 \pi

D) 33

Show the worked solution

Answer: A

Explanation

For y=acos(bx)y = a\cos(bx) the period is

2πb=2π3\frac{2\pi}{b} = \frac{2\pi}{3}

The 5 outside sets the amplitude and has no effect on the period at all.

Note the direction: a bigger number inside gives a shorter period, because it squeezes the wave. Three full cycles now fit where one used to.

Why each wrong option is wrong:

  • B, 2π2 \pi — the period of cosx\cos x, ignoring the coefficient.
  • C, 6π6 \pi — multiplied by 3 rather than dividing, stretching the wave when it should be squeezed.
  • D, 33 — gave the coefficient.

Takeaway: period is 2πb\frac{2\pi}{b}, where bb is the number multiplying xx. Outside sets height, inside sets width — and inside works in reverse.


Q18Medium

Topic: The amplitude and midline of a trigonometric graph

A temperature model is y=4sin(x)+20y = 4\sin(x) + 20. What are the maximum and minimum values of yy?

F) (24, 16)\left( 24, \ 16\right)

G) (20, 4)\left( 20, \ 4\right)

H) (4, 4)\left( 4, \ -4\right)

J) (80, 20)\left( 80, \ 20\right)

Show the worked solution

Answer: F

Explanation

sinx\sin x runs between 1-1 and 11. Multiplying by 4 makes it run between 4-4 and 44, and adding 20 lifts the whole thing:

maximum=20+4=24minimum=204=16\text{maximum} = 20 + 4 = 24 \qquad \text{minimum} = 20 - 4 = 16

The 20 is the midline — the level the wave oscillates about — and the 4 is the amplitude, the swing either side of it.

Why each wrong option is wrong:

  • G, (20, 4)\left( 20, \ 4\right) — gave the two coefficients rather than the values they produce.
  • H, (4, 4)\left( 4, \ -4\right) — ignored the +20+20, so the wave sits about zero.
  • J, (80, 20)\left( 80, \ 20\right) — multiplied the two numbers.

Takeaway: for y=asin(x)+cy = a\sin(x) + c, the graph runs from cac - a to c+ac + a. The constant is the midline; the coefficient is the swing.


Q19Medium

Topic: The Pythagorean identity rearranged

For an acute angle θ\theta, cosθ=725\cos\theta = \dfrac{7}{25}, and sinθ\sin\theta is required. What is sinθ\sin\theta?

A) 2425\frac{24}{25}

B) 1825\frac{18}{25}

C) 257\frac{25}{7}

D) 724\frac{7}{24}

Show the worked solution

Answer: A

Explanation

sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1

sin2θ=1(725)2=149625=576625\sin^2\theta = 1 - \left(\tfrac{7}{25}\right)^2 = 1 - \tfrac{49}{625} = \tfrac{576}{625}

sinθ=2425\sin\theta = \tfrac{24}{25}

The angle is acute, so the sine is positive. This is the 7-24-25 triangle.

Why each wrong option is wrong:

  • B, 1825\frac{18}{25} — subtracted the ratio without squaring it.
  • C, 257\frac{25}{7} — inverted the cosine, giving a value above 1, which no sine can take.
  • D, 724\frac{7}{24} — gave the tangent, 247\frac{24}{7} inverted.

Takeaway: square, subtract from 1, root. And use "no more than 1" to reject an answer before doing any arithmetic.


Q20Hard

Topic: Simplifying a trigonometric expression

An expression in a model reduces to sinθcosθ×cosθ\dfrac{\sin\theta}{\cos\theta} \times \cos\theta, where cosθ0\cos\theta \ne 0. Which of the following is the expression equal to?

F) sinθ\sin\theta

G) cosθ\cos\theta

H) tanθ\tan\theta

J) 11

Show the worked solution

Answer: F

Explanation

The fraction is the definition of the tangent:

sinθcosθ=tanθ\frac{\sin\theta}{\cos\theta} = \tan\theta

So the expression is tanθ×cosθ\tan\theta \times \cos\theta. Writing the tangent back out shows the cancellation:

sinθcosθ×cosθ=sinθ\frac{\sin\theta}{\cos\theta} \times \cos\theta = \sin\theta

Check at θ=30°\theta = 30°: tan30°0.577\tan 30° \approx 0.577 and cos30°0.866\cos 30° \approx 0.866, and their product is 0.5=sin30°0.5 = \sin 30°. ✓

Why each wrong option is wrong:

  • G, cosθ\cos\theta — cancelled the sine instead of the cosine.
  • H, tanθ\tan\theta — stopped at the tangent without multiplying.
  • J, 11 — cancelled everything, which would need the numerator and denominator to be identical.

Takeaway: tan=sincos\tan = \frac{\sin}{\cos} works in both directions. When an expression mixes the three ratios, rewrite the tangent as a fraction and look for what cancels.


Q21Hard

Topic: An identity used to simplify a sum

An expression reduces to sin2θ+cos2θ+3\sin^2\theta + \cos^2\theta + 3. What is the value of the expression?

A) 33

B) 55

C) 44

D) 66

Show the worked solution

Answer: C

Explanation

sin2θ+cos2θ=1for every θ\sin^2\theta + \cos^2\theta = 1 \qquad \text{for every } \theta

So the expression is

1+3=41 + 3 = 4

no matter what θ\theta is. That is what makes it an identity rather than an equation: it holds everywhere, so no value of θ\theta needs to be found.

Why each wrong option is wrong:

  • A, 33 — treated the two squared terms as zero.
  • B, 55 — counted each term as 1, giving 1+1+31 + 1 + 3. The identity says their sum is 1, not each of them.
  • D, 66 — doubled the constant.

Takeaway: spot sin2+cos2\sin^2 + \cos^2 and replace the pair with 1 immediately. It is the most useful identity on the paper.


Q22Medium

Topic: Finding the hypotenuse with a trigonometric ratio

A cable runs from the top of a 12-metre mast to a point on the ground, making an angle of 40°40° with the ground. Which expression gives the cable's length, in metres?

F) 12sin40°\dfrac{12}{\sin 40°}

G) 12sin40°12\sin 40°

H) 12cos40°\dfrac{12}{\cos 40°}

J) 12tan40°12\tan 40°

Show the worked solution

Answer: F

Explanation

Sketch it. The mast (12 m) is opposite the 40°40° angle, and the cable is the hypotenuse. Opposite over hypotenuse is the sine:

sin40°=12cablecable=12sin40°\sin 40° = \frac{12}{\text{cable}} \quad\Longrightarrow\quad \text{cable} = \frac{12}{\sin 40°}

Here the unknown is on the bottom, so the equation is rearranged by dividing rather than multiplying — the opposite of the usual case.

Sanity check: a hypotenuse is always the longest side, so the answer must exceed 12. Dividing by sin40°0.64\sin 40° \approx 0.64 gives about 18.7 metres. ✓

Why each wrong option is wrong:

  • G, 12sin40°12\sin 40° — multiplying gives about 7.7 metres, shorter than the mast it is attached to. Impossible for a hypotenuse.
  • H, 12cos40°\dfrac{12}{\cos 40°} — cosine uses the adjacent side, which is the ground distance and is not given.
  • J, 12tan40°12\tan 40° — tangent does not involve the hypotenuse at all.

Takeaway: when the unknown is the hypotenuse, it ends up on the bottom of the ratio, so you divide. Check the answer is the longest side.