Printable · GCSE Foundation · ages 14-16
Exact trigonometric values worksheet — GCSE Foundation
Fifteen questions on "exact trigonometric values" — DfE statement G21. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Exact trigonometric values worksheet — GCSE Foundation
MathsUKwww.geekhero.co.uk
- 1.Write down the exact value of tan 45°.
- 2.Write down the exact value of sin 0°.
- 3.A kite string makes an angle of 30° with the ground. The kite is flying at a height of 6 m directly above a point on the ground. Using the exact value of sin 30°, work out the exact length of the kite string.
- 4.Write down the exact value of cos 45°.
- 5.Three of these are rational numbers and one is irrational. Write down the one that is irrational.
- 6.Write down the exact value of cos 60°.
- 7.Work out the exact value of sin 30° + cos 60°.
- 8.Work out the exact value of tan 30° + tan 30°.
- 9.A cable supporting a flagpole is anchored to the ground 5 m from the base of the pole. The cable makes an angle of 60° with the ground. Using the exact value of tan 60°, work out the exact height of the flagpole.
- 10.Work out the exact value of cos 0° − sin 30°.
- 11.A tent's sloping side makes an angle of 45° with the ground. The sloping side is 3√2 m long. Using the exact value of sin 45°, work out the exact height of the tent.
- 12.Work out the exact value of (cos 45°)² + (sin 45°)².
- 13.Here are four trigonometric ratios of special angles. Write down the one that does not have a value.
- 14.A right-angled triangle has a hypotenuse of 14 cm. One of the other angles is 30°. Work out the exact length of the side opposite the 30° angle.
- 15.Each of these has an exact value. Write down the one whose value is the greatest.
Answer key
- (d) 1 — At 45° the right-angled isosceles triangle has both shorter sides equal, so tan 45° = opposite ÷ adjacent = 1 ÷ 1 = 1. 0 is the value of tan 0° (mixing up the two angles). √3 is the value of tan 60°. √3/3 (equivalent to 1/√3) is the value of tan 30°.
- (b) 0 — At 0°, the opposite side of the right-angled triangle has shrunk to zero length, so sin 0° = 0. 1 is the value of sin 90° (mixing up the two angles). 1/2 is the value of sin 30°. 'Undefined' is what happens for tan 90°, not sin 0° — sin 0° has a perfectly good exact value.
- (b) 12 m — sin 30° = opposite ÷ hypotenuse, where the opposite side is the height (6 m) and the hypotenuse is the string. So string = height ÷ sin 30° = 6 ÷ (1/2) = 12 m. The distractor 3 m comes from multiplying by sin 30° instead of dividing (6 × 1/2 = 3). The distractor 6√3 m comes from using tan 30° = 1/√3 instead of sin 30° (6 ÷ (1/√3) = 6√3). The distractor 4√3 m comes from using cos 30° = √3/2 instead of sin 30° (6 ÷ (√3/2) = 12/√3 = 4√3).
- (b) √2/2 — cos 45° comes from a right-angled isosceles triangle with both shorter sides 1 and hypotenuse √2, giving cos 45° = 1/√2 = √2/2. 1/2 is the value of cos 60° (mixing up the two angles). √2 is the hypotenuse length itself, not divided by it (forgetting to divide by the hypotenuse). √3/2 is the value of cos 30° (using the wrong special triangle).
- (c) cos 45° — cos 45° = √2/2, and √2 cannot be written as an exact fraction, so this value is irrational. tan 0° = 0, sin 90° = 1 and sin 30° = 1/2 are all rational, since each can be written as an exact whole number or fraction.
- (c) 1/2 — Method: cos 60° comes from half of an equilateral triangle. Cut an equilateral triangle of side 2 down its middle: each half is right-angled, with hypotenuse 2, base 1 and height √3. Working: cosine is the adjacent side over the hypotenuse, the side alongside the 60° angle is the base of length 1, and the hypotenuse is 2, so cos 60° = 1 ÷ 2. Answer: cos 60° = 1/2. A candidate who assumes the cosine falls steadily from cos 0° = 1 to cos 90° = 0 puts 60° two thirds of the way along and writes 1/3; the cosine does not change at a constant rate. Reading the start of the cosine row gives cos 0° = 1, and reading its end gives cos 90° = 0.
- (c) 1 — sin 30° = 1/2 and cos 60° = 1/2, so sin 30° + cos 60° = 1/2 + 1/2 = 1. 0 comes from subtracting the two values instead of adding them (1/2 − 1/2). 1/2 comes from writing down only one of the two exact values and forgetting to add the other. √3 comes from swapping the two angles and working out sin 60° + cos 30° = √3/2 + √3/2 = √3.
- (a) 2√3/3 — tan 30° = √3/3, so tan 30° + tan 30° = 2 × √3/3 = 2√3/3. √3 comes from wrongly treating tan 30° + tan 30° as tan(30° + 30°) = tan 60° = √3 — adding angles is not the same as adding ratios. √3/3 comes from forgetting to double the value and just writing down tan 30° on its own. 2√3 comes from doubling the numerator of √3/3 but forgetting to keep the denominator of 3.
- (c) 5√3 m — The cable, the pole and the ground form a right-angled triangle: the ground distance (5 m) is adjacent to the 60° angle, and the height of the pole is opposite it, so height = 5 × tan 60° = 5 × √3 = 5√3 m. 5√3/2 m comes from using sin 60° = √3/2 instead of tan 60°. 5/√3 m comes from using tan 30° = 1/√3, the reciprocal-angle value, instead of tan 60°. 10√3 m comes from doubling the correct height by mistake.
- (a) 1/2 — cos 0° = 1 and sin 30° = 1/2, so cos 0° − sin 30° = 1 − 1/2 = 1/2. 1 comes from writing down cos 0° alone and forgetting to subtract sin 30°. 3/2 comes from adding the two values instead of subtracting. −1/2 comes from working out sin 30° − cos 0°, the two terms the wrong way round.
- (a) 3 m — Height = sloping length × sin 45° = 3√2 × √2/2 = (3 × 2)/2 = 3 m, since √2 × √2 = 2. 3√2 m comes from forgetting to multiply by sin 45° at all. 3√2/2 m comes from using sin 30° = 1/2 instead of sin 45° = √2/2. 6 m comes from using √2 instead of √2/2 for sin 45°, dropping the denominator of the exact value: 3√2 × √2 = 6.
- (c) 1 — cos 45° = √2/2 and sin 45° = √2/2. Squaring each gives (√2/2)² = 2/4 = 1/2, so (cos 45°)² + (sin 45°)² = 1/2 + 1/2 = 1. The distractor √2 comes from adding cos 45° + sin 45° directly without squaring first (√2/2 + √2/2 = √2). The distractor 2 comes from squaring the top of the fraction, (√2)² = 2, but then dividing by 2 instead of 4 for each term, giving 1 + 1 = 2. The distractor 1/2 comes from squaring only cos 45° and forgetting to add the sin 45° term.
- (b) tan 90° — Method: write each ratio as one side of a right-angled triangle divided by another, and look for the one whose bottom line can be zero. Working: sine and cosine are both a side divided by the hypotenuse, and a hypotenuse is never zero, so sin 90° = 1 and cos 90° = 0 are both perfectly good values — a value of zero is still a value. Tangent is the opposite side divided by the adjacent side. As the angle opens out towards 90° the adjacent side shrinks to nothing, and a division by zero has no result, while at the other end of the row the adjacent side is the long one and tan 0° = 0. Answer: tan 90°.
- (b) 7 cm — The side opposite the 30° angle is found using sin 30° = opposite/hypotenuse, so opposite = 14 × sin 30° = 14 × 1/2 = 7 cm. 7√3 cm comes from using cos 30° = √3/2 instead of sin 30° (mixing up the opposite and adjacent sides). 14/√3 cm comes from using tan 30° = 1/√3 instead of sin 30°. 28 cm comes from dividing 14 by sin 30° instead of multiplying by it.
- (c) tan 45° — Method: replace each ratio by its exact value, then compare. Working: a right-angled triangle with a 45° angle is isosceles, so its opposite and adjacent sides are equal and the tangent of 45° is exactly 1. The others are cos 30° = √3/2, about 0.87; sin 45° = √2/2, about 0.71; and cos 60° = 1/2. Answer: tan 45°, the only one of the four that reaches 1. Reading √3/2 as though it were √3, about 1.73, makes cos 30° look the largest, but the division by 2 is part of the value. Ranking by the size of the angle also fails here, because the cosine of an angle falls as the angle grows.
Build your own mix at the worksheet builder.