IMO Practice Test — Trigonometry
6 Questions • 15 min • Olympiad level
15:00
Question 1 of 6
hard
Evaluate the following continuous product chain: tan 1° × tan 2° × tan 3° × ... × tan 88° × tan 89°.
0
√3
1
Not Defined
Explanation: Pair terms like tan 1° × tan 89° = tan 1° × cot 1° = 1. tan 45° in the middle is 1. All terms multiply to 1.
Question 2 of 6
hard
If sin theta + sin² theta = 1, find the value of the expression (cos² theta + cos⁴ theta).
0
1
2
1/2
Explanation: From the given info, sin theta = 1 - sin² theta = cos² theta. Substitute this into the target expression: sin theta + sin² theta = 1.
Question 3 of 6
hard
An observer looks up at the top of a tower at an angle of 45°. If the observer moves 10 meters further away from the tower base, the angle shifts to 30°. Find the height of the tower.
5(√3 - 1) m
5(√3 + 1) m
10(√3 + 1) m
10 / (√3 - 1) m
Explanation: Let height be h. Base 1 = h × cot 45° = h. Base 2 = h × cot 30° = h√3. h√3 - h = 10 -> h(√3 - 1) = 10 -> h = 10 / (√3 - 1).
Question 4 of 6
hard
Eliminate the angle parameter theta from the pair of parametric equations: x = a × cos theta and y = b × sin theta.
x² + y² = a²b²
x²/a² + y²/b² = 1
x²/b² + y²/a² = 1
x/a + y/b = 1
Explanation: Rearrange to x/a = cos theta and y/b = sin theta. Square and add them: (x/a)² + (y/b)² = cos² theta + sin² theta = 1.
Question 5 of 6
hard
If sec theta + tan theta = p, determine the value of sec theta in terms of the variable p.
(p² - 1) / 2p
(p² + 1) / 2p
(p² + 1) / p
(1 - p²) / 2p
Explanation: Since sec²x - tan²x = 1, sec x - tan x = 1/p. Add this to the given equation: 2 × sec x = p + 1/p = (p² + 1)/p -> sec x = (p² + 1)/2p.
Question 6 of 6
hard
Evaluate the numerical value of the expression: 2 × (sin⁶ theta + cos⁶ theta) - 3 × (sin⁴ theta + cos⁴ theta) + 1.
1
-1
0
2
Explanation: Let sin theta = 1 and cos theta = 0 (using angle 90°). Expression becomes: 2 × (1 + 0) - 3 × (1 + 0) + 1 = 2 - 3 + 1 = 0. True for all angles.