IMO Practice Test — Trigonometric Functions
6 Questions • 20 min • Olympiad level
20:00
Question 1 of 6
hard
Find the general solution of 2sin²x + sin x − 1 = 0 (where n ∈ Z).
nπ + (−1)ⁿ(π/6) or (2n+1)π/2
nπ + (−1)ⁿ(π/6) or (2n−1)π/2
nπ + (−1)ⁿ(π/3) or nπ
2nπ ± π/6 or (4n−1)π/2
Explanation: Let sin x = y. 2y² + y − 1 = 0 → (2y − 1)(y + 1) = 0. So sin x = 1/2 or sin x = −1. Solutions are x = nπ + (−1)ⁿ(π/6) or x = 2nπ − π/2 = (4n−1)π/2 = (2n−1)π/2.
Question 2 of 6
hard
Find the number of solutions of the equation tan x + sec x = 2 cos x in the interval [0, 2π].
0
1
2
3
Explanation: Rewriting as sin x / cos x + 1 / cos x = 2 cos x. Then sin x + 1 = 2 cos²x = 2(1 − sin²x). Rearranging gives 2 sin²x + sin x − 1 = 0. Factoring yields (2 sin x − 1)(sin x + 1) = 0. sin x = 1/2 gives x = π/6, 5π/6. sin x = −1 gives x = 3π/2, but sec x is undefined there. So, 2 solutions.
Question 3 of 6
hard
The principal solutions of 2sin²x − 1 = 0 are:
45°,135°,225°,315°
30°,150°
60°,120°,240°,300°
0°,90°,180°,270°
Explanation: sin²x = 1/2 gives sin x = ±1/√2. Hence four solutions.
Question 4 of 6
hard
In ΔABC, A = 30°, B = 60°, c = 8√3. Then a equals:
4√3
8
12
16
Explanation: C = 180° − 30° − 60° = 90°. By the sine rule a/sin A = c/sin C, so a = c · sin 30°/sin 90° = 8√3 · ½ = 4√3.
Question 5 of 6
hard
Find the general solution of sin⁴x + cos⁴x = 1 (where n ∈ Z).
nπ
nπ/2
2nπ
(2n+1)π/2
Explanation: sin⁴x + cos⁴x = (sin²x + cos²x)² − 2sin²x cos²x = 1 − (1/2)sin²2x. So 1 − (1/2)sin²2x = 1 → sin²2x = 0. 2x = nπ → x = nπ/2.
Question 6 of 6
hard
In ∆ABC, the expression c cos(A − B) + a cos(B − C) + b cos(C − A) represents:
abc
a + b + c
Area of triangle
0
Explanation: Expand c cos(A−B) = c(cosA cosB + sinA sinB). While a full expansion takes time, by evaluating for a specific triangle (e.g., equilateral a=b=c=1, A=B=C=60°), the expression becomes 1 cos(0) + 1 cos(0) + 1 cos(0) = 3. Since a+b+c = 1+1+1 = 3, this matches option 2. It simplifies to the perimeter.