Pure 2 radians, trigonometry and parametric equations: mixed practice

The perimeter includes two radii.

Arc length 6(5π/6) = 5π cm; perimeter 12 + 5π cm; area ½(6²)(5π/6) = 15π cm².

The inner angle lies in [−π/6, 11π/6). Find all of its roots there.

2x − π/6 = π/6 or 5π/6. Hence x = π/6 or π/2. These are the complete roots in the stated interval.

cos u ≈ 1 − u²/2.

1 − cos 4x ≈ (4x)²/2 = 8x², so the limiting value is 8. The identity 1 − cos 4x = 2 sin²2x gives the same result using sin 2x / (2x) → 1.

Use sec²x = 1 + tan²x and retain x ≠ π/2.

tan²x − 3 tan x − 2 = 0, so tan x = (3 ± √17)/2. The positive root gives x = arctan((3 + √17)/2); the negative root gives x = π + arctan((3 − √17)/2). Both are in the required domain.

The numerator is sin²x.

The expression equals sin x only where sin x ≠ 0, meaning x ≠ nπ for integer n. The simplified sine expression alone has a wider domain.

Start from cos x = 0 and the range of sec x.

Asymptotes x = π/2 and 3π/2. Range y ≤ −3 or y ≥ 1. The boundary values occur at x = π and x = 0, respectively.

arcsin outputs [−π/2, π/2]; arccos outputs [0, π].

sin(7π/6) = −1/2, so arcsin gives −π/6. cos(7π/6) = −√3/2, so arccos gives 5π/6.

The angle lies in [0, π/2), so sine is non-negative.

Let θ = arccos(1/x). Then sin θ = √(1 − 1/x²) = √(x² − 1)/x since x ≥ 1. Dividing by cos θ = 1/x gives √(x² − 1). At x = 1 both sides equal 0.

Use π/12 = π/4 − π/6.

sin(π/4)cos(π/6) − cos(π/4)sin(π/6) = (√6 − √2)/4.

Factor cos x rather than dividing by it.

cos x(2 sin x − 1) = 0. Hence x = π/2, 3π/2, π/6 or 5π/6. The cosine-zero roots would be lost by division.

Subtract and use sin 3x − sin x = 2 cos 2x sin x.

sin x = 0 gives 0, π. cos 2x = 0 gives π/4, 3π/4, 5π/4, 7π/4. All six are distinct and valid.

Square cos²x = (1 + cos 2x)/2 and reduce cos²2x.

cos⁴x = (3 + 4 cos 2x + cos 4x)/8.

Add x and y; then use the parameter restriction.

x + y = 1, so y = 1 − x. The restrictions are 0 < x ≤ 1 and 0 ≤ y < 1. The point (0, 1) is approached but never reached.

t = ln x.

y = ln(1 + ln x), with 1 ≤ x ≤ e². The range is 0 ≤ y ≤ ln 3. The physical interval is stricter than the algebraic logarithm domain x > e⁻¹.

Cosine is non-positive throughout the interval.

x²/4 + y²/9 = 1, but the selected branch is y = −3√(1 − x²/4), −2 ≤ x ≤ 2. The whole ellipse would add an untraced branch.

Factor t³ − t² − t + 1.

(t − 1)²(t + 1) = 0, giving t = 1 or −1. Both produce (0, 0), so there is one distinct point, reached twice. A repeated parameter root alone is not enough to classify the local geometry of both branches.

Substitute y = 2x + c and require a repeated root.

t² − 2t + 2 − c = 0 has discriminant 4c − 4. Thus c = 1, t = 1, and the tangent y = 2x + 1 touches at (1, 3).

The radius is 3 and the angular sweep is π/2.

Endpoints are (1 + 3√3/2, −7/2) and (5/2, −2 + 3√3/2). Arc length is 3π/2 coordinate units. The interval traces a quarter-circle, not a whole circumference.

Ground landing uses y = 0. Complete the square for the peak.

y = 8 − 2(t − 1)². Ground roots are −1 and 3, so landing is at t = 3 s with horizontal range 15 m. Maximum ground height is 8 m at t = 1 s, a rise of 2 m above launch.

Use separate parameters for crossings and compare their arrival times.

t = 2s and t² = 4s give t = 0 or 2, with s = 0 or 1. The paths meet at (0, 0) and (2, 4). They share the start, but arrive at (2, 4) at different times, so there is no later collision under these rules.