A Level Mathematics: Trigonometry, questions with answers

Using cos²θ = 1 − sin²θ: 8sin⁴θ + 6sin²θ − 5 = 0, i.e. (4sin²θ + 5)(2sin²θ − 1) = 0.
sin²θ = 1/2 (sin²θ = −5/4 is impossible), so sinθ = ±1/√2.
θ = π/4, 3π/4, 5π/4, 7π/4 (0.785, 2.36, 3.93, 5.50).

Replace cos² by 1 − sin² to get a quadratic in sin²θ, factorise, and reject the impossible root.

(a) sin²2x = 4sin²x cos²x, and cos2x = cos²x − sin²x. So sin²2x + 4cos²x cos2x = 4sin²x cos²x + 4cos²x(cos²x − sin²x) = 4sin²x cos²x + 4cos⁴x − 4cos²x sin²x = 4cos⁴x, as required.
(b) By part (a), the LHS ≡ 4cos⁴x + 5. Since cos⁴x ranges over [0, 1], the LHS ranges over [5, 9]. So the equation has no real solutions for k < 5 or k > 9.
(c) Using the identity from (a) with x = t/2 (so 2x = t): sin²t + 4cos²(t/2)cos t = 4cos⁴(t/2), so the integrand [sin²t + 4cos²(t/2)cost]/[2cos²(t/2)] = 2cos²(t/2). By the double-angle formula, 2cos²(t/2) = 1 + cos t. Integrating: ∫(1 + cos t)dt = t + sin t. …

The full model answer is in the app.

(a) 3√2 sin(x+45°) = 3√2(sinx cos45° + cosx sin45°) = 3(sinx+cosx). Adding cosx: total = 3sinx + 4cosx.
Write as Rcos(x−α) = R cosα cosx + R sinα sinx. So R cosα = 4, R sinα = 3, giving R = 5, α = arctan(3/4) = 36.9°.
So 3√2 sin(x+45°)+cosx = 5cos(x − 36.9°).
(b) With x = 3θ: 5cos(3θ − 36.9°) = −4 → cos(3θ − 36.9°) = −0.8 → 3θ − 36.9° = ±143.13° + 360°k.
For 0°<θ<180° (so 3θ ranges over (0°,540°)): 3θ−36.9°=143.13° → θ=60.0°; 3θ−36.9°=216.87° (i.e. −143.13°+360°) → θ=84.6°. No other solutions lie in the interval.

Combine the sine and cosine terms into a single R cos(x−α) form using the compound-angle expansion, then substitute x = 3θ and solve for θ within the given range.

(a) cos(θ+30°)cos(θ+60°) = (cosθcos30° − sinθsin30°)(cosθcos60° − sinθsin60°) = (½√3cosθ − ½sinθ)(½cosθ − ½√3sinθ)
= ¼√3cos²θ − ¾sinθcosθ − ¼sinθcosθ + ¼√3sin²θ = ¼√3(cos²θ+sin²θ) − sinθcosθ = ¼√3 − ½sin2θ, as required.
(b) Using the identity from (a) with θ = 2α: cos(2α+30°)cos(2α+60°) = ¼√3 − ½sin4α. So 5(¼√3 − ½sin4α) = 1 → (5√3)/4 − (5/2)sin4α = 1 → sin4α = ½√3 − 2/5 = 0.466 (3 s.f.).
Solving for 4α in (0°, 360°): 4α = 27.8° or 4α = 180° − 27.8° = 152.2°. So α = 6.9° or α = 38.1°.
(c) Using the identity from (a): with θ = −10°, cos20°cos50° = ¼√3 − ½sin(−20°) = ¼√3 + ½sin20°. With θ = 10°, cos40°cos70° = ¼√3 − ½sin20°. …

The full model answer is in the app.

(a) Multiply every term by sin x, using tan x = sin x / cos x:
sin x(4 sin x) + sin x(5/tan x) + sin x(2/sin x) = 0
4 sin²x + 5 cos x + 2 = 0
Using sin²x = 1 − cos²x:
4(1 − cos²x) + 5 cos x + 2 = 0
4 − 4cos²x + 5 cos x + 2 = 0
−4cos²x + 5 cos x + 6 = 0
Multiplying by −1:
4cos²x − 5cos x − 6 = 0
So a = 4, b = −5, c = −6.
(b) Factorise: 4cos²x − 5cos x − 6 = 0
(4cos x + 3)(cos x − 2) = 0
cos x = 2 (rejected, outside the range of cosine) or cos x = −3/4
cos x = −3/4 gives x = 138.6° (to 1 d.p.), and by symmetry (cos negative in second and third quadrants) the other solution in 0°–360° is x = 360° − 138.6° = 221.4°.
So x = 138.6° or x = 221.4°.

Multiply through by sin x and use tan x = sin x/cos x to get an equation in sin x only, then substitute sin²x = 1 − cos²x to reach a quadratic in cos x; factorise and solve, rejecting any value of cos x outside [−1, 1].

(a) Combine the LHS over a common denominator (sinθ − 1)(1 + sinθ):
Numerator = sin³θ(1 + sinθ) − sin²θ(sinθ − 1) = sin³θ + sin⁴θ − sin³θ + sin²θ = sin⁴θ + sin²θ = sin²θ(1 + sin²θ)
Denominator = (sinθ − 1)(1 + sinθ) = sin²θ − 1 = −cos²θ
So LHS = sin²θ(1 + sin²θ)/(−cos²θ) = −(sin²θ/cos²θ)(1 + sin²θ) = −tan²θ(1 + sin²θ) = RHS ✓
(b) By part (a), the LHS of the given equation equals −tan²θ(1 + sin²θ). So the equation becomes:
−tan²θ(1 + sin²θ) = tan²θ(1 − sin²θ)
−tan²θ(1 + sin²θ) − tan²θ(1 − sin²θ) = 0
−tan²θ[(1 + sin²θ) + (1 − sin²θ)] = 0
−2tan²θ = 0
tan²θ = 0 → tanθ = 0
Within 0 < θ < 2π, tanθ = 0 only at θ = π (θ = 0 and θ = 2π are excluded as the interval is open).
θ = π

Combine the two fractions on the left over a common denominator and simplify using sin²θ + cos²θ = 1 to reach the given identity, then reuse that identity to reduce the equation in part (b) to tan²θ = 0.

Using the compound angle formula, tan(x+45°) = (tanx + 1)/(1 - tanx). Since cotx = 1/tanx, the equation becomes (tanx+1)/(1-tanx) = 2/tanx. Cross-multiplying: tanx(tanx+1) = 2(1-tanx) → tan²x + tanx = 2 - 2tanx → tan²x + 3tanx - 2 = 0. By the quadratic formula: tanx = [-3 ± √(9+8)]/2 = (-3 ± √17)/2, giving tanx = 0.5616 or tanx = -3.5616. So x = 29.3° (first solution) or x = 180° - 74.28° = 105.7° (second solution, tangent negative in range 0°–180°).

Expand tan(x+45°) using the compound angle formula, rewrite cotx as 1/tanx, and solve the resulting quadratic in tanx.

(a) Expand both sides using the compound angle formulas:
cos(x − 30°) = cos x cos30° + sin x sin30° = (√3/2)cos x + (1/2)sin x
2 sin(x + 30°) = 2(sin x cos30° + cos x sin30°) = √3 sin x + cos x
Set equal:
(√3/2)cos x + (1/2)sin x = √3 sin x + cos x
Multiply both sides by 2:
√3 cos x + sin x = 2√3 sin x + 2 cos x
sin x − 2√3 sin x = 2 cos x − √3 cos x
sin x(1 − 2√3) = cos x(2 − √3)
Divide both sides by cos x(1 − 2√3):
tan x = (2 − √3)/(1 − 2√3) (as required)
(b) (2 − √3)/(1 − 2√3) = 0.267949/(−2.464102) = −0.108757
x = tan⁻¹(−0.108757) → basic (reference) angle = 6.2° (1 d.p.)
Since tan x is negative, x lies in the 2nd and 4th quadrants:
x = 180° − 6.2° = 173.8° …

The full model answer is in the app.