In a precision laser alignment system, the deviation ratio DDD for a small angle of incidence θ\thetaθ (measured in radians, where θ≠0\theta \neq 0θ=0) is modeled by the function:
D(θ)=3θsin(4θ)1−cos(5θ) D(\theta) = \frac{3\theta \sin(4\theta)}{1 - \cos(5\theta)} D(θ)=1−cos(5θ)3θsin(4θ)Using small angle approximations, show that for small values of θ\thetaθ, D(θ)≈KD(\theta) \approx KD(θ)≈K where KKK is a constant to be determined.
317 exam-style questions on OCR (MEI) A Level Maths 1.7 Trigonometry, covering 1.7.1 Solve right-angled triangles, 1.7.2 Definitions of sin, cos and tan for any angle, 1.7.3 Graphs of sin, cos and tan, 1.7.4 Exact values of trig functions (degrees), 1.7.5 Area of a triangle, 1.7.6 Sine and cosine rules, 1.7.7 Identity tan = sin/cos, 1.7.8 Identity sin^2 + cos^2 = 1, 1.7.9 Solve simple trigonometric equations, 1.7.10 Exact values of trig functions (radians) (A-level only), 1.7.11 Inverse trigonometric functions (A-level only), 1.7.12 Radians and degree conversion (A-level only), 1.7.13 Arc length and area of a sector (A-level only), 1.7.14 Small angle approximations (A-level only), 1.7.15 Sec, cosec and cot functions (A-level only), 1.7.16 Graphs of reciprocal trig functions (A-level only), 1.7.17 Pythagorean identities for sec and cosec (A-level only), 1.7.18 Compound angle formulae (A-level only), 1.7.19 Double angle identities (A-level only), 1.7.20 Expressions for a cos θ ± b sin θ (A-level only), 1.7.21 Use identities to solve equations (A-level only), 1.7.22 Proofs involving trigonometric functions (A-level only), 1.7.23 Trig to solve problems in context (A-level only), and 1.7 Trigonometry. Each one has a worked solution and a mark scheme showing where the marks go.