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EXTENSION 1 REVISION OF FORMULAE AND RESULTS Co-ordinate Geometry When solving asinθ + bcosθ = c we can solve by writing in the form Rsin(θ + α) = c where: Dividing an interval in the ratio m:n mx2 + nx1 my2 + ny1 , m+n m+n Trigonometric Ratios Sum and Difference Results sin(A + B) = sinA cosB + cosA sinB sin(A – B) = sinA cosB – cosA sinB cos(A + B) = cosA cosB – sinA sinB cos(A – B) = cosA cosB + sinA sinB tan(A + B) = tan(A – B) = The parametric equations for the parabola x2 = 4ay are x = 2at and y = at2 All other formulae in this subject are not to be committed to memory but students must know how they are derived. Polynomials 1– tanAtanB P(x) = pnx2 + pn-1xn-1 + ... p2x2 + p1x + p0 1+ tanAtanB Double Angle Results 2tanA tan2A = 1 – tan2A θ The ‘t’ Formulae where t = tan2 a tanA – tanB b Parameters tanA + tanB sin2A = 2sinA cosA cos2A = cos2A – sin2A cos2A = 1 – 2sin2A cos2A = 2cos2A – 1 R = a2 + b 2 and tan α = Acute angle between two lines (or tangents) m1 − m2 tanθ = 1 + m1 m2 Subsidiary Angle Method (Rsin(θ + α)) A real polynomial is in the form: p1, p2, p3, ....., pn are coefficients and are real numbers, usually integers. The degree of the polynomial is the highest power of x with non-zero coefficient. A polynomial of degree n has at most n real roots but may have less. The result of a long division can be written in the form P(x) = A(x) . Q(x) + R(x) The remainder theorem states that when P(x) is divided by (x – a) the remainder is P(a). The factor theorem states that if x = a is a factor of P(x) then P(a) = 0. If , , , , ... are the roots of a polynomial then 2t sin x = 1+ t2 cos x = 1 − t2 1+ t2 2t tan x = 1 − t2 b c d = − a, = a, = − a, = e a Numerical Estimation of the Roots of an Equation Inverse Trigonometric Functions Halving the Interval Method Newton’s Method If x = x0 is an approximation to a root of P(x) = 0 then x1 = x0 – P(x0 ) P' (x0 ) y = sin-1x y = cos-1x is generally a better approximation. y = tan-1x Be familiar with the conditions under which this method fails. Mathematical Induction Step 1: Prove result true for n = 1 (It is sometimes necessary to have a different first step.) Step 2: Assume it is true for n = k and then prove true for n = k + 1 Step 3: Conclusion as given in class Integration 1 1 cos = 2 1 + cos2θ −2 ≤ y ≤ − 2 Domain: –1 x 1 Range: 0≤ y≤ Domain: all real x Range: −2 ≤ y ≤ − 2 π π π π Properties: General Solutions of Trigonometric Equations: Derivatives: d 2 Range: if sin = q, then = n + (–1)n sin-1q if cos = q, then = 2n cos-1q if tan = q, then = n + tan-1q sin2 θ dθ and cos2 θ dθ can be solved using the substitutions: sin2 = 2 1 − cos2θ –1 x 1 sin-1(-x) = -sin-1x cos-1(-x) = – cos-1x tan-1(-x) = –tan-1x π sin-1x + cos-1x = 2 sin(sin-1x) = x cos(cos-1x) = x tan(tan-1x) = x Domain: dx d dx d Integration by first making a substitution. dx sin-1 x = cos-1 x = 1 1− x2 −1 1− x2 1 tan-1 x = 1 + x2 Table of Standard Integrals as provided in HSC Integrals: 1 𝑑𝑥 = sin-1 a2 – x2 1 1 𝑑𝑥 = tan-1 2 2 a +x a x x = −cos-1 a a x a

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