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Queen Margaret Academy Higher Maths Traffic Light Summary Unit 3 A.Allen Queen Margaret Academy Higher Mathematics o o Be able to solve trig equations, e.g. a) Cos2x = b) 3sin2x + 7sinx – 6 = 0 c) cos22x = 1 Be able to use the addition formulae to solve trig equations. Cos(A + B) = CosACosB – SinASinB Cos(A - B) = CosACosB + SinASinB Sin(A + B) = SinACosA + CosASinB Sin(A - B) = SinACosA - CosASinB e.g Find the exact value of cos(A+ B): Cos(A + B) = CosACosB – SinASinB = √ = = x √ √ - x Find the missing sides of the triangles using Pythagoras. 1 Sin A = Cos A = Sin B = √ A √ Cos B = √ - √ √ 4 √ B 4 o Be able to prove complex trig identities using addition formulae. e.g. Int 2/Nat 5 Trig Prove that cos(90 + x)o = -sinx: Identities you must LHS: Cos(90 + x)o = cos90ocosxo – sin90osinxo o o remember: = 0 x cosx – 1 x sinx o = -sinx Therefore LHS = RHS Prove that LHS: = tanα + tanβ: = + = = o A.Allen + = tanα + tanβ Therefore LHS = RHS I can use the double angle formulae and decide which version is best to use. Sin2A = 2SinACosA Cos2A = Cos2A – Sin2A = 2Cos2A – 1 = 1 – 2Sin2A Green Red Learning Statement Amber Unit 3: Trig Formulae o I can solve trigonometric equations using double angle formulae. o Tackle past paper questions on trig formulae. A.Allen Queen Margaret Academy Higher Mathematics o Use the remainder theorem to divide a polynomial f(x) by (x – h) e.g. Show that(x – 3) is a factor of x3 + 2x2 – 14x – 3 3 1 1 2 -14 -3 3 15 3 5 1 0 Remainder is zero therefore (x – 3) is a factor. o Be able to state the answer in the form f(x) = (ax – b)Q(x) + R o Be able to fully factorise a polynomial using the remainder theorem e.g. Show that (x – 3) is a factor of 2x3 + 5x2 – 28x – 15 and factorise fully: 3 2 5 -28 -15 6 2 11 33 15 5 0 Remainder is zero therefore (x – 3) is a factor. (x – 3)(2x2 + 11x + 5) (x – 3)(x + 5)(2x + 1) o Be able to solve any polynomial equation (find the roots): e.g. Show that (x – 3) is a factor of 2x3 + 5x2 – 28x – 15 and factorise fully: 3 2 5 -28 -15 6 2 11 33 15 5 0 (x – 3)(2x2 + 11x + 5) (x – 3)(x + 5)(2x + 1) x = 3, x = -5, x = A.Allen Remainder is zero therefore (x – 3) is a factor. Green Amber Learning Statement Red Unit 3: Polynomials o Find a polynomials unknown coefficients using the factor theorem, e.g. If (x + 3) is a factor of 2x4 + 6x3 + px2 + 4x – 15, find the value of p. -3 2 6 p -6 0 2 0 4 -3p -15 9p - 12 p -3p + 4 9p - 27 Since (x + 3) is a factor the remainder must be zero, therefore 9p – 27 = 0 9p = 27 P=3 Remember if you have two unknown coefficients then you must do the remainder must do the remainder theorem with both factors then solve the equations left in both remainders simultaneously. o Be able to establish the equation of a polynomial from its graph or when given its roots and another point. e.g. Find an expression for f(x) from the graph opposite: y = k(x – a)(x – b)(x – c) y = k(x + 2)(x – 1)(x – 4) y – intercept (0, 6) 6 = k(0 + 2)(0 – 1)(0 – 4) 6 = 8k K= Y = (x + 2)(x – 1)(x – 4) o o A.Allen Be able to sketch the graph of a polynomial by: - Finding the coordinates of the y-intercept. - Finding the coordinates of the x-intercepts (the roots) - Differentiating to find the stationary points and their nature. Finding the behavior of the curve for large positive and negative x. Tackle past paper questions involving polynomials. Queen Margaret Academy Higher Mathematics o Know that a quadratic function y = ax2 + bx + c = 0: - has a max turning point if a > 0 - has a min turning point a < 0 - that it has a y-intercept (0, c) - can find the roots of a function by factorising and solving ax 2 + bx + c = 0 o Can write the equations y= ax2 + bx + c in the form y = a(x + p)2 + q and be able to state the axis of symmetry is x = -p, the turning point is at (-p, q) o Can sketch quadratic functions using the information gained from the above skills. o Solve quadratic equations by: - Graphically - Factorising to get the roots - Completing the square, y = a(x + p)2 + q - Using the quadratic formula You will have to learn this: √ o Know that the discriminant of ax2 + bx + c is b2 – 4ac o Use the discriminant to determine the nature of the roots of a quadratic: b2 – 4ac > 0, has two real unequal roots. b2 – 4ac = 0 has two real equal roots. b2 – 4ac < 0 has non real roots. o Use the discriminate to find an unknown constant in a quadratic equations when given the nature of its roots: e.g. Find p given that 2x2 + 4x + p = 0 has real roots. a = 2, b = 4 and c = p A.Allen b2 – 4ac ≥ 0 (4)2 – 4 x 2 x p ≥ 0 16 – 8p ≥ 0 Green Amber Learning Statement Red Unit 3: Qudaratics o o 16 ≥ 8p 2≥p P≤2 Be able to determine whether a line cuts, touches or does not meet a curve by substituting the equation of a line into the curve. e.g. y = x2 + 3x + 2 and y = x + 2 meet where? x2 + 3x + 2 = x + 2 x2 + 2x = 0 a = 1, b = 2, c = 0 b2 – 4ac (2)2 – 4 x 1 x 0 = 4 therefore the two roots are real and unequal, two distinct oints of intersection between curve and line. Know the conditions for tangency. b2 – 4ac > 0, the line intersects the curve at two places. b2 – 4ac = 0, the line touches the curve at one place, therefore a tangent to the curve. b2 – 4ac < 0, the line does not intersect the curve. o A.Allen Tackle past paper questions on quadratics. Queen Margaret Academy Higher Mathematics o o Be able to express acosθ + bsinθ in the form kcos(x ± α) or ksin(x ± α), where k is the amplitude and α the phase angle. Be able to apply the wave function formula to multiple angles, e.g. kcos(3x ± α) or ksin(3x ± α), e.g. 2cos2x – 3sin2x = kcos(2x – α) = kcosαcos2x + ksinαsin2x Kcosα = 2 S A Ksinα = -3 = tanα = α = 304˚ k=√ =√ T C o Can state the max and min values and when they occur, of a of the acosx + b sinx by expressing it as a single trig function, e.g. 4cosx + 3sinx = 5cos(x – 36.9), where 0 ≤ x ≤ 360. Max Value is 5 Max value occurs when……………5cos(x – 36.9) = 5 cos(x – 36.9) = 1 (x – 36.9) = 0, 360 X = 36.9, 396.9 Min Value is -5 Min Value occurs when……………..5cos(x – 36.9) = -5 cos(x – 36.9) = -1 (x – 36.9) = 270 X = 306.9 o Be able to solve equations involving acosx + bsinx by using the wave function formula, e.g. √ cosθ + sinθ = √ , for 0 ≤ θ ≤ 2π √ cosθ + sinθ = kcos(θ – α) = kcosθcosα + ksinθsinα K=√ √ Kcosα = √ Ksinα = 1 = tanα = √ α= rad So, √ cosθ + sinθ = 2cos(θ - ) 2cos(θ - ) = √ Cos(θ - ) = √ (θ - ) = 45, 315 θ= 75, 345 θ= , radians o A.Allen Be able to tackle past paper questions on the wave function. =2 Green Red Learning Statement Amber Unit 3: The Wave Function Queen Margaret Academy Higher Mathematics o Be able to re-arrange an algebraic expression into a form to allow you to integrate, eg. o Know that ∫ o Be able to evaluate Definite integrals, eg. ∫ dx = + C, where n ≠ -1 and C is the constant of integration. =[ ] =[ =[ ] ]-[ [ ] ] Note: you must show the substitution of zero = o Be able to write a shaded area as a definite integral. ∫ o Know how to work out the shaded area above and below the x-axis. Remember the equation of the x-axis is y = 0 Above x-axis: ∫ o Below x – axis ∫ Know that the limits of the integral are the x-coordinates where the two curves or line and curve meet. o Be able to find the limits if they do not give you the x-coordinates you can find them by setting the two functions equal to each other, re-arrange, factorise and solve for x. o Know how to work out the shaded area between two curves, or a curve and a straight line. ∫ o Solve Differential equations in the form. e.g. = 4x + 1, when x = 1 and y = 2 y=∫ y= y=2 2=2 +1+c 2=3+c C = -1 Therefore y = 2 + x - 1 o Be able to tackle past paper questions on Integration A.Allen Green Red Learning Statement Amber Unit 3: Integration A.Allen