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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