Download Extension 1 Results Summary

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