
![]()
Note: The area of the sector in the diagram
is given by: ![]()
Questions:2003 II 4a, 2001 II 4a, 2000 II 4a, 1994 II 5a.
Harder Questions:
2005 II 4c, 2004 II 4c, 2003 II 4c, 2002 II 4c(iii),
RIGHT-ANGLED TRIANGLES
SOHCAHTOA
and Pythagoras’ Theorem.
Questions:
2002 II 4c(i), (ii), 1994 II
5c.
The unit
circle, x = (cosA, sinA) and ASTC
AREA FORMULA
![]()
Questions: 2005 II 5a, 2002 II 5a, 2001 II 5b (i), 1999 II 5a,
Harder Questions: 1998 II 4c(i),
THE GOLDEN RULE
![]()
This equation is used constantly to change
to
and vice versa.
It can be proved using a right-angled triangle with a hypotenuse of length one and Pythagoras’ theorem.
Questions: 1998 II 5b, 1996 II 5a
GIVEN
SINA FIND COSA ETC.
EXAMPLE given
find
. (NB check if A is
acute or obtuse!)
Draw a right-angled triangle with opposite = 2 and hypotenuse = 7. Use
Pythagoras’ theorem to find the adjacent. Use SOHCAHTOA to write down the ratio
for
.
Questions 2004 II
4a(i), 1998 II 5a
SIN COS AND TAN OF
SPECIAL ANGLES.
(a) ![]()

Use the triangle and SOHCAHTOA.
(b) ![]()

Use the triangle and SOHCAHTOA.
(c) ![]()
e.g. to find
in surd form.
![]()
Note: page 9 of the tables is invaluable here.
(d)
etc.
Use the double angle formula, e.g.
(page 9, Tables).
Questions: 2003 II 5a, 2002 II 5b (ii), 1999 II 4a, 1999 II 4c (i) (ii),
CHANGING SUMS TO PRODUCTS AND VICE VERSA.
Questions: 2001 II 5c, 1999 II 4a, 1999 II 5c,
PROOFS
Prove that
2005 II 4b(i)
Prove
2003 II 5c(i)
Extension: Prove ![]()
Prove that
2002 II 5b(i), 1999
II 4c
Extension: Hence how
that ![]()
Extension: Find, in the
form ![]()
(i) tan 75o (ii)
tan 15o.
Prove that
2004 II 5a
Extension: Show that
simplifies to a
constant.
![]()
Prove that
2001 II 4c
Extension : Show that
Derive the formula
2000 II 4c
Extension : Show that ![]()
Show that
2000 II 5b
Extension : Find
values of the integers l and k so that
![]()
Cosine Rule:, 1997 II 5b,
ONE TERM
TRIGONOMETRIC EQUATIONS
e.g
where ![]()
Find an acute angle A that solves the equation
Then check
and
Only give those
answers that satisfy the original equation.
Questions: 2002 II 4a, 1998
II 4a, 1997 II 4a
TRIGONOMETRIC
EQUATIONS.
The basic idea of
equations, finding a product and solving FIRST FACTOR = 0 and SECOND FACTOR = 0 will work here.
Look out for chances to use the double
angle formulae, e.g. if one angle is x
and the other is 2x or
. Quite often these
formulae will change a trigonometric equation into a quadratic.
If all the angles are the same, e.g. x,
it may be possible to use the golden rule
to express
in terms of
.
Let
and factorise the
resulting quadratic. Solve for y and
hence x.
Questions:
2005 II 4b(ii)
2004 II 4b(ii)
2003 II 4b
2002 II 4b
2001 II 4b
(Change sum to
product.) 2001 II 5c and ![]()
2000 II 4b
1999 II 4b
1999 II 5c
1998 II 4b
1997 II 5b
1996 II 5c 
1995 II 5b
1994 II 5b.
SOLVING TRIANGLES
WITH SINE AND COSINE RULES.
Questions: 2005 II 5b(i), 2001 II 5b(ii), 2000 II 5c, 1999 II 5b, 1998 II 4c(ii), 1997 II 4b, 1996 II 4a, 1996 II 5b, 1994 II 4b.
THREE DIMENSIONAL
PROBLEMS
Questions: 2005
II 5c, 2003 II 5b 2002 II 5c, 1996 II 4c, 1995 II 4c, 1994 II 4c.
TRIGONOMETRIC
LIMITS
![]()
NB: The angle must be in radians for this formula.
Questions: 2005 II 4a, 2001 II 5a, 2000 II 5a, 1997 II 5a, 1995 II 4a