STEP Support Programme STEP 2 Trigonometry Questions: Solutions - Maths
step
STEP Support Programme STEP 2 Trigonometry Questions: Solutions
1 We have
cos 3x = cos 2x cos x - sin 2x sin x
= (2 cos2 x - 1) cos x - (2 sin x cos x) sin x
= 2 cos3 x - cos x - 2 cos x sin2 x
= 2 cos3 x - cos x - 2 cos x(1 - cos2 x)
= 2 cos3 x - cos x - 2 cos x + 2 cos3 x
= 4 cos3 x - 3 cos x
Since the answer is given, you do need to show every step. Remember "One equal sign per line, all equal signs aligned"!
Similarly, using sin 3x = sin 2x cos x + cos 2x sin x leads to sin 3x = 3 sin x - 4 sin3 x.
(i) Using 4 sin3 x = 3 sin x - sin 3x gives us:
7 sin x - 8 sin3 x dx = (7 sin x - 2(3 sin x - sin 3x)) dx
0
0
= (sin x + 2 sin 3x) dx
0
=
- cos x -
2 3
cos 3x
0
=
-
cos
-
2 3
cos
3
-
-1
-
2 3
=
-c
-
2 3
(4c3
-
3c)
+
5 3
=
-
8 3
c3
+
c
+
5 3
I() = 0 when c = 1 (this can be easily found, as when = 0 the integral is of the
form
0 0
f(x)
dx
=
0
giving
cos(0)
=
1).
(ii) If we call Eustace's attempt J() we have:
J() =
7 sin x - 8 sin3 x dx
0
=
7 2
sin2 x - 2 sin4 x
0
=
7 2
sin2
-
2 sin4
=
7 2
(1
-
cos2 )
-
2(1
-
cos2
)2
=
-2c4
+
1 2
c2
+
3 2
You
can
then
substitute
c
=
cos()
=
-
1 6
and
show
that
I()
=
J(),
but
this
is
a
little fiddly (there are fractions with denominator 648 involved). Another way is to
show that 6c + 1 is a factor of the equation you get by solving I() = J(). Remember
that for a "Show that" you must fully support your answer.
STEP 2 Trigonometry: Solutions
1
step
Equating I() = J() gives:
-
8 3
c3
+
c
+
5 3
=
-2c4
+
1 2
c2
+
3 2
12c4 - 16c3 - 3c2 + 6c + 1 = 0
(*)
If
you
have
already
substituted
c
=
-
1 6
into
I()
and
J()
and
shown
that
this
is
a
solution then you can factorise out (6c + 1) without further explanation of why you
can do it.
Otherwise, you can:
?
substitute
c
=
-
1 6
into
()
and
show
that
this
is
a
solution
(this
is
not
too
bad,
especially if you don't evaluate 63)
? use long division to show that (6c + 1) is a factor (remember to show that the
remainder is 0 in this case)!
? factorise out (6c + 1) by inspection and then expand the brackets to show that
this works.
In
each
case
you
do
need
to
end
by
stating
something
like
"Hence
c
=
cos()
=
-
1 6
gives Eustace the correct value of I()".
The
equation
fully
factorises
to
give
(6c + 1)(2c + 1)(c - 1)2
=
0,
so
we
have
cos
=
-
1 6
,
cos
=
-
1 2
and
cos
=
1.
There
are
no
given
restrictions
on
,
so
we
need
the
general
solutions1. These are:
cos = 1 = = 2n
cos
=
-
1 2
=
=
2 3
+
2n
or
=
-
2 3
+
2n
cos
=
-
1 6
= = cos-1
-
1 6
+ 2n
or
= - cos-1
-
1 6
+ 2n
Another method would be to show that = 0 = c = 1 is a solution first and then
factorise out (c - 1). You could keep factorising until you had fully factorised and then
show
that
c
=
-
1 6
is
a
solution.
1See the topic notes.
STEP 2 Trigonometry: Solutions
2
step
2 There are two basic approaches for the first identity. Starting on the LHS gives:
tan
4
-
x 2
tan
4
1 + tan
- tan
x 2
4
tan
x 2
1 - tan
1 + tan
x 2 x 2
cos
cos
x 2 x 2
- sin + sin
x 2 x 2
Multiplying top and bottom by
cos
x 2
cos cos
x 2 x 2
- sin + sin
x 2 x 2
cos
x 2
- sin
x 2
cos
x 2
- sin
x 2
1 - 2 sin
cos2
x 2
x 2
cos
- sin2
x 2 x 2
1 - sin x
cos x
sec x - tan x
Alternatively
you
can
start
on
the
RHS
and
use
the
t
=
tan
1 2
A
formulae
from
the
formula
book (with A = x):
1 + t2 2t sec x - tan x 1 - t2 - 1 - t2
(1 - t)2
(1 - t)(1 + t)
1-t
1+t
1 - tan
1 + tan
x 2 x 2
tan
4
-
x 2
(i)
Setting
x=
4
into
the
identity
from
the
stem
gives:
tan
4
-
1 2
?
4
= sec
4
- tan
4
=
2-1.
Then
note
that
11 24
=
1 8
+
1 3
which
gives:
tan
1 8
+
1 3
=
tan
1 8
1 - tan
+ tan
1 8
tan
1 3
1 3
2-1+ 3
=
1 - ( 2 - 1) 3
3+ 2-1
=
as required.
3- 6+1
(ii) Since the answer is given here, you canmultiply across and verify2 that 3 - 6 + 1 2 + 2 + 3 + 6 = 3 + 2 - 1.
2In this case verify means "expand the brackets and show it is true". You MUST show all the working here as it is a "show that". A table might be a nice clear way of showing your working.
STEP 2 Trigonometry: Solutions
3
step
Alternatively you can rationalise the denominator3:
3 + 2 - 1 ( 3 + 2 - 1)( 3 + 1 + 6)
=
3 - 6 + 1 ( 3 + 1 - 6)( 3 + 1 + 6)
2+4 2+2 3
=
2 3 - 2
= (1 +2 2 + 3)( 3 + 1)
( 3 - 1)( 3 + 1)
4+2 2+2 3+2 6
=
2
=2+ 2+ 3+ 6
You should show a few more lines of working than presented here.
(iii)
Htaenre1214usi=ng2x+=12214
gives
us tan
1 48
+ 3 + 6 and
= we
sec
11 24
-
also have
tan sec2
11 24
. =
We 1+
know that tan2 . combining
these
gives us:
1
2
tan = 1 + 2 + 2 + 3 + 6 - 2 + 2 + 3 + 6 .
48
You then need to carefully expand the squared bracket (I would recommend a table for doing this) to get to the required result.
3If you had not been given the answer this is probably the approach you would have had to take. The first approach probably has less room for error though!
STEP 2 Trigonometry: Solutions
4
step
3 Using the given substitution we have:
a2
+
1 a2 tan2
?
a
sec2
d
=
1 d
a
1 = ?+c
a
1
x
= arctan + c .
a
a
Since this result is in the "stem" of the question, you should expect to use it at least once (and probably more often) in the following question parts.
(i) (a) Using the substitution t = sin x gives us:
1 2
cos x
1 cos x dt
0 1 + sin2 x dx = 0 1 + t2 cos x
11 = 0 1 + t2 dt
=
arctan t
1 0
=
4
using the stem result
(b)
Using
the
suggested
substitution
(which
means
that
dt dx
=
1 2
sec2
1 2
x)
gives
us:
1 1 - t2 0 1 + 6t2 + t4 dt =
2
0
1
-
tan2
1 2
x
1
+
6
tan2
1 2
x
+
tan4
1 2
x
?
1 2
sec2
1 2
x
dx
=
1 2
2
0
1
-
tan2
1 2
x
1
+
6
tan2
1 2
x
+
tan4
1 2
x
cos2
1 2
x
dx
=
1 2
2
0
1
-
sin2 cos2
1 2
x
1 2
x
cos2
1 2
x
+
6 sin2
1 2
x
+
sin4 cos2
1 2
x
1 2
x
dx
=
1 2
2
0
cos2
1 2
x
-
sin2
1 2
x
cos4
1 2
x
+
6 sin2
1 2
x
cos2
1 2
x
+
sin4
1 2
x
dx
=
1 2
2
0
cos2
1 2
x
+
cos2 sin2
1 2
x
1 2
x
-
sin2
1 2
x
2 + 4 sin2
1 2
x
cos2
1 2
x
dx
=
1 2
2
0
cos x 1 + sin2 x dx =
1 2
I
The last step came from using cos 2A = cos2 A - sin2 A, sin 2A = 2 sin A cos A and cos2 A + sin2 A = 1.
STEP 2 Trigonometry: Solutions
5
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