Download C3.4 Trigonometry 2

A-Level Maths:
Core 3
for Edexcel
C3.4 Trigonometry 2
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Contents
The addition formulae
The addition formulae
The double angle formulae
Expressions of the form a sin θ + b cos θ
Examination-style questions
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The addition formulae
In trigonometry we often have to solve problems involving
compound angles. For example,
sin( + 30 )
It is important to note that trigonometric functions are not
distributive over addition. In other words,
sin( + 30 )  sin + sin30
In fact, for two angles A and B,
sin( A + B )  sin A cos B + cos A sin B
This is known as one of the addition formulae (or compound
angle formulae) and it should be learnt.
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Proof of sin(A + B) = sin A cos B + cos A sin B
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The addition formulae involving sin
To find an identity for sin(A – B) we can replace B in
sin( A + B )  sin A cos B + cos A sin B
by –B to give:
sin( A  B )  sin A cos(  B ) + cos A sin(  B )
Remember that
cos (–B) = cos B
and
sin (–B) = –sin B
So,
sin( A  B)  sin A cos B  cos A sin B
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Proof of cos(A + B) = cos A cos B – sin A sin B
To find the addition formula for cos(A + B) we can use the fact
that cos θ = sin(90 – θ), to give
cos( A + B) = sin(90  ( A + B)
= sin((90  A)  B)
= sin(90  A)cos B  cos(90  A)sin B )
Using the fact that
sin(90  A) = cos A
and
cos(90  A) = sin A
It follows that the addition formula for cos(A + B) is
cos( A + B)  cos A cos B  sin A sin B
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The addition formulae involving cos
To find an identity for cos(A – B) we can replace B in
cos( A + B)  cos A cos B  sin A sin B
by –B to give:
cos( A  B )  cos A cos(  B )  sin A sin(  B )
Using the fact that
cos (–B) = cos B
and
sin (–B) = –sin B,
we have:
cos( A  B)  cos A cos B + sin A sin B
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The addition formulae involving tan
Addition formulae involving tan can be derived from the
addition formulae involving sin and cos.
sin( A + B )
tan( A + B ) 
cos( A + B )

sin A cos B + cos A sin B
cos A cos B  sin A sin B
Dividing through by cos A cos B gives:
sin A cos B cos A sin B
+
tan( A + B )  cos A cos B cos A cos B
cos A cos B sin A sin B

cos A cos B cos A cos B
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The addition formulae involving tan
sin A sin B
+
tan( A + B )  cos A cos B
sin A sin B
1
cos A cos B
This can be written in terms of tan as:
tan A + tan B
tan( A + B ) 
1  tan A tan B
If we replace B by –B we get:
tan A  tan B
tan( A  B ) 
1+ tan A tan B
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Using the addition formulae
Use the identity for sin(A + B) to find the exact value of sin 75°.
We don’t know the exact value of 75° but we do know the sine
and cosine of 30° and 45°.
We can write sin 75° as sin (30° + 45°) to give
sin(30 + 45 ) = sin30 cos 45 + cos30 sin 45
1
2
3
2
= ×
+
×
2 2
2
2
2+ 2 3
=
4

2 1+ 3
=
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
4
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Using the addition formulae
Calculate the exact value of
tan 56  tan 2
1+ tan 56 tan 2
tan A  tan B
Using the identity tan( A  B ) 
1+ tan A tan B
tan 56  tan 2
1+ tan 56 tan 2
= tan( 56  2 )
= tan 3
= 3
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Using the addition formulae
Given that sin A = 1 and cos B = 54 , where A and B are acute,
3
find the exact value of cos (A + B).
cos( A + B)  cos A cos B  sin A sin B
We can find the value of cos A and sin B using Pythagoras:
sin A = 31
So cos A =
cos B = 54
3
8
3
1
So sin B
A
3
=5
5
B
4
?8
Using these values, cos( A + B ) =
=
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?
3
8 4 1 3
×  ×
3 5 3 5
4 8 3
8 2 3

=
15 15
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Using the addition formulae
Solve cos(45   ) = sin(30 +  ) for  180    180
Using the addition formulae:
cos 45 cos + sin45 sin = sin30 cos + cos30 sin
2
2
1
3
cos +
sin = cos +
sin
2
2
2
2
2 cos + 2 sin = cos + 3 sin
( 2  3)sin = (1  2)cos
sin
1 2
=
cos
2 3
1 2
tan =
2 3
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Using the addition formulae
Now, using a calculator,
 = tan1
1 2
2 3
= 52.5°
Tan is positive in the first and third quadrants, so the second
solution in the required range is
θ = –180° + 52.5°
52.5°
S
A
T
C
= –127.5°
So the solution set for –180° ≤ θ ≤ 180° is
–127.5°
θ = –127.5°, 52.5°
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Contents
The double angle formulae
The addition formulae
The double angle formulae
Expressions of the form a sin θ + b cos θ
Examination-style questions
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Double angle formulae
We can use the addition formulae to derive formulae involving
sin 2A, cos 2A and tan 2A.
Using the formula
sin( A + B )  sin A cos B + cos A sin B
where A is equal to B gives:
sin( A + A)  sin A cos A + cos A sin A
sin2 A  2sin A cos A
This is the first double angle formula and it should be learnt.
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Double angle formulae
Using the addition formula
cos( A + B)  cos A cos B + sin A sin B
where A is equal to B gives:
cos( A + A)  cos A cos A  sin A sin A
cos2 A  cos2 A  sin2 A
We can use the identity sin2A + cos2A = 1 to write this in two
more ways:
cos2 A  2cos2 A  1
cos2 A  1  2sin2 A
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Double angle formulae
Using the addition formula
tan A + tan B
tan( A + B ) 
1  tan A tan B
where A is equal to B gives:
tan A + tan A
tan( A + A) 
1  tan A tan A
2 tan A
tan2 A 
1  tan2 A
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Using double angle formulae to prove identities
We can use the double angle formulae to prove other identities
involving multiple angles. For example:
Use the compound and double angle formulae to prove that
cos3  4cos3   3cos
LHS = cos3
= cos( + 2 )
= cos cos2  sin sin2
= cos (2cos2   1)  2sin sin cos
= 2cos3   cos  2sin2  cos
= 2cos3   cos  2cos (1  cos2  )
= 2cos3   cos  2cos  2cos3 
= 4cos3   3cos = RHS
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Using double angle formulae to prove identities
Prove that
sin2 x
 tan x
1+ cos2 x
sin2 x
1  cos2 x
2sin x cos x
=
1+ 2cos2 x  1
2sin x cos x
=
2cos2 x
sin x
=
cos x
= tan x = RHS
LHS =
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Solving equations
We can also use the double angle formulae to solve equations
involving double angles. For example,
Solve cos = 3cos2 + 2 for 0    360
cos = 3(2cos2   1) + 2
cos = 6cos2   3 + 2
6cos2   cos  1= 0
(2cos  1)(3cos +1) = 0
cos  =
1
2
or
θ = 60°, 300°
cos  =  31
θ = 109.5°, 250.5°
The complete solution set is θ = 60°, 300°, 109.5°, 250.5°.
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Contents
Expressions of the form a sin θ + b cos θ
The addition formulae
The double angle formulae
Expressions of the form a sin θ + b cos θ
Examination-style questions
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Functions of the form a cos θ + b sin θ
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Expressions of the form a cos θ + b sin θ
As we have seen, the graph of f(θ) = a cos θ + b sin θ can be
obtained by stretching a sine or cosine curve vertically and
translating it horizontally.
Expressions of the form a cos θ + b sin θ can therefore be
expressed in the form R cos (θ ± α) or R sin (θ ± α).
Suppose we want to express a cos θ + b sin θ in terms of cos
only.
If we divide and multiply a cos θ + b sin θ by
we have
a 2 + b2


a
b
a cos + b sin = a + b 
cos +
sin 
2
2
2
2
a +b
 a +b

2
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Expressions of the form a cos θ + b sin θ
Let cos  =
a
a 2  b2
Now, using a right-angled triangle containing the acute angle α
we have:
a 2 + b2
b
sin =
α
b
a b
2
2
and
b
tan  =
a
a


a
b
So a cos + b sin = a + b 
cos +
sin 
2
2
2
2
a +b
 a +b

2
2
= a 2 + b2  cos cos + sin sin 
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Expressions of the form a cos θ + b sin θ
But cos α cos θ + sin α sin θ = cos(θ – α), so:
a cos + b sin = a 2 + b2 cos(   )
= R cos(θ – α)
where R = a 2 + b2 and α = tan–1 ba .
Now use the following right-angled triangle to show that
a cos θ + b sin θ can be written in the form R sin(θ + α)
where R = a 2 + b2 and α = tan–1 ba .
a 2 + b2
α
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b
a
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Expressions of the form a cos θ + b sin θ
In general, for positive values of a, b and R and 0 < α < 90°, it
can be shown that:
a cos θ + b sin θ = R cos (θ – α)
and
a cos θ – b sin θ = R cos (θ + α)
where R = a 2 + b 2
and  = tan1 ba
and
a cos θ + b sin θ = R sin (θ + α)
and
– a cos θ + b sin θ = R sin (θ – α)
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where R = a 2 + b 2
and  = tan1 ba
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Equations of the form a cos θ + b sin θ = c
Express 3 cos θ + 4 sin θ in the form R cos (θ – α).
Start by writing this as an identity:
3cos + 4sin  R cos(   )
Using the addition formula for cos(A – B) gives:
3cos + 4sin  R cos cos + R sin sin
Equating the coefficients of cos θ and sin θ :
3
3 = R cos   cos  =
R
4
4 = R sin  sin =
R
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Equations of the form a cos θ + b sin θ = c
Using the following right-angled triangle:
R = 32 + 42 = 5
R
4
 = tan1 34
= 53.1 (to 3 s.f.)
α
3
So, using these values:
3 cos θ + 4 sin θ = 5 cos (θ – 53.1°)
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Equations of the form a cos θ + b sin θ = c
a) Express 12 sin θ – 5 cos θ in the form R sin(θ – α).
b) Solve the equation 12 sin θ – 5 cos θ = 8 in the
interval 0 < θ < 360°.
12sin – 5cos  R sin( –  )
a)
Using the addition formula for sin(θ – α) gives:
12sin  5cos  R sin cos  R cos sin
Equating the coefficients of cos θ and sin θ :
5
5 = R sin   sin =
R
12
12 = R cos   cos  =
R
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Equations of the form a cos θ + b sin θ = c
Using the following right-angled triangle:
R = 52 +122 = 13
R
5
α
12
5
 = tan1 12
= 22.6 (to 3 s.f.)
So, using these values
12 sin θ – 5 cos θ = 13 sin (θ – 22.6°)
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Equations of the form a cos θ + b sin θ = c
b) Using the form found in part a) we can write the equation
12 sin θ – 5 cos θ = 8 as
13sin(  22.6) = 8
8
sin(  22.6 ) = 13
(Using a calculator set to degrees:)
8 = 38.0 (to 3 s.f.)
sin1 13
So
This is the solution
in the 2nd quadrant.
θ – 22.6° = 38.0° or 142°
θ = 60.6° or 164.6° (to 3 s.f.)
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Contents
Examination-style questions
The addition formulae
The double angle formulae
Expressions of the form a sin θ + b cos θ
Examination-style questions
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Examination-style question 1
The function f is defined by
f ( x ) = 4cos(60 + x)  cos x
a) Show that when f(x) = 0, cos x = 2 3 sin x .
b) Hence solve f(x) = 0 in the interval –180° < x < 180°.
4cos(60 + x )  cos x = 0
a)
4cos60 cos x  4sin60 sin x  cos x = 0
 
 
4 21 cos x  4 23 sin x  cos x = 0
2cos x  2 3 sin x  cos x = 0
cos x  2 3 sin x = 0
cos x = 2 3 sin x
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Examination-style question 1
b) When f(x) = 0, cos x = 2 3 sin x
Dividing through by cos x gives:
1= 2 3 tan x
tan x = 1
2 3
(Using a calculator set to degrees:)
tan1 1 = 16.1 (to 3 s.f.)
2 3
In the interval –180° < x < 180°:
S
A
16.1°
–163.9°
T
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x = 16.1° or –164° (to 3 s.f.)
C
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Examination-style question 2
a) Express 2 sin 2θ + 3 cos 2θ in the form R cos (2θ – α),
where R > 0 and 0 < α < 90.
b) Use the double angle formulae to express
4 sin θ cos θ + 6 cos2 θ in the form a cos 2θ + b sin 2θ + c
where a, b and c are constants to be determined.
c) Use your answer to part a) to find the maximum value of
4 sin θ cos θ + 6 cos2 θ and find the smallest positive value
of θ at which it occurs.
a) 2sin2 + 3cos2  R cos(2   )
Using the addition formula for cos(2θ – α) gives:
2sin2 + 3cos2  R cos2 cos + R sin2 sin
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Examination-style question 2
Equating the coefficients of cos θ and sin θ :
2 = R sin 
sin =
3 = R cos   cos 
2
R
= R3
Using the following right-angled triangle:
R = 22 + 32 = 13
R
2
α
 = tan1 32
= 33.7 (to 3 s.f.)
3
So, as required
2sin2 + 3cos2  13 cos(2  33.7 )
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Examination-style question 2
b) We can write 4sin cos + 6cos2  as:
2(2sin cos ) + 3(2cos2  )
Using the double angle formulae, sin2 = 2sin cos
we can write this as:
cos2 = 2cos2   1
2sin2 + 3(cos2 +1)
2sin2 + 3(cos2 +1) = 2sin2 + 3cos2 + 3
 4sin cos + 6cos2   2sin2 + 3cos2 + 3
c) Using the answer to part a) we can write:
2sin2 + 3cos2 + 3  13 cos(2  33.7) + 3
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Examination-style question 2
Therefore, 4 sin cos + 6cos2   13 cos(2  33.7 ) + 3
This reaches its maximum value when
cos(2  33.7 ) = 1
Therefore the maximum value of 4sin cos + 6cos2  is
13 + 3 = 6.61 (to 3 s.f.)
The smallest positive value of θ at which this maximum occurs
is given when
2  33.7 = 0
2 = 33.7
 = 16.8 (to 3 s.f.)
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