Download further trigonometric identities

C3 Chapter 7
FURTHER TRIGONOMETRIC IDENTITIES
ADDITION FORMULAE
Remember that:
sin(90 – ) = cos 
cos(90 – ) = sin 
It is easily verified that sin (A + B) ≠ sin A + sin B
One counter-example is all that is needed
e.g. Let A = 30º and B = 45º
Then sin(A + B) = sin 75º = 0.9659…
and sin 30º + sin 45º = 1.2071…
so sin(30º + 45º) ≠ sin30º + sin45º
In fact:
sin (A + B) = sin A cos B + cos A sin B
sin (A – B) = sin A cos B – cos A sin B
cos (A + B) = cos A cos B – sin A sin B
cos (A – B) = cos A cos B + sin A sin B
(You do not need to know how to derive these formulae)
You can deduce the results from each other
Example 1
In sin(A + B) = sin A cos B + cos A sin B,
replace A with (90º - A)
sin([90 – A] + B) = sin(90 – A)cosB + cos(90 – A)sinB
sin(90 – [A – B]) = cosAcosB + sinAsinB
cos(A – B) = cosAcosB + sinAsinB
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C3 Chapter 7
tan(A + B) =
sin(A + B)
cos(A + B)
=
sinAcosB + cosAsinB
cosAcosB – sinAsinB
Divide top & bottom by cosAcosB
tan(A + B) =
tanA + tanB
1 – tanAtanB
Replace B by –B
tan(A – B) =
tanA + tan( – B)
1 – tanAtan( – B)
Hence we have:
tan(A + B) =
tan(A – B) =
=
tanA – tanB
1 + tanAtanB
tanA + tanB
1 – tanAtanB
tanA – tanB
1 + tanAtanB
Example 2
2
Prove the identity
2
sin(A + B)sin(A – B) = sin A – sin B
sin(A + B)sin(A – B) = (sinAcosB + cosAsinB)(sinAcosB – cosAsinB)
2
2
2
2
2
2
= sin A cos B + sinAcosAsinBcosB – sinAcosAsinBcosB – cos A sin B
2
2
= sin A cos B – cos A sin B
2
2
2
2
= sin A (1 – sin B) – (1 – sin A)sin B
2
2
2
2
2
2
2
2
= sin A – sin A sin B – sin B + sin A sin B
= sin A – sin B
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QED
2
C3 Chapter 7
Example 3
Given that sin A =
3
5
and A is acute and cos B = –
5
13
and B is obtuse
Find the exact value of sin(A + B)
sin(A + B) = sinAcosB + cosAsinB
A is acute
5
 cosA =
3
4
5
A
4
B is obtuse
 2nd Q  sinB is positive
 sinB = +
13
12
12
13
B
5
 3
5 
 4 12 
sin(A + B) = 
+
 –




5
13
5
13




= –
15
65
+
48
65
=
33
65
Ex 7A p 99
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C3 Chapter 7
DOUBLE ANGLE FORMULAE
In the formula sin (A + B) = sin A cos B + cos A sin B, let A = B
Therefore, sin (A + A) = sin A cos A + cos A sin A
sin 2A = 2 sin A cos A
Hence the sine double angle formula:
sin 2A = 2 sin A cos A
In the formula cos (A + B) = cos A cos B – sin A sin B, let A = B
Therefore, cos (A + A) = cos A cos A – sin A sin A
cos 2A = cos2A – sin2A
Since cos2A = 1 – sin2A, cos 2A = (1 – sin2A) - sin2A = 1 - 2 sin2A
Or sin2A = 1 – cos2A, cos 2A = cos2A – (1 – cos2A) = 2 cos2A – 1
Hence the 3 versions of the cosine double angle formula:
cos 2A = cos2A – sin2A
or 1 - 2 sin2A
or 2 cos2A – 1
In the formula
tan(A + B) =
tan(A + A) =
tanA + tanB
1 – tanAtanB
tanA + tanA
1 – tanAtanA
=
, let A = B
2tanA
2
1 – tan A
Hence the tangent double angle formula:
tan 2A =
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2tanA
2
1 – tan A
4
C3 Chapter 7
Example 1
Write the following as a single trigonometric ratio:
1
2
2cos 37 – 1
2
2cos 37 – 1 = cos (2  37) = cos 74
1
1
=
= sec 74

2
2cos 37 – 1 cos 74
Example 2
Without using a calculator, find the value of
2
(cos 60 – sin 60)
2
2
2
(cos 60 – sin 60) = cos 60 – 2cos 60 sin 60 + sin 60
2
2
= (cos 60 + sin 60) – 2 cos 60 sin 60
= 1 – sin 120
= 1 – sin 60 = 1 –
3
2
=
2 –
3
2
Example 3
Given that sin θ = ¼ and that θ is obtuse, find the exact value
of sin 2θ
4
1
Pythagoras  x =
15
Obtuse   cos  is – ve

x
sin 2 = 2sin  cos  = 2 
cos  = –
15
4
1

15 
15
=
–
  –

4
4
8


Ex 7B p 103
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C3 Chapter 7
USING THE DOUBLE ANGLE FORMULAE TO PROVE
IDENTITIES
Example
Prove the identity
cot 2 =
 LHS =
=
2
1
tan 2
1
sin 
1
sin 
cosec θ – 2 cot 2θ cos θ = 2 sin θ
1 – tan 
=
2tan 
2
–
2(1 – tan )
2 tan 
2
–
2(1 – tan )cos 
2 sin 
2
=
 cos 
2
1 – (1 – tan ) cos 
sin 
 cos 
2
=
2
1 – cos  + sin 
sin 
2
=
2 sin 
sin 
= 2 sin 
QED
Ex 7C Q 1, 2, 7-13 p 106
TRIPLE ANGLE FORMULAE
sin 3A = sin(2A + A) = sin 2A cos A + cos 2A sin A
2
= 2 sin A cos A cos A + (1 – 2 sin A) sin A
2
2
= 2 sin A (1 – sin A) + (1 – 2 sin A) sin A
3
3
= 2 sin A – 2 sin A + sin A – 2 sin A
3
= 3 sin A – 4 sin A
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C3 Chapter 7
cos 3A = cos(2A + A) = cos 2A cos A – sin 2A sin A
2
= (2 cos A – 1) cos A – 2 sin A cos A sin A
2
2
= (2 cos A – 1) cos A – 2 cos A (1 – cos A)
3
3
= 2 cos A – cos A – 2 cos A + 2 cos A
3
= 4 cos A – 3 cos A
HALF-ANGLE FORMULAE
2
2
cos 2A = 2 cos A – 1  cos 2A + 1 = 2 cos A
2
 cos A =
1 + cos 2A
2
2
2
cos 2A = 1 – 2 sin A  1 – cos 2A = 2 sin A
2
 sin A =
1 – cos 2A
2
Replacing A with
2
cos
x
2
=
x
2
gives:
1 + cos x
2
and
2
sin
x
2
=
1 – cos x
2
These are useful in integration.
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C3 Chapter 7
USING THE DOUBLE ANGLE FORMULAE TO SOLVE
EQUATIONS
Example 1
Solve the equation
3 sin 2θ = 4 tan θ for 0º ≤ θ ≤ 360º
3 sin 2 = 4 tan   6sin  cos  =
2
4 sin 
cos 
2
 6 sin  cos  = 4 sin   sin  (3 cos  – 2) = 0
 sin  = 0 or cos  = 
2
3
sin  = 0   = 0, 180 , 360
cos  = 
2
  has a solution in all 4 quadrants
3
  = 35·3, 144·7, 215·3, 324·7
Hence  = 0, 35·3, 144·7, 180, 215·3, 324·7, 360
Ex 7C Q 3, p 107
Example 2
Eliminate θ from the equations:
2
x = sin , y = 1 + cos 2
2
2
y = 1 + cos 2 = 1 + (1 – 2sin ) = 2 – 2 sin 
y = 2 – 2x
Note: θ is known as a parameter
Ex 7C Q 4-6, p 107
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C3 Chapter 7
THE FORM “a cos θ ± b sin θ”
If you draw the graph of a cos θ + b sin θ it will have the same
form as a sine or cosine graph.
It can be expressed in the form R sin (θ ± α) or R cos (θ ± α)
Example 1
Express 3 sin x + 2 cos x in the form R sin (x +  )
where R > 0 and 0 <  <

2
Let 3 sin x + 2 cos x = R sin (x +  ) = R(sin x cos  + cos x sin  )
Equate coefficients of sin x  R cos  = 3 (1)
Equate coefficients of cos x  R sin  = 2 (2)
(2)
(1)
(1)
2

R sin 
R cos 
+ (2)
2
=
2
3
2
 tan  =
2
2
2
3
  = 33·7
2
 R cos  + R sin  = 3
2
 R = 9 + 4 = 13  R =
Hence 3 sin x + 2 cos x =
Remember:
2
+ 2
2
13
13 sin (x + 33·7)
Divide the equations to find tan α
Square and add the equations to find R2
Ex 7D Q 1-7 p 111
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C3 Chapter 7
This method can be used to find maximum or minimum values of
expressions and to solve certain types of trig equations.
Example 2
Find the maximum value of 2 cos 2θ – 5 sin 2θ and the smallest
positive value of θ for which it occurs.
Let 2 cos 2 – 5sin 2 = Rcos(2 +  ) = R(cos 2 cos  – sin 2 sin  )
Equate coefficients of cos 2  R cos  = 2 (1)
Equate coefficients of sin 2  R sin  = 5
(2)
(1)
2

r sin 
R cos 
(1) + (2)
2
=
2
5
2
 tan  =
5
2
  = 68·2
2
2
 R = 2 + 5 = 29  R =
Hence 2 cos 2 – 5 sin 2 =
(2)
29
29 cos (2 + 68·2)
29
Hence maximum value =
This occurs when 2 + 68·2 = 0, 360, etc
Smallest + ve   2 + 68·2 = 360  2 = 291·8   = 145·9
Note; if necessary use a sketch to determine the max/min
values
e.g.
y
x
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cos  graph
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C3 Chapter 7
Example 3
Solve the equation
4 cos x + 3 sin x = 1 for 0º ≤ x ≤ 360º
Let 4 cos x + 3 sin x = R cos (x –  ) = R (cos x cos  + sin x sin  )
Equate coefficients of cos x  R cos  = 4 (1)
Equate coefficients of sin x  R sin  = 3 (2)
(2)
(1)
(1)
2
 tan  =
+ (2)
2
3
4
2
  = 36·87
 R = 4
2
+ 3
2
= 25  R = 5
 5 cos(x – 36·87) = 1
cos (x – 36·87) = 0·2
+ ve cosine  1st or 4th quadrant
 x – 36·87 = 78·46, 281·54
 x = 115·3 or 318·4
Ex 7D Q 8-15 p 111
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C3 Chapter 7
FACTOR FORMULAE
sin(A + B) = sin A cos B + cos A sin B
sin(A – B) = sin A cos B – cos A sin B
Adding gives:
sin(A + B) + sin(A – B) = 2 sin A cos B
Let P = A + B and Q = A – B
 A =
P + Q
2
and B =
P – Q
2
 P + Q 
 P – Q 
cos
 sin P + sin Q = 2 sin 



2
2




Replacing Q with –Q gives:
 P + ( – Q) 
 P – ( – Q) 
sin P + sin ( – Q) = 2 sin 
cos



2
2




 P – Q 
 P + Q 
cos
 sin P – sin Q = 2 sin 



2
2




Hence:
sin P + sin Q = 2 x sin (semi-sum) cos (semi-difference)
sin P – sin Q = 2 x sin (semi-difference) cos (semi-sum)
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C3 Chapter 7
cos(A + B) = cos A cos B – sin A sin B
cos(A – B) = cos A cos B + sin A sin B
Adding gives cos(A + B) + cos(A – B) = 2 cos A cos B
Let P = A + B and Q = A – B
 A =
P + Q
2
and B =
P – Q
2
 P + Q 
 P – Q 
cos
cos P + cos Q = 2 cos 



2
2




Replacing Q with –Q gives exactly the same result, so this time
we subtract the two starting equations:
cos(A + B) – cos(A – B) = – 2 sin A sin B
 P + Q 
 P – Q 
Hence, cos P – cos Q = – 2 sin 
 sin 

2
2




Hence:
cos P + cos Q = 2 x cos (semi-sum) cos (semi-difference)
cos P – cos Q = - 2 x sin (semi-sum) sin (semi-difference)
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C3 Chapter 7
Hence the 4 factor formulae:
 P + Q 
 P – Q 
sin P + sin Q = 2 sin 
cos



2
2




 P + Q 
 P – Q 
sin P – sin Q = 2 cos 
sin



2
2




 P + Q 
 P – Q 
cos P + cos Q = 2 cos 
cos



2
2




 P + Q 
 P – Q 
cos P – cos Q = – 2 sin 
sin



2
2




Note the minus sign in the last formula
Example 1
Express as a sum or difference of sines:
 P + Q 
 P – Q 
sin P + sin Q = 2 sin 
cos



2
2




Use
Let
sin 4x cos 6x
P + Q
2
= 4x and
P – Q
2
Adding gives
P = 10x
Subtracting gives
Q = – 2x
= 6x
 sin 10x + sin ( – 2x) = 2 sin 4x cos 6x
 sin 4x cos 6x =
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sin 10x – sin 2x
2
14
C3 Chapter 7
Example 2
Solve the equation
cos 2θ + cos 3θ = 0
cos 2 + cos 3 = 2 cos
 cos
5
2
= 0 
5
2
  =
and cos

2
= 0 

2
=
=
5
2
cos

2
for 0 ≤ θ ≤ 2π
= 0
 3 5 7 9 11
, etc.
,
,
,
,
,
2
2 2 2 2 2
 3 5 7 9
,
,
,
,
5 5 5 5 5
 3
, etc.
,
2 2
  = 
Hence
 =
7 9
 3
,
, ,
,
5 5
5 5
Ex 7E p 115
Mixed Ex 7F p 116
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