Compound Angle Formula
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Transcript Compound Angle Formula
Compound Angle Formulae
1. Addition Formulae
cos( A B) cos A cos B sin A sin B
cos( A B) cos A cos B sin A sin B
sin( A B) sin A cos B cos A sin B
cos( A B) sin A cos B cos A sin B
Example:
sin( x 30)o sin x cos30 cos x sin 30
3 sin x cos x
2
2. Formulae Involving Double Angle (2A)
sin 2 A 2sin Acos A
cos 2 A cos 2 A sin 2 A
2cos 2 A 1
1 2sin 2 A
Two further formulae derived from the cos2 A formulae.
cos 2 A 12 (1 cos 2 A)
sin 2 A 12 (1 cos 2 A)
Mixed Examples:
4
Given that A is an acute angle and tan A , calculate sin 2 A and cos 2 A.
3
sin A 4
cos A 3
sin 2 A cos 2 A 1
sin 2 A ( 43 sin A)2 1
sin A
Similarly:
sin 2 A 2sin A cos A
cos A
16
25
3
5
4
5
Substitute form the tan
(sin/cos) equation
+ve because A is acute
3-4-5 triangle
!!!
24
25
cos 2 A cos2 A sin 2 A
9 16
25
7
25
A is greater than 45
degrees – hence 2A is
greater than 90
degrees.
Find the exact value of sin 75o.
2
sin(75o ) sin(45 30)
sin(75o ) sin 45cos30 cos 45sin30
45
1
o
1
Prove that
sin( )
tan tan
cos cos
sin cos cos sin
cos cos
sin sin
cos cos
tan tan
30 o
1
1 3 1 1
1 3
2 2
22
2 2
sin( )
cos cos
2
Q.E.D.
3
For the diagram opposite show that cos LMN
5
.
5
M
cos LMN cos( )
3 2
Length of LM
18 3 2
Length of MN
10
1 3
1 1
2 10
2 10
2
2
20
4 5
5
5
10
3
3
cos( ) cos cos sin sin
1
5
L
1
N
(A Higher Question)
Show that, for the triangle ABC in the diagram, a
a
b
c
sin a sin b sin c
b sin
.
cos( )
The sine
rule
b
From the diagram:
C
a
b
sin sin( [ 2 ])
a
b
sin( 2 [ ])
The sum of the angles of
a triangle=180
sin( 2 ) cos
b
cos( )
b sin
cos( )
As required
A
2
c
a
B
Prove that,
cos 4 sin 4 cos 2 .
x 2 y 2 ( x y)( x y )
cos4 sin 4 (cos2 )2 (sin 2 )2
(cos2 sin 2 )(cos2 sin 2 )
cos 2 sin 2 1
cos 2 sin 2
cos( ) cos cos sin sin
cos2
TRIGONOMETRIC EQUATIONS
Double angle formulae (like cos2A or sin2A) often occur in trig equations.
We can solve these equations by substituting the expressions derived in
the previous sections.
Use
sin2A = 2sinAcosA
when replacing sin2A
cos2A = 2cos2A – 1
cos2A = 1 – 2sin2A
if cosA is also in the equation
if sinA is also in the equation
when replacing cos2A
Solve:
cos 2 x o 4sin x o 5 0 for 0 x 360o.
cos2x and sin x, so substitute 1-2sin2
(1 2sin 2 x) 4sin x 5 0
6 4sin x 2sin 2 x 0
cp. w. 6 4 z 2 z 2 0
(6 2sin x)(1 sin x) 0
sin x 1 or sin x 3
x 90o
0 sin x 1 for all real angles
Solve:
5cos 2 x o cos x o 2
for 0 x 360o.
cos 2x and cos x, so substitute 2cos2 -1
5(2cos2 x 1) cos x 2
10cos 2 x cos x 3 0
(5cos x 3)(2cos x 1) 0
3
1
cos x or cos x
5
2
s
a
t
c
All S_ Talk C*&p ??
x 51.3 or
x 90 60 150o or
x 360 51.3
x 270 60 210
o
308.7 o
o
2
y 0.6
y0
y 0.5
3
2
The diagram shows the graphs of
f ( x) a sin bx o
and g ( x) c sin x o
for 0 x 360o.
yy
4
y f ( x)
2
360o
0
-2
xx
y g ( x)
-4
Three problems concerning this graph follow.
i)
State the values of a, b and c.
yy
4
f ( x) a sin bxo
The max & min values of asinbx
are 3 and -3 resp.
The max & min values of sinbx
are 1 and -1 resp.
a3
2
g ( x) c sin x o
The max & min values of csinx are
2 and -2 resp.
c2
360o
0
-2
-4
f(x) goes through 2 complete cycles from 0 – 360o
b2
y f ( x)
y g ( x)
xx
ii) Solve the equation
f ( x) g ( x) algebraically.
From the previous problem we now have:
f ( x) 3sin 2 x
and
g ( x) 2sin x
Hence, the equation to solve is:
3sin 2 x 2sin x
Expand sin 2x
3(2sin x cos x) 2sin x
6sin x cos x 2sin x 0
Divide both sides by 2
3sin x cos x sin x 0
Spot the common factor in the terms?
sin x(3cos x 1) 0
Is satisfied by all values of x for which:
sin x 0 or
cos x
1
3
iii) find the coordinates of the points of intersection of the graphs for 0 x 360o.
From the previous problem we have:
sin x 0 or
sin x 0
cos x
cos x
1
3
Hence:
x
0o
or
x
70.5o
x
x
180o
360o
or
x
x
(360 70.5)o
289.5o
or
1
3