Transcript Trigonometric Addition Identities

Trigonometric Addition Identities
OP  1
OQ  cos A
PQ  sin A
OR  cos A cos B
QR  cos A sin B
QS  sin A cos B
PS  sin A sin B
P
S
cos( A  B)  OT
 OR  PS
 cos A cos B  sin A sin B
sin A
1
UA
cos
sin( A  B)  PT
 QS  QR
 sin A cos B  cos A sin B
B
Q
A
B
O
T
R
The tan addition formula
Q What is tan in terms of sin and cos ?
sin 
A tan  
cos
Q What is tan( A  B), and how are you going to continue?
A tan( A  B) 
sin( A  B) sin A cos B  cos A sin B

cos( A  B) cos A cos B  sin A sin B
Q How are you going to express this in terms of tan A and tan B ?
A Divide top and bottom by cos A cos B
sin A cos B cos A sin B

cos
A
cos
B
cos A cos B  tan A  tan B
tan( A  B) 
cos A cos B sin A sin B 1  tan A tan B

cos A cos B cos A cos B
The circular definitions of trigonometric functions
The right angled triangle definition
X
cos  , so X  R cos ,
R
Y
sin   , so Y  R sin 
R
R
Y

X
1
Scale the triangle, to make the hypoteneuse  1
x  cos
y

x
y  sin 
y sin 
tan   
x cos
P(x,y)
1
Place this triangle in a unit radius
circle, with centre at the origin
We now turn things round and DEFINE
( cos ,sin  ) as the coordinates of P,
where OP makes angle  with the x axis.
y

O
x
Deductions from the circular definition
To show the angles, Polar
coordinates have been used
P1[1, ] means OP1  1, and
OP1 makes angle  with the
P2 [1,180]
P1[1,]
(cos ,sin )
(cos ,sin )
positive x axis
You should be able to see
how to write the cos or sin
of any of  , 180   , 360  
in terms of  the cos or sin
P3 [1,180]
P4 [1,360 ]
(cos ,sin )
(cos ,sin )
of 
cos(180   )   cos
cos(180   )   cos
cos(360   ) 
cos
cos( )  cos(360   )  cos
sin(180   ) 
sin 
sin(180   )   sin 
sin(360   )   sin 
sin( )  sin(360   )   sin 
Trigonometric Subtraction Identities
We could modify the addition identities argument by using the
cicular definitions, and replacing lengths by signed lengths, and
will assume this can be done, but not here.
cos( A  B)  cos( A  ( B))  cos A cos( B)  sin A sin( B)  cos A cos B  sin A sin B
sin( A  B)  sin( A  ( B))  sin A cos( B)  cos A sin( B)  sin A cos B  cos A sin B