Sum & Difference Identities for Sine & Cosine
Derive the difference identity: cos
(a - B) = cos a
cos
B + sin a sin B
-
Draw a unit circle on a rectangular coordinate system. Mark off angle a
in quadrant II
and angle b in quadrant I,
so that a > b.
Mark the angle a - b.
Let P be the point
where angle a intersects the
unit circle and Q be the point where angle b
intersects the
unit circle. Draw segment PQ.
-
Draw another unit circle on a rectangular coordinate system.
Place
the angle a - b
in standard position. Let b
be the point where this angle intersects the unit circle.
Let R be the point where the x-axis
intersects
the unit circle.
- Now, PQ = DR
-
The coordinates where P intersects the unit circle are (cos
a,
sin a) and the
coordinates where Q intersects the unit circle are (cos
b
,
sin b ). These come
rom the fact that x = r cos q
and y = r sin q. In
this
case, r = 1.
, using the
distance formula.-
Likewise, the coordinates where D intersects the unit
circle
are
[cos (a - b ),
sin (a - b )]
and the coordinates where
R intersects the unit circle
are (1,0).
-
=
-
Square both sides:
[(cos a - cos b
)2
+ (sin a - sin b
)2]=
[(cos (a - b
) -
1)2 + (sin (a - b
)]2
- Use FOIL:
cos2a - 2 cos a
cos b + cos2b
+ sin2a - 2sin a
sin b + sin2b
= cos2(a
- b ) - 2 cos (a
- b ) + 1 + sin2(a
- b )
- Group all the cos2q
+ sin2q
together:
cos2a + sin2a
+ cos2b + sin2b
- 2 cos a cos b
-
2 sin a sin b
= cos2(a
- b ) + sin2(a
- b ) - 2 cos (a
- b ) + 1
- Change all the cos2q
+ sin2q
to 1:
1 + 1 - 2 cos a cos b
- 2 sin a sin b
=
1 - 2 cos (a - b )
+ 1
- Combine the ones:
2 - 2 cos a cos b
- 2 sin a sin b
=
2 - 2 cos (a - b )
- Solve for cos (a - b
):
-2 cos a cos b
- 2 sin a sin b
= -2 cos (a - b )
(Subtract
2 from each side)
cos a cos b
+ sin a sin b
= cos (a - b ) (Divide
both sides by -2)
- So cos (a -
b )
= cos a cos b + sin
a
sin b
Derive the sum identity: cos
(A
+ B) = cos A cos B - sin A sin B
-
cos (A + B) = cos (A - (-B))
-
Using the first identity:
cos (A - (-B) = cos A cos (-B) + sin A sin (-B)
- Since cos ( -q ) = cos q
and
sin ( -q ) = - sin q
,
cos (A - (-B) = cos A cos B - sin A sin B
cos (A + B) = cos A cos B - sin A sin B
Derive the difference identity:
sin
(A - B) = sin A cos B - cos A sin B
-
Since sin
0 = cos ( 90o - q
)
and cos q = sin (90o
- q ), let q
= A - B, so that
sin (A - B) = cos [ 90o - (A - B)]
-
= cos [( 90o - A) + B]
-
= cos ( 90o - A) cos B - sin ( 90o - A)
sin
B
-
= sin A cos B - cos A sin B
So sin (A - B) = sin A cos B - cos A sin B
Derive the sum identity: sin
(A
+ B) = sin A cos B + cos A sin B
-
Using the difference identity:
sin (A + B) = sin (A - (-B)
-
= sin A cos (-B) - cos A sin (-B)
-
= sin A cos B + cos A sin B
Thus, sin (A + B) = sin A cos B + cos A sin B
Derive the difference identity:
tan (A - B) = tan A - tan B
1 + tan A tan B
Problems
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