2 = 
2 + 
2 – 2 
cos 
Therefore
2 = 
2 + 
2 – 2 
cos 
02. USED TO FIND THE PROJECTION OF ONE VECTOR ONTO ANOTHER VECTOR
– Suppose the projection of a onto b
i.e
Cos Q = 
Cos Q = 
proj 
= 
cos Q ———– i
Also
. 
= 
cos Q
= 
cos Q ———–ii
Equalizing i and ii as follows;
proj 
= 
cos Q = 
proj 
= 
similary
proj 
= 
Where;
Proj 
= projection of 
onto 
Proj 
= projection of 
onto 
VECTOR PROJECTION
This is given by
V proj
=
. 
And
V. proj
=
. 
Where;
V. proj 
= vector projection of 
onto 
V. proj 
= vector projection of 
onto 
03. TO FIND THE WORKDONE
– Consider the diagram below
Force applied (F) = component of tone
cos Q = 
Cos Q = 
F = 
Cos Q
Also
Distance d = 
d = 
….ii
Hence
Work done = Force applied (F) x distance (d)
W.D = F x d
W. D = 
cos Q x 
= 
cos Q
= 
. 
W.D = 
Note
i) Force F in the direction of vector 
Force applied = F.
ii) Distance in the direction of vector 
Displacement 
= d.
Individual
i) 
= 2i + 
+ 2
= i + 
+ 
W.D = 
04. TO PROVE COMPOUND ANGLE FORMULA OF COSINE
i.e cos (A + B) = cos A cos B – sin A sin B
– consider the vector diagram below.
Diagram
= (cos A) 
+ (sin A) 
= (cos B) 
– (sin B) 
Hence
= 
cos (A + B)
but
. 
= 
= cos A cos B + -sin A sin 0
= cos A cos B – sin A sin B
Also
= 
= 
= 
= 1
= 
= 
= 
= 1
Therefore
Cos A cos B – sin A sin B = (1)(1) cos (A + B)
Cos (A + B) = cos A cos B – sin A sin B
Proved
Pg. 2 drawing
= (cos A) 
+ (sin A) j
= (cos B) I + (sin B) 
Hence
. 
= 
cos (A – B)
But
. 
= 
. 
= cos A cos B + sin A sin B
Also
= 
= 
= 
= 1
= 
= 
= 
= 1
There
Cos A cos B + sin A sin B = (1) (1) cos (A – B)
Cos (A – B) = cos A cos B + sin A sin B
Proved