Trig
posted by sally .
A. Find simpler, equivalent expressions for the following. Justify your answers.
(a) sin(180 + è) (b) cos(180 + è) (c) tan(180 + è)
B. Show that there are at least two ways to calculate the angle formed by the vectors
[cos 19, sin 19] and [cos 54, sin 54].

I will do a)
sin(180 + è)
= sin180cos è + cos180sin è
= 0 + (1)sin è
= sin è
for the other two, you will have to know the expansion for cos(180+ è) and tan(180+ è)
B) [cos 19, sin 19]•[cos 54, sin 54] = [cos 19, sin 19][cos 54, sin 54]cos Ø, where Ø is the angle between
cos19cos54 + sin19sin54 = 1x1cosØ
cos(1954) = cosØ
Ø = 1954
= 35°
second way: vector [cos 19, sin 19] makes an angle P with the xaxis such that tan P = sin19/cos19
tan P = tan 19
P = 19
similarly the second vector makes an angle of 54° with the xaxis
so the angle between them is 35° 
A. use sum of angles formulae:
sin(180+α)
=sin(180)cos(α)+cos(180)sin(α)
=0.cos(α)+(1)sin(α)
=sin(α)
cos(180+α)
=cos(180)cos(α)sin(180)sin(α)
=(1)cos(α)(0)sin(α)
=cos(α)
for tan(180+α)
use (tan A + tan B)/(1  (tan A)(tan B))
B.
Since both vectors are unit vectors, the cosine of the included angle is simply the dot product of A(cos(a),sin(a)) and B(cos(b),sin(b)):
cos(θ)
=(cos(a)cos(b)+sin(a)sin(b))/(A B)
=(cos(ab))/(1 1)
=cos(ab)
therefore θ=ab or (ab)
The magnitude of the cross product of the two vectors represent the positive area of a parallelogram formed by the two vectors as adjacent sides, namely,
Area=ABsin(θ)=A x B
sin(θ)
=cos(a)sin(b)cos(b)sin(a)
=sin(ab)
therefore
θ=ab
Also, since the terminal points of the vectors A and B represent points on the unit circle at angles a and b from the xaxis, we conclude that the angle between the vectors is ab.