Saturday

April 19, 2014

April 19, 2014

Posted by **sally** on Saturday, April 17, 2010 at 5:58pm.

(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].

- Trig -
**Reiny**, Saturday, April 17, 2010 at 7:35pmI 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(19-54) = cosØ

Ø = |19-54|

= 35°

second way: vector [cos 19, sin 19] makes an angle P with the x-axis such that tan P = sin19/cos19

tan P = tan 19

P = 19

similarly the second vector makes an angle of 54° with the x-axis

so the angle between them is 35°

- Trig -
**MathMate**, Saturday, April 17, 2010 at 7:49pmA. 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(a-b))/(|1| |1|)

=cos(a-b)

therefore θ=a-b or -(a-b)

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(a-b)|

therefore

θ=|a-b|

Also, since the terminal points of the vectors A and B represent points on the unit circle at angles a and b from the x-axis, we conclude that the angle between the vectors is |a-b|.

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