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C3 Rcos(x-a) question

Okay so I have learned this a few months back and trying to do a question now it has completely got me.

My first question is why does

Rcos(x-a) = R(cosxcosa + sinxsina)

I cannot see why there is a change of sign and even why the is two sins and cos'?
Reply 1
Avatar for f1mad
f1mad
13
It's an identity:

cos(A+B)= cosAcosB-sinAsinB.
Reply 2
Avatar for jimbo lad
jimbo lad
OP
2
Original post by f1mad
It's an identity:

cos(A+B)= cosAcosB-sinAsinB.


let me guess , its in the formulae book and I havent bothered or even thought to look in there? haha
Reply 3
Avatar for TenOfThem
TenOfThem
18
Original post by jimbo lad
Okay so I have learned this a few months back and trying to do a question now it has completely got me.

My first question is why does

Rcos(x-a) = R(cosxcosa + sinxsina)

I cannot see why there is a change of sign and even why the is two sins and cos'?


This is a bit like asking why does

3+4 = 8-1

It just does
Reply 4
Avatar for f1mad
f1mad
13
Original post by jimbo lad
let me guess , its in the formulae book and I havent bothered or even thought to look in there? haha


It is.
Reply 5
Avatar for bananarama2
bananarama2
Original post by TenOfThem
This is a bit like asking why does

3+4 = 8-1

It just does


Except the former can be proved....
Reply 6
Avatar for TenOfThem
TenOfThem
18
Original post by wcp100
Except the former can be proved....


only by using the latter (oe)
Reply 7
Avatar for ztibor
ztibor
10
Original post by jimbo lad
Okay so I have learned this a few months back and trying to do a question now it has completely got me.

My first question is why does

Rcos(x-a) = R(cosxcosa + sinxsina)

I cannot see why there is a change of sign and even why the is two sins and cos'?


This identity is come from the rotation of the unit vectors of the coordinate system bby angle of x then by angle of a. In the plane the relation beween the coordinates of the rotated vectors and that of in the original coordinate system may be described with
a rotational matrix ( this is a linear transformation). Multiplying this matric by a vector you get the coordinates of the rotated vector.
For rotating by angle x the rotational matrix is
∣cosx−sinxsinxcosx∣\begin {vmatrix} cosx & -sinx \\sinx &cosx \end {vmatrix}
For rotating subsequently by x then by a means the multiplication of the rotation matrice.
so
∣cosx−sinxsinxcosx∣⋅∣cosa−sinasinacosa∣=∣cos(x+a)−sin(x+a)sin(x+a)cos(x+a)∣\begin {vmatrix} cosx & -sinx\\ sinx & cosx \end {vmatrix} \cdot \begin {vmatrix}cos a & -sina \\ sin a & cos a \end {vmatrix} =\begin {vmatrix} cos(x+a) & -sin(x+a) \\ sin(x+a) & cos(x+a) \end {vmatrix}
with matrix multipilication we get
∣cosx⋅cosa−sinx⋅sina−cosx⋅sina−sinx⋅cosasinx⋅cosa+cosx⋅sina−sinx⋅sina+cosx⋅cosa∣=\begin {vmatrix} cosx\cdot cos a -sinx \cdot sina & -cosx \cdot sina - sinx \cdot cosa \\sinx\cdot cosa +cosx\cdot sina & -sinx\cdot sina +cosx\cdot cosa \end {vmatrix}=
=∣cos(x+a)−sin(x+a)sin(x+a)cos(x+a)∣=\begin {vmatrix} cos(x+a) & -sin(x+a)\\ sin(x+a) & cos(x+a) \end {vmatrix}
From this you can read down the identities
For e.g cos(x-a) consider cos(x+(-a)) and that cos(-a)=cosa and sin(-a)=-sina
(edited 11 years ago)
Reply 8
Avatar for bananarama2
bananarama2
Original post by TenOfThem
only by using the latter (oe)


That's true. I fear this could lead to a very philosophical debate :smile:
Reply 9
Avatar for TenOfThem
TenOfThem
18
Original post by wcp100
That's true. I fear this could lead to a very philosophical debate :smile:


:biggrin:

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