Solving for p and q: A Graph of 5 = cos 0 and 1 = cos 180

[thomas49th]

Homework Statement

I have a graph of 5 = cos 0 and 1 = cos 180. The question is find the values for integers p and q
y = p + q cos x

p and q are integers

How do I go about doing this

Thx

 

[Gib Z]

Excuse me? I don't think anyone will understand that...5= cos 0? Not only is that not true..how you graph something like that is out of the question, similarly to the 1=cos 180. State the question more clearly please.

 

[thomas49th]

sorry. I mean the graph cos x starts at y = 1 when x = 0 and goes into a bucket like shape. The graph I'm given is the same as y = cos x EXCEPT that it's when x = 0 y = 5 and when x = 180 y = 1... so the lowest value of y is 1 and it's highest is 5, if you see what i mean?

Thx

 

[VietDao29]

sorry. I mean the graph cos x starts at y = 1 when x = 0 and goes into a bucket like shape. The graph I'm given is the same as y = cos x EXCEPT that it's when x = 0 y = 5 and when x = 180 y = 1... so the lowest value of y is 1 and it's highest is 5, if you see what i mean?

Thx

Ok, so when x = 0, then y = 5, it means that:
5 = p + q cos(0^o) = p + q

When x = 180^o, y = 1, that means:
1 = p + q cos(180^o) = p - q

So you have 2 equations:
[tex]\left\{ \begin{array}{ccc} 5 & = & p + q \\ 1 & = & p - q \end{array} \right.[/tex]
From the system of equations above, can you solve for 2 unknowns, namely, p, and q?
Can you go from here? :)

 

[thomas49th]

similtaneous equations
p = 1 + q

so 5 = 1 + q + q
so q = 2
therefore p = 3

right?

 

[VietDao29]

similtaneous equations
p = 1 + q

so 5 = 1 + q + q
so q = 2
therefore p = 3

right?

Perfectly correct. Congratulations. :)