Sketch Graph: Tips for \frac{cos x}{x + \frac{\pi}{2}}

[Chewy0087]

Homework Statement

Sketch \frac{cos x}{x + \frac{\pi}{2}}

Homework Equations

The Attempt at a Solution

hey guys, I just need help sketching this graph, I can work out the points of intercept with the y-axis however having trouble doing more than that.

On a more general note, I generally find sketching graphs such as this difficult, are there any tips you can give me? I suppose I could try to work out turning points by differentiating however is there a better way?

Thanks in advance

 

[marmoset]

It might help to make a substitution u=x+\frac{1}{2}, then you can sketch \frac{sin(u)}{u}, which is easier, then translate it to get \frac{cos(x)}{x+\frac{\pi}{2}}.

 

[Chewy0087]

Ah, i get the sin bit, that's clever =P, (sorry we havn't been taught any of this yet, took me ages to see it).

But as far as sketching sin (u) / u how would you even sketch that? :s

 

[marmoset]

I guess first think about whether it is even or odd, then think about what happens for large values of x, to get a general idea. Then you could consider what happens for x approaching 0 (try using the small angles approximation for sin, or you could use l'hopital's rule).

 

[Chewy0087]

I think I see, yeah =o, one final thing though, how would you translate from sin(u)/u to get to the original? Just move it pi/2 to the left? However i don't understand why having x + pi/2 will translate it like that, because it's it effectively;

f(x) = sin (u) / u so to translate to cos (x) / x + pi/2 wouldn't it be like...

f(x - pi/2) or something? I don't get it >.<

Sureley it's an f(x + pi/2) translation but that would mean that the graph you need...eh i don't get this ><

 

[marmoset]

If you've just sketched \frac{sin(u)}{u}, call this f(u), then you can easily sketch \frac{sin(u+\frac{\pi}{2})}{u+\frac{\pi}{2}}, because this is just f(u+\frac{\pi}{2}).

But sin(u+\frac{\pi}{2})=cos(u), so

\frac{sin(u+\frac{\pi}{2})}{u+\frac{\pi}{2}} = \frac{cos(u)}{u+\frac{\pi}{2}}

This is just the function you were trying to sketch, but with the variable called u instead of x.

 

[Chewy0087]

Oh i see, that's great thanks, however would it not be -sin(u) / u then? because at the start substituting x + pi/2 into x for cos would move it <-- by pi/2 making it -sin then?

 

[marmoset]

When you substitute u = x + pi/2, you are actually translating the curve to the pi/2 to the right.

Keeping everything in terms of x, you started with \frac{cos x}{x + \frac{\pi}{2}}, call this f(x), then you sketched \frac{sin(x)}{x}, which is \frac{cos (x-\frac{\pi}{2})}{(x-\frac{\pi}{2}) + \frac{\pi}{2}}, i.e. f(x-\frac{\pi}{2}), i.e. f(x) translated \frac{\pi}{2} to the right.

Overall you took your function, translated it to the right, sketched it, and then translated it back to the left.

 

[Chewy0087]

Got it, that's great thanks!

 

[marmoset]

No problem. For reference here's a pic of the actual curve : )

 

[Chewy0087]

Cheers =D - btw you should use google chrome :PP