# Sketching Graphs

## Homework Statement

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

## 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?

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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 $\frac{cos(x)}{x+\frac{\pi}{2}}$.

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

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).

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 ><

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.

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?

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.

Got it, that's great thanks!

No problem. For reference here's a pic of the actual curve : ) Cheers =D - btw you should use google chrome :PP