VCE Specialist 3/4 Question Thread!

[Mr. Study]

Dis-regard my last question.

Just this one.

Express sin(3x), in terms of x.

I used the general formula to get this: sin(x)cos(2x)+cos(x)sin(2x). I am probably wrong here but could someone show me how it's done?

Thanks

[kamil9876]

now apply the same thing to cos(2x) and sin(2x)  (which turns out to be the double angle formula)

[paulsterio]

[brightsky]

Express sin(3x), in terms of x. --> it's already in terms of x?

[TrueTears]

lol maybe he meant in terms of sinx

[iirene]

I request some help please
Not having much luck with this...

Simplify the following expression:
cos(4x)sin(4x)

Answer: 1/2 sin(8x)

[b^3]

(double angle formula)
Which means

[trinh]

Referring to the double angle formulae,

More generally,

So... for your question , it can be seen that



Just for the sake of clarity, referring back to the general formula , it is evident that for our specific question, , and thus

Thus

and

Edit: beaten

[iirene]

Thank you! Was stuck on that since last night.

[Mr. Study]

Ugh, This question is not very well explained.

Find the asymptotes for y-1/x^2 = x^2.

I moved the -1/x^2 over, to make it +1/x^2 and then sketched the +1/x^2 and the -x^2 on the same axis but with this book, it just wants me to find the asymptotes WITHOUT sketching.

Could someone explain how?

Thanks.

Note: Not really relevant but does anyone know of an online calculator? Similar to the CAS? My CAS calculator was nicked.

[brightsky]

y=x^2 + 1/x^2
you notice that when x --> +-infinity, y=x^2, so you have an oblique asymptote there. then we have another vertical asymptote at x = 0 for obvious reasons.

[Mr. Study]

Thanks for that.

Just this last spesh question for today.

Volume of a solid cylinder is 128pi cm^3.
Show that the total surface are, A cm^2, is A=2pir^2 + 256pi/r, where r>0.

[paulsterio]

Solve for h



Now sub into A = 2pi.r(r+h)

[b^3]

and r>0 as the radius can't be negative or 0

EDIT: Half beaten, working above

[paulsterio]

Thanks b^3, i know how to write pi in LaTeX now

(quoted your post)