-
Finished your HSC? Check out our University Discussion, Non-School forums or help out next year's HSC students! -
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page
HSC Mathematics Marathon (1 Viewer)
- Thread starter ohexploitable
- Start date
ohexploitable
hey
Drongoski, are those identities in the HSC course?
I'm talking about,
 \cos\left( \frac{\theta \mp \varphi}{2} \right))
etc.
I'm talking about,
etc.
ninetypercent
ninety ninety ninety
^ yes and no. typical exam questions will have part (i) to show the identity and then in part (ii) you use it
ohexploitable
hey
Anyway, new question:
"hscishard" and "bored of fail", you're not allowed to answer this
"hscishard" and "bored of fail", you're not allowed to answer this
hscishard
Active Member
Lol. Good luck to whoever attempts that question ^
4025808
Well-Known Member
it used to be in the syllabus but they took it off...Drongoski, are those identities in the HSC course?
I'm talking about,
etc.
nevertheless it's still useful to learn esp in trig expansion
Yes = not directly. But if I were doing Ext 2 I will try to add it to my arsenal. Useful in some of the messier derivations in conics (see P89, Cambridge 4U book)
I'll give it a crack. I remember seeing your solution/hscishard's solution (whichever one of you uploaded the handwritten answer) but can't really remember it LOL. Might not end well.Anyway, new question:
"hscishard" and "bored of fail", you're not allowed to answer this
ohexploitable
hey
It was mine
Anyway, new question:
"hscishard" and "bored of fail", you're not allowed to answer this
Last edited:
Factorise: 1-(sinx)^10=(1-(sinx)^2)(1+(sinx)^2+(sinx)^4+(sinx)^6+(sinx)^8)Find
=(cosx)^2 (1+(sinx)^2+(sinx)^4+(sinx)^6+(sinx)^8)
integ (cosx)(1+(sinx)^2+(sinx)^4+(sinx)^6+(sinx)^8) dx
let u=sinx etc.
integ (1+u^2+u^4+u^6+u^8) du etc.
hscishard
Active Member
Why are uni students dominating this thread?
ohexploitable
hey
I've noticed that too...
Nice work...though I was hoping someone would pick it up more in the lines of a geometric series
}{\cos^2x}\\\\=\dfrac{\cos x(1-(\sin^2x)^5)}{1-\sin^2x}\\\\=\cos x+\cos x\sin^2x+\cos x\sin^4x+\cos x\sin^6x+\cos x\sin^8x)
Trebla
Interesting question & beautiful solution. I thought I knew how to integrate. Ends up similar to jyu's.
UnrealAnchovies
Member
- Joined
- Apr 3, 2010
- Messages
- 777
- Gender
- Male
- HSC
- 2011
What's the likelihood of that appearing in the 4U HSC exam?Find
Probably like 15% I'd say...