c3 question..unbelievably hard

guuys im panicking.did jan 07 paper..aced it easily.....i need to get an A 2moro big time..i just started this solomon paper and havent done a question right up till now and im on quesiton 4!!!!
iim panicking so much i cant think...this is crazy..what if something like this comes up 2moro..im screwed!!!!

also this question i cant even do with the answers!!!..........
sin5x+sinx=0...
how am i suposed to do that..they seem to get some sort of a double angle formulae out but wont sin5x in double angle equate to 5sinxcosx so its 5sinxcosx+ sin x =0!!.

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Reply 1

TheTallOne

Which solomon paper?

Try split the sin(5x) up - check the formula book for C3

Reply 2

Swayum

Look in the formula book. There's a formula for sin P + sin Q which puts it into one expression.

Reply 3

username62978

there are formulae for splitting up sin(A+B)x and sin(A-B)x.

Are there numbers A and B such that A+B=5 and A-B=1?

Spoiler

Reply 4

krisstassen

not a bad idea..i thought i could open the sin5x ...like a double angle...

Reply 5

Swayum

You can open it as a double angle, but that would be retarded. It certainly wouldn't be 5sinxcosx. It'd be 2sin(2.5x)cos(2.5x). A better way would be to say

sin5x= sin(2x + 3x)

And use the sin(A+B) formula on that. And then to keep using it only you're left with terms just in terms of sinx and cosx.

Reply 6

username62978

Swayam

sin5x= sin(2x + 3x)

And use the sin(A+B) formula on that. And then to keep using it only you're left with terms just in terms of sinx and cosx.

It would be easier to use it just once on both sin5x and sinx. (with the latter using the subtraction version). For example, how would you do:

sin(619x)+sinx=0

?

Reply 7

Swayum

username62978

It would be easier to use it just once on both sin5x and sinx. (with the latter using the subtraction version). For example, how would you do:

sin(619x)+sinx=0

?

I think you missed the point of my post. I was just explaining that you wouldn't do what the OP was suggesting. I wasn't suggesting to actually do sin(2x+3x) in order to solve the question.

(edited 5 years ago)

Reply 8

generalebriety

I should point out that these questions are technically no longer on the C3 syllabus (even if they do come up occasionally in things like M4)...

Reply 9

krisstassen

Swayam ur a genius mate..

EVERYONE ELSE: Id like to point out that I got 90 on jan 2007 c3..i went to school for a revision session , got that paper and started doing that wrong as well cos my confidence was so low and im in panic mode now!!....
I think the moral of the story is that unless ur very very clever..like a 100 percentstudent..dont go anywhere near solomon press!!...i told my teacher and my other pals (expecting As too)...and they said they dont bother ...as solomon press is a waste of time cos they are sooo difficult....

Im now just gona do some past papers and get my confidence back and then rip that c3 apart 2moro!!..i hope!!

Reply 10

Doji

generalebriety

I should point out that these questions are technically no longer on the C3 syllabus (even if they do come up occasionally in things like M4)...

Since when? ...i think i've seen one in a recent paper, so just making sure when they stopped it. :smile:

Reply 11

generalebriety

Doji

Since when? ...i think i've seen one in a recent paper, so just making sure when they stopped it. :smile:

No idea, I just remember being told so. And I've never seen one in a recent paper. :s-smilie:

Reply 12

</gibberish>

username62978

It would be easier to use it just once on both sin5x and sinx. (with the latter using the subtraction version). For example, how would you do:

sin(619x)+sinx=0

?

0 = Sin(619x) + sin(x) = 2Sin(310x)Cos(309x) = 0
(Not sure about this lol)

0^2 = [2Sin(310x)Cos(309x)]^2

0 = 4Sin(310x)^2Cos(309x)^2

0 = 4Sin(310x)^2(1-Sin(309x)^2)

therefore A) 4Sin(310x)^2 = 0 &/or B)Sin(309x)^2 = 1
A) Sin(310x)^2 = 0, Sin(310x)=0, 310x = 0 x=0

B)

Sin(309x) = ±\sqrt1

etc

I need to start revising maths now though:P

(edited 5 years ago)

Reply 13

username62978

ChrisRH

0 = Sin(619x) + sin(x) = 2Sin(310x)Cos(309x) = 0
(Not sure about this lol)

0^2 = [2Sin(310x)Cos(309x)]^2

0 = 4Sin(310x)^2Cos(309x)^2

0 = 4Sin(310x)^2(1-Sin(309x)^2)

therefore A) 4Sin(310x)^2 = 0 &/or B)Sin(309x)^2 = 1
A) Sin(310x)^2 = 0, Sin(310x)=0, 310x = 0 x=0

B)

Sin(309x) = ±\sqrt1

etc

I need to start revising maths now though:P

Yes, although you could have just gone straight from 2sin(310x)cos(309x)=0 to your solutions :rolleyes:

. Be wary of squaring both sides: you have created two values where there should only have been one (only sin(309x)=1, not-1).

Reply 14

karate_kat

I wouldn't worry about the solomon press papers, they are nothing like the type of questions they will ask you. If you aced the Edexcel paper, then you'll probably be fine tomorrow. I consistently get high marks in edexcel papers but low marks in solomon press papers, so I don't do the solomon press papers any more. :p:

Reply 15

marshmallows

The weird thing I found with Solomon papers is that the core is really hard, yet the applied modules are piss easy. I managed to do m2 in 20 (TWENTY!) mins and get 100% - and I'm crap at M2.

Reply 16

maths-enthusiast

ok since threadstarter already got the solution, i would only advise him not to panic, that'd be the best piece of advice i can think of.

Reply 17

Ferchichi

generalebriety

No idea, I just remember being told so. And I've never seen one in a recent paper. :s-smilie:

Isnt the double angle formula shizzle in the c3 book?

Reply 18

generalebriety

Ferchichi

Isnt the double angle formula shizzle in the c3 book?

I'm talking about the sin P + sin Q = sin(P+Q)/2coswhatever stuff.

Reply 19

allysizzle

anyone trying to expand sin5x is insane...

sin5x + sinx =0

try: 2sin[(5x+x)/2]cos[(5x-x)/2] = 0

2sin3xcos2x = 0

so sin3xcos2x = 0

hence sin3x = 0 and/or cos2x = 0