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Mega A Level Maths Thread MK II

Why do all my friends know my examination timetable better than me LOL, they're like 'GOOD LUCK FOR MATHS!!!' and i'm like urm..so whens your next exam? :doh:

Reply 2301

nm786

2

Original post by Xx4L3x

5cosθ − 2sinθ = 0,

prove tanx = 2.5

divide through by cosθ bring -2tanθ to the other side and finally divide everything by 2.

(edited 10 years ago)

Reply 2302

justinawe

19

Original post by Xx4L3x

5cosθ − 2sinθ = 0,

prove tanx = 2.5

tanx or tanθ?

People seem to have been posting this same question a lot lately :redface:

I'd have thought it was fairly simple

5\cos \theta - 2 \sin \theta = 0

5\cos \theta = 2\sin \theta

\dfrac{\sin \theta}{\cos \theta} = \dfrac{5}{2}

Reply 2303

tigerz

5\cos \theta = 2\sin \theta , \tan \theta =\frac{sin}{cos} \therefore 2\frac{sin\theta }{cos\theta }=5

\rightarrow 2tan\theta =5 \therefore tan\theta =\frac{5}{2}

(edited 10 years ago)

Reply 2304

L'Evil Fish

OP

18

Original post by Xx4L3x

5cosθ − 2sinθ = 0,

prove tanx = 2.5

Add 2sin theta to both sides
Divide by cos theta
Divide by 2

Tan.theta = 2.5

Reply 2305

MathsNerd1

17

When I get home I'm gonna try all these questions out! I've missed doing Maths quite a lot for the past 4 days but I'm now back! :biggrin:

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Reply 2306

Jkn

11

Original post by Felix Felicis

Old but gold. :awesome:

Spoiler

Hmm, interesting result...

Reply 2307

Jkn

11

Original post by Felix Felicis

Old but gold. :awesome:

Spoiler

Let

S=-1^2+2^2-3^2+4^2-...-(n-1)^2+n^2

, where n is arbitrarily chosen to be even.

\Rightarrow S=(2^2-1^2)+(4^2-3^2)+...(n^2-(n-1)^2)[br]\Rightarrow S=(1+2)+(3+4)+...((n-1)+n)[br]\Rightarrow S=1+2+3+4+...+(n-1)+n[br]\Rightarrow S+1=1+2+3+4+...+(n-1)+n-(n+1)+(n+2)[br]\Rightarrow S+a=1+2+3+4+...+(n-1)+n-(n+1)+(n+2)-(n+3)+...-(n+2a-1)+(n+2a)

,
so all S can be constructed by altering the signs of the original series of differences of squares.

I'm too tired and busy to be rigorous and consider/write-up different case as well as part 2! :lol:

But is that the bulk of it? (i.e. have I done something stupid? I almost-definitely have)

Reply 2308

Chloe_P

1

Original post by GCSE-help

Are we all anticipating a hard C2 paper tomorrow?

Can C2 ever be made hard? :s-smilie:

Reply 2309

joostan

13

Original post by Chloe_P

Can C2 ever be made hard? :s-smilie:

Would you like to find out? :evil:

Reply 2310

tigerz

Original post by Chloe_P

Can C2 ever be made hard? :s-smilie:

Original post by joostan

Would you like to find out? :evil:

I suggest you don't ask that, especially in this forum :innocent:

Reply 2311

Scientific Eye

Original post by Jkn

[br]\Rightarrow S+1=1+2+3+4+...+(n-1)+n-(n+1)+(n+2)[br]\Rightarrow S+a=1+2+3+4+...+(n-1)+n-(n+1)+(n+2)-(n+3)+...-(n+2a-1)+(n+2a)

,
so all S can be constructed by altering the signs of the original series of differences of squares.

I'm too tired and busy to be rigorous and consider/write-up different case as well as part 2! :lol:

But is that the bulk of it? (i.e. have I done something stupid? I almost-definitely have)

I don't think doing that works because you have pairs of numbers switching their signs, not just one number as you seem to suggest above. So by changing the signs of the differences of