The 2013 STEP Prep Thread!

I am trying question 4 on STEP 1994, but I can't seem to get the correct answer.

You have to show that tan(0.5x)=(1-cosx)/sinx
So cos(0.5x)=sqrt(0.5(1+cosx)) and sin(0.5x)=sqrt(0.5(1-sinx))
dividing sine by cos, you get
tan(0.5x)=sqrt((1-sinx)/(1+cosx)). I then multiplied top and bottom by sqrt(1-cosx)
this then gives me tan(0.5x)=(sqrt((1-sinx)(1-cosx)))/sinx
This is not the given result so can someone tell me where I have gone wrong. I don't really know how to use latex properly, so I can clarify on some points if you can't understand it.

Don't worry, I know where I went wrong. I used the wrong formula for sin(0.5x). :facepalm:

(edited 11 years ago)

Reply 583

OP

Original post by brittanna

I am trying question 4 on STEP 1994, but I can't seem to get the correct answer.

You have to show that tan(0.5x)=(1-cosx)/sinx
So cos(0.5x)=sqrt(0.5(1+cosx)) and sin(0.5x)=sqrt(0.5(1-sinx))
dividing sine by cos, you get
tan(0.5x)=sqrt((1-sinx)/(1+cosx)). I then multiplied top and bottom by sqrt(1-cosx)
this then gives me tan(0.5x)=(sqrt((1-sinx)(1-cosx)))/sinx
This is not the given result so can someone tell me where I have gone wrong. I don't really know how to use latex properly, so I can clarify on some points if you can't understand it.

\cos \dfrac{x}{2}= \sqrt{\dfrac{1+\cos x}{2}}

is not always true and

\sin \dfrac{x}{2}\neq \sqrt{\dfrac{1-\sin x}{2}}

Where are you getting these identities from? :curious:

(edited 11 years ago)

Reply 584

ian.slater

12

Original post by brittanna

I am trying question 4 on STEP 1994, but I can't seem to get the correct answer.

You have to show that tan(0.5x)=(1-cosx)/sinx
So cos(0.5x)=sqrt(0.5(1+cosx)) and sin(0.5x)=sqrt(0.5(1-sinx))
dividing sine by cos, you get
tan(0.5x)=sqrt((1-sinx)/(1+cosx)). I then multiplied top and bottom by sqrt(1-cosx)
this then gives me tan(0.5x)=(sqrt((1-sinx)(1-cosx)))/sinx
This is not the given result so can someone tell me where I have gone wrong. I don't really know how to use latex properly, so I can clarify on some points if you can't understand it.

The substitution t = tan(x/2) is an interesting one. Ignoring the present question for the minute, you might try to express cos(x) in terms of t. Then do sin(x). Then draw yourself a triangle.

Reply 585

Original post by Lord of the Flies

\cos \dfrac{x}{2}\neq \sqrt{\dfrac{1+\cos x}{2}}

(same for sin)

Where are you getting that from? :curious:

My sine one was wrong, it should have been 1-cosx instead of 1-sinx

If you use the double formulae, you can get them.

So cos2x=2cos^2(x) - 1
Now let x=x/2
Then you get cosx=2cos^2(x/2) - 1
If you rearrange for cos(x/2), you get cos(x/2)=sqrt(0.5(1+cos(x)))
You can do the same for sine as well, except it is 1-cosx instead of 1+cosx

Reply 586

OP

Original post by Aurum

It's because

\sin \frac{x}{2} \not= \sqrt{\frac{1- \sin x}{2}}

, it's equal to

\sqrt{\frac{1- \cos x}{2}}

Only for

0\leq x\leq 2\pi\pod{4\pi}

.

(edited 11 years ago)

Reply 587

OP

Original post by brittanna

My sine one was wrong, it should have been 1-cosx instead of 1-sinx

If you use the double formulae, you can get them.

So cos2x=2cos^2(x) - 1
Now let x=x/2
Then you get cosx=2cos^2(x/2) - 1
If you rearrange for cos(x/2), you get cos(x/2)=sqrt(0.5(1+cos(x)))
You can do the same for sine as well, except it is 1-cosx instead of 1+cosx

What I meant was: the cos is not always true. It depends on x. You can't take the square root without saying something about x... (see above)

(edited 11 years ago)

Reply 588

music lover

I'm trying to do part 3 of question 11 in the Advanced Mathematics booklet: I got up to the 2nd last part I.E.

-\int^{\frac{1}{b}}_\frac{1}{a} \frac{1}{{\sqrt(1-xa)(xb-1)}}dx

. The square root is over the whole Denominator.

I can't seem to get to the next step

-\int^{\frac{1}{b}}_\frac{1}{a} \frac{1}{\sqrt(ab)(x-a^{-1})(b^{-1}-x)}dx=\frac{\pi}{\sqrt ab}

Reply 589

Original post by Lord of the Flies

Only for

0\leq x\leq 2\pi\pod{4\pi}

.

I'm probably being stupid, but where does that range come from in relation to this question. Thank you.

Reply 590

OP

Original post by brittanna

I'm probably being stupid, but where does that range come from in relation to this question. Thank you.

What I am pointing out is, it is wrong to write:

\cos x=2\cos^2 \dfrac{x}{2}-1\Rightarrow \cos\dfrac{x}{2}=\sqrt{\dfrac{ \cos x +1}{2}}

This is equivalent to writing

a^2=b^2\Rightarrow a=b

In the post you quoted, I specified for what range the sine equality is true. For any x not in that range, it isn't true, because sin is negative and therefore you would have to add a minus in front of the square root. Clear? :smile:

(edited 11 years ago)

Reply 591

Original post by Lord of the Flies

What I am pointing out is, it is wrong to write:

\cos x=2\cos^2 \dfrac{x}{2}-1\Rightarrow \cos\dfrac{x}{2}=\sqrt{\dfrac{ \cos x +1}{2}}

This is equivalent to writing

a^2=b^2\Rightarrow a=b

In the post you quoted, I specified for what range the sine equality is true. For any x not in that range, it isn't true, because sin is negative and therefore you would have to add a minus in front of the square root. Clear? :smile:

Yeah, thank you. :smile:

Reply 592

Appeal to reason

14

After some halo to distract me, I came back to it, and I *think* I've done it.
Will scan it up later, I'm not good enough with tex to type it all in.

Reply 593

M4ths

Original post by Lord of the Flies

Updated. I find it amusing they compare a Distinction to a 1 though.

They don't want to be considered an insurance for Imperial I guess (this is what I was told in a one to one conversation with the admissions tutor). This may not necessarily be the reason of the increase in entry requirements.

Reply 594

OP

Original post by shamika

...

Any chance you could upload some of that stuff you have for bored people? :smile:

Reply 595

31541

2

Original post by Blazy

Nerve wracking, 5 people had an offer for Cambridge Maths in the year above and only one made it! Slightly scared at the prospect of applying now aha...

How were their academics?

Reply 596

und

Original post by Blazy

Nerve wracking, 5 people had an offer for Cambridge Maths in the year above and only one made it! Slightly scared at the prospect of applying now aha...

Do you go to my school?! Six people had an offer and one made it in mine.

Reply 597

Oromis263

2

Original post by und

Do you go to my school?! Six people had an offer and one made it in mine.

Ignoring the possible sarcasm, I think you should consider if maths is the best thing for you.

5 =/= 6 :wink:

That being said, I was trying to work out my A2 maths + fm ums combined percentage, and apparently (98+100)/2 is 97% :colonhash:

Reply 598

Blazy

8

Original post by CD315

How were their academics?

Very good by the sounds of things, 90+ in most their modules.

Original post by und

Do you go to my school?! Six people had an offer and one made it in mine.

Nope, don't live in London :smile:

- teacher tried to persuade me to apply Oxford instead!

Reply 599

und