Prelim 2016 Maths Help Thread (2 Viewers)

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leehuan

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I did not think of doing sin^2(θ)-cos^2(θ)
Thank you so much :)
Would this be better (sin(2θ)+1)/(cos(2θ))=(cos(θ)+sin(θ))/(cos(θ)-sin(θ))?
Legit question, How do i get better at further trig? I dont know why I am so bad...:/
Also yes those brackets are MUCH better

That one around the cos(2θ) wasn't necessary though cause it was the ONLY thing on the denominator

Rathin, I'm sure I'm not the first to mention this but you REALLY need to use brackets.

That expression is too ambiguous for me to decipher instantaneously.

That being said...



And the result follows through from cancellation.

And of course also the cosine double angle identity
 

Green Yoda

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lol soz for this but I am gonna post a couple more questions today which I have found hard..but I'll go one by one
Prove:
(1-cos(x))/sin(x) = tan(x/2)
My working out so far:
(1/sin(x))-(cos(x)/sin(x))
=cosec(x)-cot(x)
but that leads to nothing..and I can't see any other way.
 

Paradoxica

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lol soz for this but I am gonna post a couple more questions today which I have found hard..but I'll go one by one
Prove:
(1-cos(x))/sin(x) = tan(x/2)
My working out so far:
(1/sin(x))-(cos(x)/sin(x))
=cosec(x)-cot(x)
but that leads to nothing..and I can't see any other way.
half angle
 

InteGrand

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lol soz for this but I am gonna post a couple more questions today which I have found hard..but I'll go one by one
Prove:
(1-cos(x))/sin(x) = tan(x/2)
My working out so far:
(1/sin(x))-(cos(x)/sin(x))
=cosec(x)-cot(x)
but that leads to nothing..and I can't see any other way.
Use the identities sin^2 (x/2) = (1/2) (1 - cos(x)) and sin(x) = 2 sin(x/2) cos(x/2). I.e. Half-angle formulas, as Paradoxica alluded to.

(Surely they didn't forbid use of these?)
 

Green Yoda

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Use the identities sin^2 (x/2) = (1/2) (1 - cos(x)) and sin(x) = 2 sin(x/2) cos(x/2). I.e. Half-angle formulas, as Paradoxica alluded to.

(Surely they didn't forbid use of these?)
Haven't coved half angle in class yet..
also I am aware that sin^2(x)=1-cos^2(x)
but how does sin^2*x/2)=1/2*(1-cos(x))?
 

KingOfActing

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Expand cos(2x), write it in terms of just sin^2(x), then just replace x with x/2

Edit: sorry for lack of LaTeX I'm using Tapatalk or whatever and it doesn't render LaTeX + it's a pain to type on my phone
 

Green Yoda

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for the question
prove:
(sin(x)+1-cos(x))/sin(x)-1+cos(x) = (1+tan(x/2))/1-tan(x/2) w
would I do the same thing?
 

InteGrand

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haven't learnt that yet..should be learning when school starts. This one is under special angle formulas exercise so It can be done without?
Yeah, you can do it via half-angle formulas (convert sin(x) and stuff into trig. functions with x/2 instead of x).
 
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Green Yoda

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Yeah, you can do it via half-angle formulas (convert sin(x) and stuff into trig. functions with x/2 instead of x).
I have tried learning the t-formula via YouTube and I am curious how would you do this question by using t formula?
 

InteGrand

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I have tried learning the t-formula via YouTube and I am curious how would you do this question by using t formula?
Using t-formulas, the RHS is (1+t)/(1-t), provided t =/= 1. You can also express the LHS in terms of t using the t-formulas, and show it simplifies to (1+t)/(1-t).
 

Green Yoda

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LHS:
((2t/1+t^2)+1-(1-t^2/1+t^2))/(2t/1+t^2)-1+(1-t^2/1+t^2) yeah??
 
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