Prelim 2016 Maths Help Thread (1 Viewer)

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Simorgh

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Since our individual Maths help threads are closing, I have created this thread instead where we can all ask for help in our 2 Unit or 3 Unit questions :)
 
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leehuan

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(x-2)=cos(theta)
(y-1)=sin(theta)

(x-2)^2+(y-1)^2=1
 

Squar3root

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How would you do this?
https://scontent-syd1-1.xx.fbcdn.net/hphotos-xta1/v/t35.0-12/12941188_1672164113047290_638544438_o.jpg?oh=fc8f7d7f97e3520741ef58de5946aa9a&oe=5703728C[//img][/QUOTE]

x = 2 + cos(Ø)
cos(Ø) = (x - 2)

y = 1 + sin(Ø)
sin(‎Ø) = (y - 1)

now sin^2(Ø) + cos^2(Ø) = 1

so (x - 2)^2 + (y - 1)^2 = 1
 

Nailgun

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Don't you make theta the subject?
not really
the only way you can remove theta from the cos is by using arccos on the x-2
doesn't help you in the slightest lol

what you can do is use the trig identity of cos^2 theta + sin^2 theta =1 as shown above by square and leehuan
 

parad0xica

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How would you eliminate theta?

Methods displayed here use the retrieval (via memory) of a trigonometric identity which is good in creativity but if you want to improve mathematical problem solving skills deliberately: increase gravity.

Example, can you do this question without explicitly using this trigonometric identity?
 

Nailgun

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Methods displayed here use the retrieval (via memory) of a trigonometric identity which is good in creativity but if you want to improve mathematical problem solving skills deliberately: increase gravity.

Example, can you do this question without explicitly using this trigonometric identity?
Observing that cosθ and sinθ are in fact ratios of sides in a right angled triangle

cosθ=adj/hyp = (x-2)/1
sinθ=opp/hyp = (y-1)/1

hence you constructed a right angled triangle with hypotenuse=1, and other sides x-2 and y-1

by pythagoreas (x-2)^2+(y-1)^2 =1

doesn't this put like random limits on it though
i.e 0<θ<pi/2, and 0<x-2<1 and 0<y-1<1
 

leehuan

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It can probably be done wlog for each quadrant because when you square the negative sign loses it's impact.

Wild guess.
 

Trebla

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It can probably be done wlog for each quadrant because when you square the negative sign loses it's impact.

Wild guess.
Yeah it can be, by considering only the magnitudes of the sides of the triangle on the number plane then applying the relevant identities of each quadrant (e.g. sin(π-θ)=sin θ)). Applying Pythagoras' theorem then cleans it up, which is in essence deriving the trig identity.
 
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Green Yoda

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I can put in in the cal but Is there a formal way of doing this?
 

leehuan

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Basically tan(theta)=1 then use ASTC - first and third quadrant
i.e. theta = 45deg, theta = 180+45deg as eyeseeyou mentioned
 

Green Yoda

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A challenge question for the 3u kids:
if secθ-tanθ= 3/5
show sinθ=8/17
 

Green Yoda

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Also need help with this question:
find the value of x when
cos2θ<=1/√2, when 0<=θ<=2π
 
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