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# Two trinity college admission questions watch

1. 1) If the two lines y=m(1)x + c(1) and y=m(2) +c(2) intersect and make an acute angle theta, show that

tan theta = modulus of m(1)m(2)/[1+m(1)m(2)]

2) Integrate (with limits of pie and 0) (xsinx)^2

help lol

i was close on the first- i can get the top but not the bottom.
2. (Original post by futureaussiecto)
1) If the two lines y=m(1)x + c(1) and y=m(2) +c(2) intersect and make an acute angle theta, show that

tan theta = modulus of m(1)m(2)/[1+m(1)m(2)]

2) Integrate (with limits of pie and 0) (xsinx)^2

help lol

i was close on the first- i can get the top but not the bottom.
For 2)
x^2.sin^2x

cos2x = 1-2sin^2x
=> sin^2x = 0.5(1-cos2x)

=> x^2sin^2x = 0.5x^2(1-cos2x) = 0.5x^2 - 0.5x^2cos2x

Does that help? Do the second bit by parts.
3. [QUOTE=Widowmaker]
cos2x = 1-2sin^2x
=> sin^2x = 0.5(1-cos2x)
[QUOTE]

if i had remembered that bit i would have got it lol
4. (Original post by futureaussiecto)
1) If the two lines y=m(1)x + c(1) and y=m(2) +c(2) intersect and make an acute angle theta, show that

tan theta = modulus of m(1)m(2)/[1+m(1)m(2)]
Are you sure that's the answer? Is it not m(1) + m(2) on the top line?

Anyway, this should help:
tan(A+B) = tanA + tanB/(1 - tanA.tanB)
5. (Original post by e-unit)
Are you sure that's the answer? Is it not m(1) + m(2) on the top line?

Anyway, this should help:
tan(A+B) = tanA + tanB/(1 - tanA.tanB)
actually its m(1)-m(2)

sowwy
6. (Original post by futureaussiecto)
actually its m(1)-m(2)

sowwy
Same thing. Do you get it now, though? You just use m1=tanθ1 and m2=tanθ2 and use an identity for tan(θ1-θ2) etc...
7. (Original post by e-unit)
Same thing. Do you get it now, though? You just use m1=tanθ1 and m2=tanθ2 and use an identity for tan(θ1-θ2) etc...
when i first saw it i remembered the addition formulae

tan = o/a = m(1)x+C(1)-m(2)x+c(2) divided by m(2)x +c2-[m(1)x + c(1) - m(2)-c(2)] the latter because the intersect is when m1+c1=m2+c2

???????
8. Look at the attached diagram

Clearly, θ = α - β

tan(θ) = tan(α - β)
= (tanα - tanβ)/(1 + tanα.tanβ)

But, m1 = tanα
and m2 = tanβ

So, tan(θ) = (tanα - tanβ)/(1 + tanα.tanβ)
= (m1 - m2)/(1 + m1m2)
9. These questions are a joke for admission into the world's best uni for maths at the most prestigious college.
10. (Original post by Nima)
These questions are a joke for admission into the world's best uni for maths at the most prestigious college.
yeh well ive hardly been doin any maths this year i have been doing physics so my maths is very rusty.

although i must admit- yes they did seem easy. heres the whole thing

http://math.mdsalih.com/Data/Oxbridg...iew%20Test.pdf

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