Hi! I'm Kyle, a 3rd year Oxford undergraduate studying Maths at Corpus Christi College. I aim to be an understanding and friendly face, whilst being thorough with the material I cover. I've acheived A* grades in Maths, Further Maths and Additional Further Maths, plus I have a decent amount of volunteering experience, a lot of which working with kids of many ages.

Hi! I'm Kyle, a 3rd year Oxford undergraduate studying Maths at Corpus Christi College. I aim to be an understanding and friendly face, whilst being thorough with the material I cover. I've acheived A* grades in Maths, Further Maths and Additional Further Maths, plus I have a decent amount of volunteering experience, a lot of which working with kids of many ages.

I'll make sure the lesson is tailored to exactly what you want to cover. Whether that be going through a concept for the first time, or going through past exam questions. I want to make sure it's the best experience possible for you! I hope to make the sessions enjoyable for both you and I! Work should be about enjoying what you study and I hope to be able to instill a little of my passion in you!

I'll make sure the lesson is tailored to exactly what you want to cover. Whether that be going through a concept for the first time, or going through past exam questions. I want to make sure it's the best experience possible for you! I hope to make the sessions enjoyable for both you and I! Work should be about enjoying what you study and I hope to be able to instill a little of my passion in you!

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28/09/20165from 207 customer reviews

Naa (Parent)

October 9 2017

Very good

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June 3 2017

all good

Martin (Parent)

February 26 2017

Really good lesson! Learnt loads, thanks Kyle!

Naa (Parent)

February 20 2017

Was a great tutor moved at a reallly fast pace. Was enthusiastic learnt loads!!!!!! :)

Using integration by parts, we can re-write the integral of ln(x) as (x*ln(x) - int(x*(1/x))) = x*ln(x) - x

Therefore, evaluating between 2 and 4 gives us (4*ln(4) - 4) - (2*ln(2) - 2) = 2ln(16/2) - 4 + 2 = ln(64) - 2. So a = 64 and b = 2

Using integration by parts, we can re-write the integral of ln(x) as (x*ln(x) - int(x*(1/x))) = x*ln(x) - x

Therefore, evaluating between 2 and 4 gives us (4*ln(4) - 4) - (2*ln(2) - 2) = 2ln(16/2) - 4 + 2 = ln(64) - 2. So a = 64 and b = 2

First we state the formula for sin(x+y)

sin(x+y) = sin(x)cos(y) + cos(x)sin(y)

Letting y = 2x

sin(x+2x) = sin(x)cos(2x) + cos(x)sin(2x)

Now sin(2x) = 2sin(x)cos(x) and cos(2x) = 1 - 2sin^2(x), substitute these into the formula gives us

sin(3x) = sin(x)(1-2sin^2(x)) + cos(x)(2sin(x)cos(x))

sin(3x) = sin(x) - 2sin^3(x) + 2sin(x)cos^2(x)

Now cos^2(x) = 1 - sin^2(x)

sin(3x) = sin(x) - 2sin^3(x) + 2sin(x)(1-sin^2(x))

sin(3x) = sin(x) - 2sin^3(x) + 2sin(x) - 2sin^3(x)

sin(3x) = 3sin(x) - 4sin^3(x)

Now letting x = 10, we get

sin(30) = 3sin(10) - 4sin^3(10)

Rearranging and evaluation sin(30) = 1/2

8sin^3(10) - 6sin(10) + 1 = 0

Hence sin(10) is a root of the cubic equation

First we state the formula for sin(x+y)

sin(x+y) = sin(x)cos(y) + cos(x)sin(y)

Letting y = 2x

sin(x+2x) = sin(x)cos(2x) + cos(x)sin(2x)

Now sin(2x) = 2sin(x)cos(x) and cos(2x) = 1 - 2sin^2(x), substitute these into the formula gives us

sin(3x) = sin(x)(1-2sin^2(x)) + cos(x)(2sin(x)cos(x))

sin(3x) = sin(x) - 2sin^3(x) + 2sin(x)cos^2(x)

Now cos^2(x) = 1 - sin^2(x)

sin(3x) = sin(x) - 2sin^3(x) + 2sin(x)(1-sin^2(x))

sin(3x) = sin(x) - 2sin^3(x) + 2sin(x) - 2sin^3(x)

sin(3x) = 3sin(x) - 4sin^3(x)

Now letting x = 10, we get

sin(30) = 3sin(10) - 4sin^3(10)

Rearranging and evaluation sin(30) = 1/2

8sin^3(10) - 6sin(10) + 1 = 0

Hence sin(10) is a root of the cubic equation