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Kyle R.

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Mathematics (Masters) - Oxford, Corpus Christi College University

5.0
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207 reviews

This tutor is also part of our Schools Programme. They are trusted by teachers to deliver high-quality 1:1 tuition that complements the school curriculum.

221 completed lessons

About me

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.  

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About my sessions

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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Personally interviewed by MyTutor

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

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Ratings & Reviews

5from 207 customer reviews
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Naa (Parent from Harpenden)

October 9 2017

Very good

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Alisiya (Student)

June 3 2017

all good

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Martin (Parent from Sandbach)

February 26 2017

Really good lesson! Learnt loads, thanks Kyle!

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Naa (Parent from Harpenden)

February 20 2017

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

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Qualifications

SubjectQualificationGrade
MathematicsA-level (A2)A*
Further MathematicsA-level (A2)A*
Additional Further MathematicsA-level (A2)A*
PhysicsA-level (A2)A
ChemistryA-level (A2)A

General Availability

Pre 12pm12-5pmAfter 5pm
mondays
tuesdays
wednesdays
thursdays
fridays
saturdays
sundays

Subjects offered

SubjectQualificationPrices
Further MathematicsA Level£30 /hr
MathsA Level£30 /hr
MathsGCSE£30 /hr
MathsIB£30 /hr
.MAT.Uni Admissions Test£30 /hr
.STEP.Uni Admissions Test£30 /hr

Questions Kyle has answered

Using Integration by Parts, find the indefinite integral of ln(x), and hence show that the integral of ln(x) between 2 and 4 is ln(a) - b where a and b are to be found

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

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2 years ago

559 views

Using the addition formula for sin(x+y), find sin(3x) in terms of sin(x) and hence show that sin(10) is a root of the equation 8x^3 - 6x + 1

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

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2 years ago

1634 views

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