PremiumKyle R. GCSE Maths tutor, IB Maths tutor, A Level Maths tutor, Uni Ad...

Kyle R.

£30 /hr

Studying: Mathematics (Masters) - Oxford, Corpus Christi College University

5.0
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.

202 reviews| 213 completed tutorials

Contact Kyle

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.  

Show more

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!

Show more

DBS Icon

Enhanced DBS Check

28/09/2016

Ratings & Reviews

5from 202 customer reviews
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.

Naa (Parent)

October 9 2017

Very good

Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.

Alisiya (Student)

June 3 2017

all good

Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.

Martin (Parent)

February 26 2017

Really good lesson! Learnt loads, thanks Kyle!

Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.
Star 1 Created with Sketch.

Naa (Parent)

February 20 2017

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

Show more reviews

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

Before 12pm12pm - 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

Show more

1 year ago

457 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

Show more

1 year ago

806 views

Arrange a free video meeting


To give you a few options, we can ask three similar tutors to get in touch. More info.

Contact Kyle

How do we connect with a tutor?

Where are they based?

How much does tuition cost?

How do tutorials work?

We use cookies to improve your site experience. By continuing to use this website, we'll assume that you're OK with this. Dismiss

mtw:mercury1:status:ok