**About Me:**

I am an engineering student at Oxford and have always enjoyed learning about maths and science. Hopefully I can bring that same enjoyment to you in my lessons. I am very patient and friendly. I have been teaching cricket since I was 14, so have a lot of experience teaching, with people as young as 8 years old. I have a younger brother who I have been tutoring for a couple of years and have helped many people learn as I have studied through school and university.

**The Sessions:**

Despite what many tutors might say the most important thing is to understand the fundamentals behind each topic and question, instead of just learning mark schemes before an exam. During the sessions, you will guide what we cover. I will use as many different ways (diagrams, words, analogies) as possible to explain a concept, until you are confident enough that you can explain it to me. I hope the sessions will be fun! A lot can be achieved in 55mins especially if it is made enjoyable - science is amazing and hopefully, if you didn’t think that before, you will by the end of the session!

**I'm applying to study engineering at university... Can you help me?!**

**Yes! **I have applied through UCAS and have helped friends to do so also, so I know how much stress it can be. Hopefully I can give you helpful advice to reduce the load. What next? If you have any questions, send me a 'Message' or book a 'Free Meeting'! (both accessible through this website). __Remember to tell me your exam board and what you're struggling with. I look forward to meeting you!__

**About Me:**

I am an engineering student at Oxford and have always enjoyed learning about maths and science. Hopefully I can bring that same enjoyment to you in my lessons. I am very patient and friendly. I have been teaching cricket since I was 14, so have a lot of experience teaching, with people as young as 8 years old. I have a younger brother who I have been tutoring for a couple of years and have helped many people learn as I have studied through school and university.

**The Sessions:**

Despite what many tutors might say the most important thing is to understand the fundamentals behind each topic and question, instead of just learning mark schemes before an exam. During the sessions, you will guide what we cover. I will use as many different ways (diagrams, words, analogies) as possible to explain a concept, until you are confident enough that you can explain it to me. I hope the sessions will be fun! A lot can be achieved in 55mins especially if it is made enjoyable - science is amazing and hopefully, if you didn’t think that before, you will by the end of the session!

**I'm applying to study engineering at university... Can you help me?!**

**Yes! **I have applied through UCAS and have helped friends to do so also, so I know how much stress it can be. Hopefully I can give you helpful advice to reduce the load. What next? If you have any questions, send me a 'Message' or book a 'Free Meeting'! (both accessible through this website). __Remember to tell me your exam board and what you're struggling with. I look forward to meeting you!__

No DBS Check

Step 1: Recognise 4cos(2x )+ 2sin(2x) must be put into the form given in the question form i.e. Rcos(2x - a) as it contains only one trigonometric function

Step 2: Calculate values of R and a

a. Expand Rcos(2x - a) using the suitable compound angle formulae: cos(A - B) = cosAcosB + sinAsinB, in our case A = 2x and B = a

Hence Rcos(2x - a) = R[cos(a)cos(2x) + sin(a)sin(2x)]

b. By identifying Rcos(a) and Rsin(a) as effective constants and matching these to the original form we can obtain two trigonometric equations in R and a: Rcos(a) = 4 {eq.1} & Rsin(a) = 2 {eq.2}

c. Dividing {eq.1} by {eq.2} & canceling R gives tan(a) = 1/2, thus arctan(1/2) = a, yielding a = 26.6°

d. Squaring {eq.1} & {eq.2} and taking the sum gives: R^2(cos^2(a) + sin^2(a)) = 4^2 + 2^2 = 20

Noting that this step has yielded cos^2(a) + sin^2(a) and can be replaced by 1 using the standard trigonometric identity cos^2(y) + sin^2(y) = 1, this leads to the result R = sqrt(20)

Step 3: Solve the equation given for x

a. From the result in step 2, 4cos(2x )+ 2sin(2x) = 1 can be written as sqrt(20)*cos(2x - 26.6) = 1

b. Dividing by sqrt(20) and taking the inverse cosine yields: arccos[1/sqrt(20)] = 2x - 26.6 From your calculator arccos[1/sqrt(20)] yields 77.08° but recognising -77.08° (282.92°, -282.92° etc.) are also solutions crucial(important)

This can be seen by looking at a sketch of y = cos(t)

c. Solving for x in the interval given can be done by rearrangement

2x - 26.6° = 77.08°, -77.08° (282.92°, -282.92°)

Hence x = 51.84° -25.24°

Note that (282.92°, -282.92°) give values out of the range defined for x

Step 4: Check on your calculator that the calculated values of x solve the original equation 4cos(2x )+ 2sin(2x) = 1Step 1: Recognise 4cos(2x )+ 2sin(2x) must be put into the form given in the question form i.e. Rcos(2x - a) as it contains only one trigonometric function

Step 2: Calculate values of R and a

a. Expand Rcos(2x - a) using the suitable compound angle formulae: cos(A - B) = cosAcosB + sinAsinB, in our case A = 2x and B = a

Hence Rcos(2x - a) = R[cos(a)cos(2x) + sin(a)sin(2x)]

b. By identifying Rcos(a) and Rsin(a) as effective constants and matching these to the original form we can obtain two trigonometric equations in R and a: Rcos(a) = 4 {eq.1} & Rsin(a) = 2 {eq.2}

c. Dividing {eq.1} by {eq.2} & canceling R gives tan(a) = 1/2, thus arctan(1/2) = a, yielding a = 26.6°

d. Squaring {eq.1} & {eq.2} and taking the sum gives: R^2(cos^2(a) + sin^2(a)) = 4^2 + 2^2 = 20

Noting that this step has yielded cos^2(a) + sin^2(a) and can be replaced by 1 using the standard trigonometric identity cos^2(y) + sin^2(y) = 1, this leads to the result R = sqrt(20)

Step 3: Solve the equation given for x

a. From the result in step 2, 4cos(2x )+ 2sin(2x) = 1 can be written as sqrt(20)*cos(2x - 26.6) = 1

b. Dividing by sqrt(20) and taking the inverse cosine yields: arccos[1/sqrt(20)] = 2x - 26.6 From your calculator arccos[1/sqrt(20)] yields 77.08° but recognising -77.08° (282.92°, -282.92° etc.) are also solutions crucial(important)

This can be seen by looking at a sketch of y = cos(t)

c. Solving for x in the interval given can be done by rearrangement

2x - 26.6° = 77.08°, -77.08° (282.92°, -282.92°)

Hence x = 51.84° -25.24°

Note that (282.92°, -282.92°) give values out of the range defined for x

Step 4: Check on your calculator that the calculated values of x solve the original equation 4cos(2x )+ 2sin(2x) = 1

Step 2: Calculate values of R and a

a. Expand Rcos(2x - a) using the suitable compound angle formulae: cos(A - B) = cosAcosB + sinAsinB, in our case A = 2x and B = a

Hence Rcos(2x - a) = R[cos(a)cos(2x) + sin(a)sin(2x)]

b. By identifying Rcos(a) and Rsin(a) as effective constants and matching these to the original form we can obtain two trigonometric equations in R and a: Rcos(a) = 4 {eq.1} & Rsin(a) = 2 {eq.2}

c. Dividing {eq.1} by {eq.2} & canceling R gives tan(a) = 1/2, thus arctan(1/2) = a, yielding a = 26.6°

d. Squaring {eq.1} & {eq.2} and taking the sum gives: R^2(cos^2(a) + sin^2(a)) = 4^2 + 2^2 = 20

Noting that this step has yielded cos^2(a) + sin^2(a) and can be replaced by 1 using the standard trigonometric identity cos^2(y) + sin^2(y) = 1, this leads to the result R = sqrt(20)

Step 3: Solve the equation given for x

a. From the result in step 2, 4cos(2x )+ 2sin(2x) = 1 can be written as sqrt(20)*cos(2x - 26.6) = 1

b. Dividing by sqrt(20) and taking the inverse cosine yields: arccos[1/sqrt(20)] = 2x - 26.6 From your calculator arccos[1/sqrt(20)] yields 77.08° but recognising -77.08° (282.92°, -282.92° etc.) are also solutions crucial(important)

This can be seen by looking at a sketch of y = cos(t)

c. Solving for x in the interval given can be done by rearrangement

2x - 26.6° = 77.08°, -77.08° (282.92°, -282.92°)

Hence x = 51.84° -25.24°

Note that (282.92°, -282.92°) give values out of the range defined for x

Step 4: Check on your calculator that the calculated values of x solve the original equation 4cos(2x )+ 2sin(2x) = 1Step 1: Recognise 4cos(2x )+ 2sin(2x) must be put into the form given in the question form i.e. Rcos(2x - a) as it contains only one trigonometric function

Step 2: Calculate values of R and a

a. Expand Rcos(2x - a) using the suitable compound angle formulae: cos(A - B) = cosAcosB + sinAsinB, in our case A = 2x and B = a

Hence Rcos(2x - a) = R[cos(a)cos(2x) + sin(a)sin(2x)]

b. By identifying Rcos(a) and Rsin(a) as effective constants and matching these to the original form we can obtain two trigonometric equations in R and a: Rcos(a) = 4 {eq.1} & Rsin(a) = 2 {eq.2}

c. Dividing {eq.1} by {eq.2} & canceling R gives tan(a) = 1/2, thus arctan(1/2) = a, yielding a = 26.6°

d. Squaring {eq.1} & {eq.2} and taking the sum gives: R^2(cos^2(a) + sin^2(a)) = 4^2 + 2^2 = 20

Noting that this step has yielded cos^2(a) + sin^2(a) and can be replaced by 1 using the standard trigonometric identity cos^2(y) + sin^2(y) = 1, this leads to the result R = sqrt(20)

Step 3: Solve the equation given for x

a. From the result in step 2, 4cos(2x )+ 2sin(2x) = 1 can be written as sqrt(20)*cos(2x - 26.6) = 1

b. Dividing by sqrt(20) and taking the inverse cosine yields: arccos[1/sqrt(20)] = 2x - 26.6 From your calculator arccos[1/sqrt(20)] yields 77.08° but recognising -77.08° (282.92°, -282.92° etc.) are also solutions crucial(important)

This can be seen by looking at a sketch of y = cos(t)

c. Solving for x in the interval given can be done by rearrangement

2x - 26.6° = 77.08°, -77.08° (282.92°, -282.92°)

Hence x = 51.84° -25.24°

Note that (282.92°, -282.92°) give values out of the range defined for x

Step 4: Check on your calculator that the calculated values of x solve the original equation 4cos(2x )+ 2sin(2x) = 1