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Osian S.

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Studying: Mathematics (Masters) - Warwick University

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12 reviews| 24 completed tutorials

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About me

Hi, I'm Osian, a Master's student at the University of Warwick. I've had fun with mathematics from a young age, and I hope to be able to instill this passion for mathematics in you, through my tutorials! Throughout my time at Warwick, and a year at the Technische Universität Berlin, I have had to explain my solutions to peers. This has lead me to being a patient listener, and taught me the dexterity of viewing problems from another person's perspective. I find that a variety of problems, analogies and (especially) diagrams can really benefit understanding. During A-level and GCSE, I found that several formulae and techniques were used, without them being fully explained. I aim to get you to fully grasp theses concepts, so you can answer questions clearly and efficiently.  I'm sure that together, we can successfully work on whichever areas you wish to excel at!Hi, I'm Osian, a Master's student at the University of Warwick. I've had fun with mathematics from a young age, and I hope to be able to instill this passion for mathematics in you, through my tutorials! Throughout my time at Warwick, and a year at the Technische Universität Berlin, I have had to explain my solutions to peers. This has lead me to being a patient listener, and taught me the dexterity of viewing problems from another person's perspective. I find that a variety of problems, analogies and (especially) diagrams can really benefit understanding. During A-level and GCSE, I found that several formulae and techniques were used, without them being fully explained. I aim to get you to fully grasp theses concepts, so you can answer questions clearly and efficiently.  I'm sure that together, we can successfully work on whichever areas you wish to excel at!

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Ratings & Reviews

5from 12 customer reviews
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Ioan (Student)

October 19 2016

Very helpful and explained the topics well.

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Ioan (Student)

June 12 2017

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Ioan (Student)

May 31 2017

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May 24 2017

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Qualifications

SubjectQualificationGrade
MathematicsA-level (A2)A*
Further MathematicsA-level (A2)A*
PhysicsA-level (A2)A*
MathematicsDegree (Masters)FIRST CLASS HONOURS

General Availability

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Subjects offered

SubjectQualificationPrices
Further MathematicsA Level£24 /hr
MathsA Level£24 /hr
Further MathematicsGCSE£22 /hr
MathsGCSE£22 /hr

Questions Osian has answered

Eleri invests £3700 for 3 years at 2% per annum compound interest. Calculate the value of her investment at the end of the 3 years. Give your answer correct to the nearest penny.

This question is a classic example of a problem which might catch some students out, if due care is not implemented.

Eleri is investing £3700 at 2% per annum compound interest. This means that the sum of the investment increases by 2% each year.

After the first year, Eleri’s £3700 investment will have grown by 2%, so our first step will be to solve 2% of £3700. Notice that 2% is equivalent to 0.02 in decimal form. Therefore, 2% of 3700 is:

3700 x (0.02) = 74

To deduce the sum of the investment at the end of the year, we add this value (74) to the initial value (3700). Therefore, the sum of the investment at the end of the year is:

3700 + 74 = 3774

We have shown that after one year, Eleri’s investment has increased from £3700 to £3774.

To discover the value of Eleri’s investment after the second year, we repeat the method that we used to discover the value of the investment after one year.

So, 2% of 3774 is:

3774 x (0.02) = 75.48

Therefore, the sum of the investment at the end of the second year is:

3774 + 75.48 = 3849.48

We use the same method to discover the value of Eleri’s investment after the third year.

 2% of 3774 is:

3849.48 x (0.02) = 76.9896

Therefore, the sum of the investment at the end of the third year is:

3849.48  x 76.9896 = 3926.4696

We have found out the value of Eleri’s investment after 3 years of compound interest!

Notice that the question asks for the answer to be correct to the nearest penny. So our final answer to the problem will be:

£3926.47

Note: Once a student is fairly confident it their understanding of a problem such as this, one can discuss methods which are quicker to use, without obscuring what is going on.

This question is a classic example of a problem which might catch some students out, if due care is not implemented.

Eleri is investing £3700 at 2% per annum compound interest. This means that the sum of the investment increases by 2% each year.

After the first year, Eleri’s £3700 investment will have grown by 2%, so our first step will be to solve 2% of £3700. Notice that 2% is equivalent to 0.02 in decimal form. Therefore, 2% of 3700 is:

3700 x (0.02) = 74

To deduce the sum of the investment at the end of the year, we add this value (74) to the initial value (3700). Therefore, the sum of the investment at the end of the year is:

3700 + 74 = 3774

We have shown that after one year, Eleri’s investment has increased from £3700 to £3774.

To discover the value of Eleri’s investment after the second year, we repeat the method that we used to discover the value of the investment after one year.

So, 2% of 3774 is:

3774 x (0.02) = 75.48

Therefore, the sum of the investment at the end of the second year is:

3774 + 75.48 = 3849.48

We use the same method to discover the value of Eleri’s investment after the third year.

 2% of 3774 is:

3849.48 x (0.02) = 76.9896

Therefore, the sum of the investment at the end of the third year is:

3849.48  x 76.9896 = 3926.4696

We have found out the value of Eleri’s investment after 3 years of compound interest!

Notice that the question asks for the answer to be correct to the nearest penny. So our final answer to the problem will be:

£3926.47

Note: Once a student is fairly confident it their understanding of a problem such as this, one can discuss methods which are quicker to use, without obscuring what is going on.

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1 year ago

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Find the values of x, where 0 < x < 360, such that x solves the equation: 8(tan[x])^2 – 5(sec[x])^2 = 7 + 4sec[x]

This question tests a variety of mathematical techniques and knowledge which is integral to A-level mathematics.

Note that the highest power of a variable function in the mentioned equation, is 2. It is therefore quite likely that we will have to solve a quadratic equation to find x. We know how to solve quadratic equations from GCSE. Therefore, if we can manipulate the equation such that we have only one trigonometric variable function, then we will be able to solve the problem.

We first notice that tan[x] = sin[x]/cos[x] and that sec[x] = 1/cos[x]

Hence the equation can be read as:

8(sin[x]/cos[x])- 5(1/cos[x])2 = 7 + 4(1/cos[x])

Solving a quadratic equation with fractions is always going to be a messy business. We notice that cos[x] is the denominator of several fractions in the equation. To get rid of the denominators, we times through by (cos[x])2:

8(sin[x])2 - 5 = 7(cos[x])+ 4cos[x]     {*}

This already looks easier to solve than the initial equation. If we could ensure that the only variable function was sin[x] or cos[x] then we could solve the equation as a quadratic! Recall that: 

(sin[x])2 + (cos[x])2 =1 
Therefore:
(sin[x])= 1 - (cos[x])2

Substituting this value into {*}, we find that:

8(1 - (cos[x])2) - 5 = 7(cos[x])+ 4cos[x] 

By rearranging appropriately, we get the quadratic equation:

15(cos[x])2 + 4cos[x] - 3 = 0                                                                               

Just as in GCSE, we find that this is equivalent to:

(3cos[x] - 1)(5cos[x] + 3) = 0
so 
(1) cos[x] = 1/3 or (2) cos[x] = -3/5

We consider cases (1) and (2) separately.

Case 1: cos[x] = 1/3

Using a calculator we see that x = cos-1[1/3] = 70.5 (to 1 decimal place). However we want every value of x such that 0 < x < 360.

Note that if cos[x] = y, then for any integer k:
x = 360k ± cos-1[y]
This is the general solution for cosine.

By using the general solution for cosine we see that 
x = 360 - cos-1[1/3] = 289.5 (to 1 decimal place)
is also a solution.

Case 2: cos[x] = -3/5

Using a calculator we see that x = cos-1[-3/5] = 126.9 (to 1 decimal place).
Using the general formula for cosine we see that 
x = 360 - cos-1[-3/5] = 233.1 (to 1 decimal place).
is also a solution in the interva 0 < x < 360.

Hence we see that the complete set of values of x which solve 8(tan[x])2 – 5(sec[x])2 = 7 + 4sec[x] in the range 0 < x < 360 is:

x = 70.5, 126.9, 233.1, 289.5

Note: In the answer we used the general solution for cosine. This may have to be justified. This can be explained intuitively with a graph of the cosine function. Graphs and diagrams are often extremely helpful in understanding and answering questions.


 


 

This question tests a variety of mathematical techniques and knowledge which is integral to A-level mathematics.

Note that the highest power of a variable function in the mentioned equation, is 2. It is therefore quite likely that we will have to solve a quadratic equation to find x. We know how to solve quadratic equations from GCSE. Therefore, if we can manipulate the equation such that we have only one trigonometric variable function, then we will be able to solve the problem.

We first notice that tan[x] = sin[x]/cos[x] and that sec[x] = 1/cos[x]

Hence the equation can be read as:

8(sin[x]/cos[x])- 5(1/cos[x])2 = 7 + 4(1/cos[x])

Solving a quadratic equation with fractions is always going to be a messy business. We notice that cos[x] is the denominator of several fractions in the equation. To get rid of the denominators, we times through by (cos[x])2:

8(sin[x])2 - 5 = 7(cos[x])+ 4cos[x]     {*}

This already looks easier to solve than the initial equation. If we could ensure that the only variable function was sin[x] or cos[x] then we could solve the equation as a quadratic! Recall that: 

(sin[x])2 + (cos[x])2 =1 
Therefore:
(sin[x])= 1 - (cos[x])2

Substituting this value into {*}, we find that:

8(1 - (cos[x])2) - 5 = 7(cos[x])+ 4cos[x] 

By rearranging appropriately, we get the quadratic equation:

15(cos[x])2 + 4cos[x] - 3 = 0                                                                               

Just as in GCSE, we find that this is equivalent to:

(3cos[x] - 1)(5cos[x] + 3) = 0
so 
(1) cos[x] = 1/3 or (2) cos[x] = -3/5

We consider cases (1) and (2) separately.

Case 1: cos[x] = 1/3

Using a calculator we see that x = cos-1[1/3] = 70.5 (to 1 decimal place). However we want every value of x such that 0 < x < 360.

Note that if cos[x] = y, then for any integer k:
x = 360k ± cos-1[y]
This is the general solution for cosine.

By using the general solution for cosine we see that 
x = 360 - cos-1[1/3] = 289.5 (to 1 decimal place)
is also a solution.

Case 2: cos[x] = -3/5

Using a calculator we see that x = cos-1[-3/5] = 126.9 (to 1 decimal place).
Using the general formula for cosine we see that 
x = 360 - cos-1[-3/5] = 233.1 (to 1 decimal place).
is also a solution in the interva 0 < x < 360.

Hence we see that the complete set of values of x which solve 8(tan[x])2 – 5(sec[x])2 = 7 + 4sec[x] in the range 0 < x < 360 is:

x = 70.5, 126.9, 233.1, 289.5

Note: In the answer we used the general solution for cosine. This may have to be justified. This can be explained intuitively with a graph of the cosine function. Graphs and diagrams are often extremely helpful in understanding and answering questions.


 


 

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1 year ago

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