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Usually we cover both subject knowledge and exam technique, although that can change depending on each individual student. Then we go through diagrams, and they ask questions, and we go from there.

Lots of students say that the classes are too big in school, or that they don't have time to ask teachers after lessons. In my tutorials, we take time to explore things in a little in a bit more detail.

I always look up the board my students are taking so the lessons are really relevant. Then we go through past papers or set texts, whatever the student finds helpful.

I use the shared whiteboard. We make diagrams together and label them, and often the student prints it off because they know it's right and they completely understand it.

After tutoring one girl went and told all her friends the new explanation I gave her. And she was so excited about what she wrote in the exam she emailed me immediately afterwards.

There was one girl who had her exam on Monday. She wanted tuition on Friday, Saturday and Sunday beforehand. It was very intense, but she said the exam went well.

The key to solving this equation is realizing that sin(2x) can be written in terms of sin(x) and cos(x) using a **double-angle formula**. (With trigonometric problems similar to this one, you should always check if any trigonometric identites like the double-angle formulae can be used, as these can often help you.)

Using your IB formula booklet, you will see that the double-angle formula for sine is:

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

Therefore, we can rewrite the given equation from:

2cos(x) = sin(2x)

to:

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

Next, we notice that both sides of the equation are multiplied by 2, so we can divide both sides by 2. This yields the equation:

cos(x) = sin(x)cos(x)

We can now divide both sides of the equation by cos(x), which leaves us with:

sin(x) = 1

Finally, we must think about the angles at which sin(x) is equal to 1. You should realize, perhaps by imagining the unit circle, that sine is equal to 1 whenever x = π/2 + n2π, where n is any integer.

However, this is not the final answer, as the problem gave us a restricted domain for x. x must be in between 0 and 3π. Therefore, the only possible values for x are π/2 and 5π/2.

So, the answer is:

x = π/2 and 5π/2

If the details of the distribution - its mean and standard deviation are known, the following transformation applies Z = (X - mean) / stdev where Z is now a random variable distributed as normal(0, 1). Now, a lookup table for values of the normal distribution can be used to find the probability you are looking for.

From the rules of logarithms, we know that:

**log(A) - log(B) = log(A/B)**

and thus:

**log(1-x) - log(x) = log[(1-x)/x]**

Therefore from the question, we know:

**log[(1-x)/x] = 1**

If we then take both sides of the equation as a power of 10:

**(1-x)/x = 10^1**

and then multiply both sides through by x:

**1-x=10x**

Solving for x:

**1=11x**

**x=1/11**

We can check our answer by inserting it into the original equation:

**log(1-x) - log(x) = log[1-(1/11)] - log[1/11]**

and using the rule **log(A) - log(B) = log(A/B):**

**l****og[1-(1/11)] - log[1/11] = log(10/11)-log(1/11)**

**= log[(10/11)/(1/11)] **

**= log(10) **

**= 1**

Thus we know x=1/11

Answered by Joshua D.

Studies Materials Science at Oxford, Corpus Christi College

This question is an example of the chain rule for differentiating.

Firstly, identify the inner function. In this case, it is x^{2 }- 3x. This function must be differentiated first:

d/dx (x^{2} - 3x) = 2x - 3

Secondly, identify the outer function. In this case, it is e^{z}, where z = x^{2} - 3x. This function must be differentiated second:

d/dz (e^{z}) = e^{z }

The final differentiated result is the derivative of the inner function multiplied by the derivative of the outer function:

dy/dx = (2x - 3)e^{z }= (2x - 3)e^{x^2 - 3x}

Answered by Ellie S.

Studies Physics at Oxford, Lady Margaret Hall

As with any question involving sines and cosines, consider complex numbers as a likely way to find the answer. For this particular kind of question, where sin/cos of a multiple of x is needed in terms of sinx/cosx, or vice versa, the complex number representation z = cosx + isinx must be expanded in two ways.

Firstly, expand using de Moivre's Theorem:

z = cosx + isinx

z^{4} = (cosx + isinx)^{4} = cos4x + sin4x

Then, expand using the binomial expansion formula to get powers of cosx:

z^{4} = (cosx + isinx)^{4} = cos^{4}x + 4icos^{3}xsinx + 6i^{2}cos^{2}xsin^{2}x + 4i^{3}cosxsin^{3}x + i^{4}sin^{4}x

= cos^{4}x + 4icos^{3}xsinx - 6cos^{2}xsin^{2}x - 4icosxsin^{3}x + sin^{4}x

Equate the real parts of both expansions to get cos equivalence:

cos4x = cos^{4}x - 6cos^{2}xsin^{2}x + sin^{4}x

Use cos^{2}x + sin^{2}x = 1 as a substitution:

cos4x = cos^{4}x - 6cos^{2}x(1-cos^{2}x) + (1-cos^{2}x)^{2}

= 8cos^{4}x - 8cos^{2}x + 1

Answered by Ellie S.

Studies Physics at Oxford, Lady Margaret Hall

The integral of sine is pretty easy to remember as it is -cos + C. However you need to be able to prove this, without using the integral of cosine. This method uses sines exponential form.

As e^{iθ} = cosθ + isinθ, sine can be expressed as sinθ = (e^{iθ}- e^{-}^{iθ}) / 2i. This can make the integration easier as the indefinite integral of e^{k}^{x} = (1/k)e^{k}^{x} and the indefinite integral of e^{-}^{k}^{x} = (-1/k)e^{-}^{k}^{x}

Thus ∫sinx dx = ∫(e^{ix}- e^{-}^{ix}) / 2i dx = (1/2i)[ ∫e^{ix}dx - ∫e^{-}^{ix} dx] = (1/2i)[e^{ix}/i + e^{-}^{ix}/i] + C = [-(e^{ix} + e^{-}^{ix}) / 2] + C.

Now just as sine can be expressed using complex numbers so can cosine such that cosθ = (e^{iθ }+ e^{-}^{iθ}) / 2.

Thus ∫sinx dx = [-(e^{ix} + e^{-}^{ix}) / 2] + C = - cosx + C

Answered by Lucile C.

Studies Mathematical Physics at Edinburgh

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