# Video: AQA GCSE Mathematics Higher Tier Pack 1 • Paper 1 • Question 29

AQA GCSE Mathematics Higher Tier Pack 1 • Paper 1 • Question 29

04:20

### Video Transcript

Write two sin of 30 degrees minus cos of 30 degrees all over three tan of 60 degrees in the form 𝑎 root three plus 𝑏 over 𝑐, where 𝑎, 𝑏, and 𝑐 are integers.

This question is testing two skills. The first is whether you know your standard trigonometric values. The second is how to work with surds. And we’ll see that in a bit.

Let’s begin now by recalling our trigonometric values. We can use this little table trick to help us remember the values for sin and cos of 30, 45, and 60 degrees. We write one, two, three then three, two, one. We then make all of those numbers into a fraction with a denominator of two. Finally, we square root the top of the fraction, the numerator.

Remember the square root of one is simply one. So we don’t need to write the square root of one for sin of 30 and cos of 60. Unfortunately, there’s no nice little trick to help us remember the values for tan. They are root three over three, one, and root three. You’ll just have to learn these.

So let’s replace what we have in the given expression with values from our table. The first is two sin of 30 degrees. We can see that sin of 30 degrees is one-half. So two sin of 30 is two multiplied by one-half. Two lots of one-half is simply one whole. So two sin of 30 degrees is one.

We also have cos of 30 degrees which we know from our table is root three over two. The denominator of the fraction is three tan 60. And we can see that tan of 60 degrees is root three. So this is three multiplied by root three, which we can write as three root three.

Let’s put these back into the original expression. We replace two sin of 30 degrees with one, cos of 30 degrees with root three over two, and three tan of 60 degrees with three root three. This is actually nothing like the form that’s required from us.

First, we have a surd on the bottom of the fraction. So we’ll need to find a way to get rid of this. In fact, remember that’s called rationalizing the denominator. To make the denominator a rational number, basically not a surd, we need to square the surd part. We need to multiply it by itself.

We can multiply the denominator by the square root of three. But we’ll also need to do that to the numerator. This works because the square root of three over the square root of three is just one. So we’re creating an equivalent fraction by multiplying through by one.

Let’s see what happens to the denominator. We get three multiplied by the square root of three multiplied by the square root of three or three multiplied by the square root of three squared. The square root of three squared is three. So the denominator becomes three multiplied by three which is nine.

For the numerator, we’ll need to multiply each part of the expression on the first fraction by the square root of three, a little like when expanding brackets. One multiplied by the square root of three is the square root of three. The square root of three over two multiplied by the square root of three is the square root of three squared over two. And we already said that the square root of three squared is three. So this part of the expression becomes negative three over two.

Now, this still doesn’t look quite like what we need since we’re told that 𝑎, 𝑏, and 𝑐 need to be integers; that’s whole numbers. And three over two is currently a fraction. We can fix this by multiplying the numerator and the denominator of our fraction by two.

Root three multiplied by two is two root three. Negative three over two multiplied by two is negative three and nine multiplied by two is 18. We have written our expression in the form 𝑎 root three plus 𝑏 over 𝑐. It’s two root three minus three all over 18.

Now, we aren’t being asked to specify the values for 𝑎, 𝑏, and 𝑐. But if we were, 𝑎 is the number of root threes we have; it’s the coefficient of root three, which we can see is two. 𝑏 is the other part of the numerator; that’s negative three. And 𝑐 is the denominator; it’s 18.