# Trig and Surds – The end of the paper. Bring caffiene.

So, if you are wondering, how they can make the last few questions on the higher paper look as terrifying as possible, you will find some ideas here (and hopefully in my next few posts).

Surds can be a pain when put into an exam question and what better way to do it then with trig?

Yes, the famous common trig values that you need to know off by heart:

$Sin\ 30\ =\ \frac{1}{2}\ \ \ \ \ \ \ \ \ \ \ Cos\ 30\ =\ \frac{\sqrt{3\ }}{2}\ \ \ \ \ \ \ \ \ \ \ Tan\ 30\ =\ \frac{1}{\sqrt{3}}$

$Sin\ 45^o\ =\ \frac{1}{\sqrt{2}}\ \ \ \ \ Cos\ 45^o\ =\ \frac{1}{\sqrt{2}}\ \ \ \ \ \ \ \ \ \ Tan\ 45^o\ =\ 1$

$Sin\ 60^o\ =\ \frac{\sqrt{3}}{2}\ \ \ \ \ Cos\ 60^o\ =\ \frac{1}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ Tan\ 60^o\ =\ \sqrt{3}$

$Sin\ 0\ =0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Cos\ 0\ =\ 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Tan\ 0\ =\ 0$

$Sin\ 90\ =\ 1\ \ \ \ \ \ \ \ \ \ \ \ \ Cos\ 90\ =\ 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Tan\ 90\ =\ undef$

You need to be comfortable with rationalising the denominator which is a surds topic.

Here is an example question:

$Simplify\ \ \ \frac{2\sin\ 45\ -\ \tan45}{4\ \tan\ 60}$

$Give\ your\ answer\ in\ the\ form\ \frac{\sqrt{a\ }\ -\ \sqrt{b\ }}{c}$

$Where\ a,\ b\ and\ c\ are\ integers$

This can be a shock at first, especially if you don’t know the trig values.

When you get stuck on a question, start by writing down what you know.

What is 2 sin 45?

$2\ \frac{1}{\sqrt{2}}\ or\ \frac{2}{\sqrt{2}}$

Tan 45 is 1

$4\ Tan\ 60\ \ is\ 4\sqrt{3}$

So we can rewrite the above question like this:

$\frac{\left(\frac{2}{\sqrt{2}}\ -\ 1\right)}{4\sqrt{3}}$

How to simplify? You need to rationalise the denominator, which is a fancy way of saying multiply the whole thing by the surds.

$\sqrt{2\ }\ \ and\ then\ \sqrt{6}\ -I\ will\ \exp lain\ below$

Multiply top and bottom by route 2 first:

$\sqrt{2}\left(\frac{2}{\sqrt{2}}\ -\ 1\ \right)=\ 2\ -\ \sqrt{2}$

$\sqrt{2\ }x\ 4\sqrt{3}\ =\ 4\sqrt{6}$

This gives:

$\frac{2-\sqrt{2}}{4\sqrt{6}}$

Then rationalise again,by mutltiplying by route 6

$\sqrt{6\ }\ \left(2\ -\ \sqrt{2}\right)\ =\ 2\sqrt{6}\ -\ 2\sqrt{3}$

$\sqrt{6}\ x\ 4\sqrt{6}\ =\ 4\ x\ 6\ =\ 24$

So far we have:

$\frac{2\sqrt{6}\ -\ 2\sqrt{3}}{24}$

Looking back at the question we can see its almost there, but the format is slightly different to what they have asked for.

If we divide by 2 we get:

$\frac{\sqrt{6}\ -\ \sqrt{3}}{12}$

Looks good!

Please shout if you need any further help with that.

Here are some more questions for you to try:

Simplify √3 × sin(60°).

Answer: √3 × sin(60°) = √3 × (1/2)

= √3/2.

Find the value of cos(45°) + √2.

$Cos\ 45\ +\ \sqrt{2}$

$Cos\left(45\right)\ +\ \sqrt{2}\ =\ \frac{1}{\sqrt{2}}\ +\sqrt{2}$

Rationalise by multiplying by route 2

$\frac{\sqrt{2\ }}{\sqrt{2}\ x\ \sqrt{2}}\ +\ \sqrt{2}\$

$\frac{\sqrt{2}}{2}\ +\ \sqrt{2}\ =\ \frac{\sqrt{2}}{2}\ +\frac{2\sqrt{2}}{2}\ =\ \frac{\sqrt{2}+2\sqrt{2}}{2}\ \$

$=\ 2.121320344\$