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:

$\sin {30}^{\circ }=\frac{1}{2}$     $\cos {30}^{\circ }=\frac{\sqrt{3}}{2}$     $\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$

$\sin {45}^{\circ }=\frac{1}{\sqrt{2}}$     $\cos {45}^{\circ }=\frac{1}{\sqrt{2}}$     $\tan {45}^{\circ }=1$

$\sin {60}^{\circ }=\frac{\sqrt{3}}{2}$     $\cos {60}^{\circ }=\frac{1}{2}$     $\tan {60}^{\circ }=\sqrt{3}$

$\sin 0=0$     $\cos 0=1$     $\tan 0=0$

$\sin {90}^{\circ }=1$     $\cos {90}^{\circ }=0$     $\tan {90}^{\circ }=\text{undef}$

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

Here is an example question:

$\text{Simplify }\frac{2\sin {45}^{\circ }-\tan {45}^{\circ }}{4\tan {60}^{\circ }}$

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

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 $2\sin {45}^{\circ }$?

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

$\tan {45}^{\circ }$ is 1

$4\tan {60}^{\circ }$ is $4\sqrt{3}$

So we can rewrite the above question like this:

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

Please shout if you need any further help with that.

Here are some more questions for you to try:

Simplify $\sqrt{3}\times \sin ({60}^{\circ })$.

Answer: $\sqrt{3}\times \sin ({60}^{\circ })=\sqrt{3}\times \frac{1}{2}$

= $\frac{\sqrt{3}}{2}$.

Find the value of $\cos ({45}^{\circ })+\sqrt{2}$.

$\cos ({45}^{\circ })+\sqrt{2}=\frac{1}{\sqrt{2}}+\sqrt{2}$

Rationalise by multiplying by $\sqrt{2}$

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

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

= $2.121320344$