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:

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

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 \(\sqrt{2}\)

\(\frac{\sqrt{2}}{\sqrt{2} \times \sqrt{2}} + \sqrt{2} \)

\(\frac{\sqrt{2}}{2} + \sqrt{2} = \frac{\sqrt{2} + 2\sqrt{2}}{2} \)

= \(2.121320344\)