Maths C3 - Trigonometry... Help??

The reason why maths is such a great thing is because there are so many ways to go about solving problems. What that RDK kid did above was just one way of solving it...

You could have easily just did

$c^2 \equiv \frac{\frac{16}{a^2}}{1-\frac{16}{a^2}}$
(common denominators, you learned this in the earlier chapters to!)$c^2 \equiv \frac{\frac{16}{a^2}}{\frac{a^2 - 16}{a^2}}$
$c^2 \equiv \frac{16}{a^2} \times \frac{a^2}{a^2 - 16}$ (You probably learned this in KS1... $\frac{\frac{a}{b}}{\frac{c}{d}} \Rightarrow \frac{a}{b} \times \frac{d}{c})$
from there it's just a simple case of cancelling the $a^2$ and multiplying.

I don't think there is a lot of gaps in your knowledge. I think you just need to practice using the skills you've acquired in previous chapters in unusual circumstances.

Yeah you're probably right. I don't think there are a lot of gaps in my knowledge but I do have trouble applying my knowledge to real situations. I mean when things are explained to me I more often than not understand. Like the workings you gave above, I understand everything, but I had trouble applying my own knowledge to it at the time!

I guess I just need to keep practicing as much as possible and try my best not to annoy everyone on TSR

Ok so questions 1,3 and 4 are meant to be worked out without a calculator.
I've read all the examples up to this set of questions but I still feel incompetent. I managed to figure out Q1(a) and (b) with the basic knowledge I already have but the others are rather daunting. I am aware that...
$arctan (x)$...is the same as... ... and similarly goes for arcsin and arccos. But I still feel like I am unable to solve these questions without the use of a calculator

Best case here is to draw these two triangles and work out the values from them which is straight forward if you know your triangle trigonometry:


The one on the left is an equilateral triangle which can be split straight down, while the one on the right is your normal 45 degree right-angled triangle.

Also know that:

$\cos(-x)=\cos(x)$

and

$\sin(-x)=-\sin(x)$

Also know that:

$\cos(-x)=\cos(x)$

and

$\sin(-x)=-\sin(x)$

Oh yeah, I had a feeling I'd have to refer to those triangles. Only ever used them with the angles as degrees though I'm sure it's the same kind of principle.

Oh wow. I've never come across these before. I shall write them down and try to remember them from memory.

Thanks again!!

These could easily be derived if you bothered to learn the unit circle definition, not directed at you RDK kid

You can also derive $\tan(-x)$ from those two.

It can also be useful to draw sketches of sine and cosine graphs for inverses if you wish.

I've never heard of the unit circle definition.

It's difficult filling gaps in your knowledge for Maths when previously you hadn't studied the subject since school 6/7 years ago
So yeah, obviously I'm not some Maths wizard. A lot of the things I have been covering during A-level Maths is completely new to me.

So you mean as in...

$tan (-x) = \frac{sin (-x)}{cos (-x)}= \frac{sin(x)}{cos(x)}$...?

When I was lacking in trigonometry, I just watched all the trig videos in the KhanAcademy playlist and done the quizzes. Trust me if you understand these videos you'll find the C3 trig easy peasy.

https://www.khanacademy.org/math/trigonometry

Yep, but...

$tan (-x) = \frac{sin (-x)}{cos (-x)} \not= \frac{sin(x)}{cos(x)}$

EDIT: You got it, saw the edit, and you can turn it back into tan now.

Ok so I decided to come back to doing those questions from the textbook after browsing the Khan Academy website and it turns out I don't really know how to use them trig triangles afterall I've used the triangle trig ratios when I'm given an angle for example... $sin(300^{\circ})$ ... but I'm not quite sure how to use it in these circumstances

Also know that $\sin(x)=\sin(x-360)=\sin(-(360-x))=-\sin(360-x)$

Just use the graphs.

Can someone please explain?

Do you want an explanation using the CAST diagram? Do you understand it now?

Yes please. I know that.. ... appears in the 2nd and 4th quadrant.

Could you also explain the Trig Ratio Triangles for this example too?

I was going to explain how to find $\sin 300$ but that would involve explaining the exact trig values and using CAST. I don't have time now to do that and don't want to give a short crap explanation.

Instead I'll be lazy and point you to two videos. Together they will show you how to find things like $\sin 300$:

https://www.youtube.com/watch?v=MoyVdhHOV_o

https://www.youtube.com/watch?v=VBhAtctYy8g

If I have time tomorrow I can help or maybe someone else will be kind enough to explain.

By the way, I know students who have got an A* in A Level maths without ever learning the exact trig values, since you're allowed a calculator in the exam. But they're a nice tool to be able to attempt certain questions that require them and really understand all the trig concepts.