Hello again everyone! Time for Part 2 of trigonometry, arguably the biggest part of the 2 Unit Course, definitely the biggest part of the Extension course. As if 2U wasn't bad enough, Extension adds about two dozen new formulas to remember, lots of new problems to tackle, and makes up a very sizeable portion of the 3 unit exam, even as BOSTES moves from trig to focus on functions and better prep you for uni courses. This guide is going to summarise all of those formulas, and walk through a few of the typical problems you could see in the exam.
Remember to register for an account and ask any questions you have below! Also, be sure to take advantage of the notes for 2/3 Unit , which go into fantastic detail and will give you that bit of extra assistance and revision, if you need it 
Okay, the first bit of extra trig you were likely to have learnt was the trig expansions, properly called the
compound angle results. Six new formulas straight off the bat. Fun! I have no tricks to remember these, though I always remembered that cos was a minus since it looked to simple otherwise, and tan makes intuitive sense as well. Any little tricks to help are great,
post yours below! }=\sin{a}\cos{b}\pm\cos{a}\sin{b} \\\\\cos{\left(a\pm b\right)}=\cos{a}\cos{b}\mp\sin{a}\sin{b} \\\\\tan{\left(a\pm b\right)}=\frac{\tan{a}\pm\tan{b}}{1\mp\tan{a}\tan{b}} )
These allow us to solve for even more angles in exact form.
Hint: Don't try and do this in a 2 unit exam, they won't care and you'll lose marks (they want it correct to however may significant figures anyway if it pops up) .
Example One: Evaluate in exact form:
We simply choose convenient angles to solve:
} \\\\=\frac{\tan{\frac{\pi}{3}}-\tan{\frac{\pi}{4}}}{1+\tan{\frac{\pi}{3}}\tan{\frac{\pi}{4}}} \\\\=\frac {\sqrt{3}-1}{1+\sqrt{3}}\\\\=\frac{2\sqrt{3}-4}{-2} \\\\=2-\sqrt{3})
These can also be used in more complex trigonometric proofs.
There are also the special cases,
double angle results. 
These are useless in solving trig equations, since we can just solve for the double angle and compensate for that in the domain, which is (usually) much simpler. However, these play a big role in extension trig proofs, in conjunction with the pythagorean identities. Here's a simple one I thought of on the spot.
 \\\\=\cos^{2}{\theta} +\sin^{2}{\theta} =1)
The most confusing of these results for me personally was the 't' results. Essentially, these allow all the trig ratios to be expressed in terms of the tangent of a half angle, denoted with 't'. The results are proven using the double angle results, it's a good thing to know, since you may be asked to prove it.
} =\frac{1-\tan^{2}{\frac{x}{2}}}{\sec^{2}{\frac{x}{2}}} =\frac{1-\tan^{2}{\frac{x}{2}}}{1+\tan^{2}{\frac{x}{2}}}=\frac{1-{t}^{2}}{1+{t}^{2}} )
Of course, just remember the formulae for most questions. I would recommend steering clear of these results unless explicitly stated, though if you are good with algebra, it eliminates the need for other trig rules in some sense.
Be warned, if you use this result to find an angle, it will not give 180 degrees as a solution. This is because tan90 is undefined. You should expect two solutions. If you only get one, check 180 degrees by substitution. These results are primarily (almost exclusively, besides proofs) used to solve equations of the form
Just convert and solve for t. However, for these, I much prefer the auxiliary angle method.
} \\\\ A=\sqrt{{a}^{2}+{b}^{2}} )
Essentially, we use the compound angle results to express the expression as a single ratio, not two, and then solve appropriately. It's probably best to just see it in action.
Now lots of textbooks talk a lot about choosing the ratio you equate to carefully to make it easier for yourself. However, choosing wrong adds only a line of working, so personally, I would just pick one of them and compensate later in the process. Choosing correctly just ensures that the auxiliary angle is acute, not a massive deal. Here is the process:
}\\\\=2\cos{\theta}\cos{\alpha}-2\sin{\theta}\sin{\alpha}\\\\\therefore\quad2\cos{\alpha}=\sqrt{3}\\\quad\quad\quad\cos{\alpha}=\frac{\sqrt{3}}{2}\\\\\quad\quad\quad\quad\quad\alpha=\frac{\pi}{6}\quad\\\\2\cos{\left(\theta+\frac{\pi}{6}\right)}=1\\\\\cos{\left(\theta+\frac{\pi}{6}\right)}=\frac{1}{2}\\\\\theta+\frac{\pi}{6}=\frac{\pi}{3},\frac{5\pi}{3}\\\\\theta=\frac{\pi}{6},\frac{3\pi}{2})
Two things to watch for in this process:
1- Be sure to adjust the range for auxiliary angle you choose
2- Don't assume the auxiliary angle is in the first quadrant, if you chose incorrectly (which is fine!), it could be in the second quadrant. Check both sine and cos in the expansion.
Again, you can use the t method for this question as well.
}{1+{t}^{2}}-\frac{2t}{1+{t}^{2}}=1\\\\\sqrt{3}-{t}^{2}\sqrt{3}-2t=1+{t}^{2}\\\\\left(1+\sqrt{3}\right)t^{2}+2t-\left(1-\sqrt{3}\right)=0\\\\t=\frac{-2\pm\sqrt{4+8}}{2+2\sqrt{3}}\\\\\tan{\frac{\theta}{2}}\quad=\frac{-2\pm2\sqrt{3}}{2+2\sqrt{3}}\\\\\quad\\\therefore\quad\theta=\frac{\pi}{6},\frac{3\pi}{2})
I omitted some algebra, purely because this question gets VERY messy with the t method. The auxiliary method was specified in the 2013 Exam, so this is expected, and shows a BIG tip:
USE THE METHOD THEY WANT YOU TO! And that is about it for extension trigonometry! The questions involving proofs are likely to chew the most of your time; look where you are going, and when in doubt, leave it until the end! Coming back later on gives you a fresh perspective on it and may trigger the answer.
Remember to register and ask any questions you have below! And check out the notes too! I'm happy to help with any queries 
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