<s>[INST] Is there a more efficient method of trig mastery than rote memorization? I would like to get alot better at trig than I am. What is the best/most efficient method?
Thanks much in advance
Joe
 [/INST] I would emphasize how to derive trigonometric identities from a few ones.  Learn:


*

*how to derive the relations between the direct functions of the same angle from the definition of the trigonometric functions and the Pythagorean formula;

*maxima, minima, zeroes and period of each function;

*if a function is odd or even;

*the trigonometric functions of 0º, 30º, 45º, 60º and 90º;

*the relations between functions of symmetric, complementary and supplementary angles;

*the relations between functions of angles whose difference is 180º;

*the relations between functions of angles whose sum is 360º;

*the inverse trigonometric functions;

*the addition formulas of sin and cos;

*how to derive the subtraction formulas of sin and cos;

*how to derive the addition and subtraction formulas of tan and cot;

*how to derive the double and half angle formulas;

*how to derive the sum to product formulas;

*how to solve some elementary trigonometric equations;

*(triangle) sin and cos laws;

*Heron's formula;

*derivatives of direct and inverse trigonometric functions. 



Added. Examples. From 
$$\sin (\alpha +\beta )=\sin \alpha \cdot \cos \beta +\cos \alpha \cdot \sin
\beta ,\tag{A}$$
if we set $\alpha =\beta =a$, we get
$$\sin 2a=2\sin a\cdot \cos a.\tag{1}$$
And from
$$\cos (\alpha +\beta )=\cos \alpha \cdot \cos \beta -\sin \alpha \cdot \sin
\beta \tag{B}$$
for $\alpha =\beta =a$, we have
$$\cos 2a=\cos ^{2}a-\sin ^{2}a.\tag{2}$$
Using the Pythagorean identity
$$\cos ^{2}a+\sin ^{2}a=1,\tag{C}$$
if $\cos a\neq 0$, then 
$$\begin{eqnarray*}
\sin 2a &=&2\sin a\cdot \cos a=2\dfrac{\sin a\cdot \cos a}{\cos ^{2}a+\sin
^{2}a} \\
&=&\dfrac{2\dfrac{\sin a\cdot \cos a}{\cos ^{2}a}}{\dfrac{\cos ^{2}a+\sin ^{2}a
}{\cos ^{2}a}}=\dfrac{2\dfrac{\sin a}{\cos a}}{1+\dfrac{\sin ^{2}a}{\cos ^{2}a}}
\\
&=&\dfrac{2\tan a}{1+\tan ^{2}a}.
\end{eqnarray*}\tag{3}$$
Similarly
$$\begin{eqnarray*}
\cos 2a &=&\cos ^{2}a-\sin ^{2}a=\dfrac{\cos ^{2}a-\sin ^{2}a}{\cos
^{2}a+\sin ^{2}a} \\
&=&\dfrac{\dfrac{\cos ^{2}a-\sin ^{2}a}{\cos ^{2}a}}{\dfrac{\cos ^{2}a+\sin
^{2}a}{\cos ^{2}a}}=\dfrac{1-\dfrac{\sin ^{2}a}{\cos ^{2}a}}{1+\dfrac{\sin ^{2}a
}{\cos ^{2}a}} \\
&=&\dfrac{1-\tan ^{2}a}{1+\tan ^{2}a}.
\end{eqnarray*}\tag{4}$$
Then
$$\tan 2a=\dfrac{\sin 2a}{\cos 2a}=\dfrac{\dfrac{2\tan a}{1+\tan ^{2}a}}{\dfrac{
1-\tan ^{2}a}{1+\tan ^{2}a}}=\dfrac{2\tan a}{1-\tan ^{2}a}.\tag{5}$$
Added 2. The linear equation in $\sin x$ and $\cos x$
$$
A\sin x+B\cos x=C\tag{6}
$$
can be solved by a resolvent quadratic equation in $\tan \frac{x}{2}$, by
writting the $\sin x$ and the $\cos x$ functions in terms of $\tan \frac{x}{2
}$ (set $x=2a$ in $(3)$ and $(4)$):
$$
\sin x=\dfrac{2\tan \dfrac{x}{2}}{1+\tan ^{2}\dfrac{x}{2}},\tag{7}
$$
$$
\cos x=\dfrac{1-\tan ^{2}\dfrac{x}{2}}{1+\tan ^{2}\dfrac{x}{2}}.\tag{9}
$$
The equation $(6)$ is equivalent to
$$\begin{eqnarray*}
A\dfrac{2\tan \dfrac{x}{2}}{1+\tan ^{2}\dfrac{x}{2}}+B\frac{1-\tan ^{2}\dfrac{x}{
2}}{1+\tan ^{2}\dfrac{x}{2}} &=&C, \\
2A\tan \dfrac{x}{2}+B-B\tan ^{2}\dfrac{x}{2} &=&C+C\tan ^{2}\dfrac{x}{2}, \\
\left( B+C\right) \tan ^{2}\dfrac{x}{2}-2A\tan \dfrac{x}{2}+C-B &=&0.
\end{eqnarray*}\tag{10}$$
</s>