Hello, math lovers! I’m Math Betty,
your slightly strict trigonometry teacher!
Today, I’ll help you memorize
the magic spells of angles —
our trigonometric identities!

Let’s start with
the angle sum formulas
— when angles get together,
their sines and cosines
make beautiful harmony.

\[
\sin(A + B) = \sin A \cos B + \cos A \sin B
\]  
\[
\cos(A + B) = \cos A \cos B - \sin A \sin B
\]  
\[
\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}
\]

\[
\sin(A - B) = \sin A \cos B - \cos A \sin B
\]  
\[
\cos(A - B) = \cos A \cos B + \sin A \sin B
\]
\[
\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}
\]

\[
\sin(A + B) = ?
\]  
\[
\cos(A + B) = ?
\]  
\[
\tan(A + B) = ?
\]

\[
\sin(A - B) = ?
\]  
\[
\cos(A - B) = ?
\]
\[
\tan(A - B) = ?
\]

Now double the fun!
Double angles look like this:

\[
\sin(2A) = 2 \sin A \cos A
\]  
\[
\cos(2A) = \cos^2 A - \sin^2 A
\]  
\[
\tan(2A) = \frac{2 \tan A}{1 - \tan^2 A}
\]

\[
\sin(2A) = ?
\]  
\[
\cos(2A) = ?
\]  
\[
\tan(2A) = ?
\]

\[
\cos(2A) = 1 - 2\sin^2 A
\]
\[
\cos(2A) = 2\cos^2 A - 1
\]

\[
\cos(2A) = ?
\]
\[
\cos(2A) = ?
\]

Or if you prefer
So many choices!

Half an angle?
A piece of cake,
I mean
a piece of pizza!

\[
\sin\left(\frac{A}{2}\right) = \pm \sqrt{\frac{1 - \cos A}{2}}
\]\[
\cos\left(\frac{A}{2}\right) = \pm \sqrt{\frac{1 + \cos A}{2}}
\]\[
\tan\left(\frac{A}{2}\right) = \pm \sqrt{\frac{1 - \cos A}{1 + \cos A}}
\]

\[
\sin\left(\frac{A}{2}\right) = ?
\]\[
\cos\left(\frac{A}{2}\right) = ?
\]\[
\tan\left(\frac{A}{2}\right) = ?
\]

\[
\tan\left(\frac{A}{2}\right) = \frac{\sin A}{1 + \cos A}
\]\[
\tan\left(\frac{A}{2}\right) = \frac{1 - \cos A}{\sin A}
\]

\[
\tan\left(\frac{A}{2}\right) = ?
\]\[
\tan\left(\frac{A}{2}\right) = ?
\]

The sign depends
on the quadrant,
my dear students!

Now boss level round.
Let’s mix sums and differences
into neat products
— perfect for simplifying
tricky equations!

\[
\sin A + \sin B = 2 \sin\left(\frac{A + B}{2}\right)\cos\left(\frac{A - B}{2}\right)
\]
\[
\sin A - \sin B = 2 \cos\left(\frac{A + B}{2}\right)\sin\left(\frac{A - B}{2}\right)
\]

\[
\sin A + \sin B = ?
\]
\[
\sin A - \sin B = ?
\]

\[
\cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right)\cos\left(\frac{A - B}{2}\right)
\]  
\[
\cos A - \cos B = -2 \sin\left(\frac{A + B}{2}\right)\sin\left(\frac{A - B}{2}\right)
\]

\[
\cos A + \cos B = ?
\]  
\[
\cos A - \cos B = ?
\]

\[
\sin A \sin B = \frac{1}{2}\big[\cos(A - B) - \cos(A + B)\big]
\]
\[
\cos A \cos B = \frac{1}{2}\big[\cos(A - B) + \cos(A + B)\big]
\]

\[
\sin A \sin B = ?
\]
\[
\cos A \cos B = ?
\]

\[
\sin A \cos B = \frac{1}{2}\big[\sin(A + B) + \sin(A - B)\big]
\]
\[
\cos A \sin B = \frac{1}{2}\big[\sin(A + B) - \sin(A - B)\big]
\]

\[
\sin A \cos B = ?
\]
\[
\cos A \sin B = ?
\]

Now, if we wanted to **prove** these,
we could start from the angle addition formulas
and apply substitutions like—

WAIT!!! No! This is
a memorization note, remember?!
No proofs today! Just memorize them!
Say it with me —

MEMO-RIZE!

You did great!
Keep these formulas
in your memory vault
— they’ll be
your secret weapon
in every trig battle.

Next time, I’ll actually
*prove* them
for you. Promise!