用三角函数歌默哀 RMG

When you first study math about 1234.
First study equation about xyzt.
It will help you think in a logical way.
When you sing sine cosine tangent.
Sine cosine tangent cotangent.
Sine cosine secant cosecant.
Let’s sing a song about trig-functions.
$\sin(2 \pi+\alpha)=\sin \alpha$
$\cos(2 \pi+\alpha)=\cos \alpha$
$\tan(2 \pi+\alpha)=\tan \alpha$
Which is induction formula $1$ and induction formula $2$.
$\sin(\pi+\alpha)=- \sin \alpha$
$\cos(\pi+\alpha)=-\cos \alpha$
$\tan(\pi+\alpha)=\tan \alpha$
$\sin(\pi-\alpha)=\sin \alpha$
$\cos(\pi-\alpha)=-\cos \alpha$
$\tan(\pi-\alpha)=-\tan \alpha.$
These are all those "name do not change",
As pi goes to half pi the difference shall be huge.
$\sin(\displaystyle \frac{\pi}{2} + \alpha) = cos \alpha$

$\sin(\displaystyle \frac{\pi}{2}-\alpha)=\cos \alpha.$

$\cos(\displaystyle \frac{\pi}{2}+\alpha)=-sin\alpha$

$\cos(\displaystyle \frac{\pi}{2}-\alpha)=sin\alpha.$

$\tan(\displaystyle \frac{\pi}{2}+\alpha)=-cot \alpha$

$\tan(\displaystyle \frac{\pi}{2}-\alpha)=cot \alpha.$
That is to say the odds will change evens are conserved,
The notations that they get depend on where they are.
But no matter where you are,
I'd like say that,
If you were my sine curve,I'd be your cosine curve.
I will be you derivative you will be my negative one.
As you change you amplitute,I change my phase.
We can oscillate freely in the external space.
As we chaange our period and constant at hand.
We tracel from the origin to infinity.
It's you sine and if you cosine,
Who make charming music around the world.
It's you tangent cotangent,
Who proclaim the true meaning of centrosymmetry.

You wanna measure width of a river height of a tower.
You scratch your head which cost you more than an hour.
You don't need to ask any "goods" or "master" for help.
These group of formulasare gonna help you solve.
$\sin(\alpha + \beta)=\sin\alpha \cdot \cos \beta + \cos \alpha \cdot \sin \beta$
$\cos(\alpha+\beta)=\cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta$
$\tan(\alpha+\beta)=\displaystyle \frac{\tan \alpha + \tan \beta}{1-\tan \alpha \cdot \tan \beta}$
$\sin(\alpha - \beta)=\sin\alpha \cdot \cos \beta - \cos \alpha \cdot \sin \beta$
$\cos(\alpha - \beta)=\cos \alpha \cdot \cos \beta + \sin \alpha \cdot \sin \beta$
$\tan(\alpha - \beta)=\displaystyle \frac{\tan \alpha - \tan \beta}{1+\tan \alpha \cdot \tan \beta}$
As you come across a right triangle you feel easy to solve.
But an obuse triangle gonna make you feel confused.
Don't worry about what you do.
There are always means to solve.
As long as you master the sine cosine law.
As this moment I've got nothing to say.
As trig-functions rain down upon me.
As this moment I've got nothing to say.
Let's sing a song about trig-functions.
Long live the trigonometric functions.