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==<span style="font-size:20px; color: blue;">Trigonometric Formulas</span>==
Note that all measurements are in radians.
===Basic Facts===
<math>\sin (-A)=-\sin A</math>

<math>\cos (-A)=\cos A</math>

<math>\tan (-A)=-\tan A</math>

<math>\sin (\pi-A) = \sin A</math>

<math>\cos (\pi-A) = -\cos A</math>

<math>\sin (-A) = -\sin A</math>

<math>\cos (-A) = \cos A</math>

<math>\tan (\pi+A) = \tan A</math>

<math>\cos (\pi/2-A)=\sin A</math>

<math>\tan (\pi/2-A)=\cot A</math>

<math>\sec (\pi/2-A)=\csc A</math>

<math>\cos (\pi/2-A) = \sin A</math>

<math>\cot (\pi/2-A)=\tan A</math>

<math>\csc (\pi/2-A)=\sec A</math>

The above can all be seen clearly by examining the graphs or plotting on a unit circle - the reader can figure that out themselves.

===Terminology and Notation===
<math>\cot A=\frac{1}{\tan A}</math>, but <math>\cot A\ne\tan^{-1} A}</math>, the former being the reciprocal and the latter the inverse.

<math>\csc A=\frac{1}{\sin A}</math>, but <math>\csc A\ne\sin^{-1} A}</math>.

<math>\sec A=\frac{1}{\sin A}</math>, but <math>\sec A\ne\cos^{-1} A}</math>.

Speaking of inverses:

<math>\tan^{-1} A=\text{atan } A=\arctan A</math>

<math>\cos^{-1} A=\text{acos } A=\arccos A</math>

<math>\sin^{-1} A=\text{asin } A=\arcsin A</math>

===Sum of Angle Formulas===
<math>\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B</math>

If we can prove this one, the other ones can be derived easily using the "Basic Facts" identities above. In fact, we can simply prove the addition case, for plugging <math>A=-B</math> into the addition case gives the subtraction case.

As it turns out, there's quite a nice geometric proof of the addition case, though other methods, such as de Moivre's Theorem, exist. The following proof is taken from ''the Art of Problem Solving, Vol. 2'' and is due to Masakazu Nihei of Japan, who originally had it published in ''Mathematics @ Informatics Quarterly'', Vol. 3, No. 2:

{{asy image|1=<asy>
pair A,B,C;
C=(0,0);
B=(10,0);
A=(6,4);
draw(A--B--C--cycle);
label("$A$",A,N);
label("$B$",B,E);
label("$C$",C,W);
draw(A--(6,0));
label("$\beta$",A,(-1,-2));
label("$\alpha$",A,(1,-2.5));
label("$H$",(6,0),S);
draw((6,0)--(5.5,0)--(5.5,0.5)--(6,0.5)--cycle);
</asy>|2=right|3=Figure 1}}

We'll find <math>[ABC]</math> in two different ways: <math>\frac{1}{2}(AB)(AC)(\sin \angle BAC)</math> and <math>[ABH]+[ACH]</math>. We let <math>AH=1</math>. We have:

<math>[ABC]=[ABH]+[ACH]</math>

<math>\frac{1}{2}(AC)(AB)(\sin \angle BAC)=\frac{1}{2}(AH)(BH)+\frac{1}{2}(AH)(CH)</math>

<math>\frac{1}{2}\left(\frac{1}{\cos \beta}\right)\left(\frac{1}{\cos \alpha}\right)(\sin \angle BAC)=\frac{1}{2}(1)(\tan \alpha)(\tan \beta)</math>

<math>\frac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta}=\frac{\sin \alpha}{\cos \alpha}+\frac{\sin \beta}{\cos \beta}</math>

<math>\sin(\alpha+\beta)=\sin \alpha \cos \beta +\sin \beta \cos \alpha</math>

<math>\mathbb{QED.}</math>

----

<math>\cos (A \pm B)=\cos A \cos B \mp \sin A \sin B</math>

<math>\tan (A \pm B)=\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}</math>

The following identities can be easily derived by plugging <math>A=B</math> into the above:

<math>\sin2A=2\sin A \cos A</math>

<math>\cos2A=\cos^2 A - \sin^2 A</math>
or
<math>\cos2A=2\cos^2 A -1</math>
or
<math>\cos2A=1- 2 \sin^2 A</math>

<math>\tan2A=\frac{2\tan A}{1-\tan^2 A}</math>

===Pythagorean identities===

<math>\sin^2 A+\cos^2 A=1</math> 

<math>1 + \tan^2 A = \sec^2 A</math>

<math>1 + \cot^2 A = \csc^2 A</math>

for all <math>A</math>.

These can be easily seen by going back to the unit circle and the definition of these trig functions.

===Other Formulas===
====Law of Cosines====
In a triangle with sides <math>a</math>, <math>b</math>, and <math>c</math> opposite angles <math>A</math>, <math>B</math>, and <math>C</math>, respectively,

<math>c^2=a^2+b^2-2bc\cos A</math>

and:
====Law of Sines====

<math>\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R</math>

====Law of Tangents====

For any <math>a</math> and <math>b</math> such that <math>\tan a,\tan b \subset \mathbb{R}</math>,
<math>\frac{a-b}{a+b}=\frac{\tan(a-b)}{\tan(a+b)}</math>

====Area of a Triangle====
The [[area]] of a triangle can be found by 

<math>\frac 12ab\sin C</math>

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