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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.
 
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:
+
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>
 
{{asy image|1=<asy>

Revision as of 22:02, 10 January 2009

Introduction | Other Tips and Tricks | Methods of Proof | You are currently viewing page 1.

Trigonometric Formulas

Note that all measurements are in radians.

Basic Facts

$\sin (-A)=-\sin A$

$\cos (-A)=\cos A$

$\tan (-A)=-\tan A$

$\sin (\pi-A) = \sin A$

$\cos (\pi-A) = -\cos A$

$\sin (-A) = -\sin A$

$\cos (-A) = \cos A$

$\tan (\pi+A) = \tan A$

$\cos (\pi/2-A)=\sin A$

$\tan (\pi/2-A)=\cot A$

$\sec (\pi/2-A)=\csc A$

$\cos (\pi/2-A) = \sin A$

$\cot (\pi/2-A)=\tan A$

$\csc (\pi/2-A)=\sec A$

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

$\cot A=\frac{1}{\tan A}$, but $\cot A\ne\tan^{-1} A}$ (Error compiling LaTeX. ! Extra }, or forgotten $.), the former being the reciprocal and the latter the inverse.

$\csc A=\frac{1}{\sin A}$, but $\csc A\ne\sin^{-1} A}$ (Error compiling LaTeX. ! Extra }, or forgotten $.).

$\sec A=\frac{1}{\sin A}$, but $\sec A\ne\cos^{-1} A}$ (Error compiling LaTeX. ! Extra }, or forgotten $.).

Speaking of inverses:

$\tan^{-1} A=\text{atan } A=\arctan A$

$\cos^{-1} A=\text{acos } A=\arccos A$

$\sin^{-1} A=\text{asin } A=\arcsin A$

Sum of Angle Formulas

$\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B$

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 $A=-B$ 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] 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]

Enlarge.png
Figure 1

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

$[ABC]=[ABH]+[ACH]$

$\frac{1}{2}(AC)(AB)(\sin \angle BAC)=\frac{1}{2}(AH)(BH)+\frac{1}{2}(AH)(CH)$

$\frac{1}{2}\left(\frac{1}{\cos \beta}\right)\left(\frac{1}{\cos \alpha}\right)(\sin \angle BAC)=\frac{1}{2}(1)(\tan \alpha)(\tan \beta)$

$\frac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta}=\frac{\sin \alpha}{\cos \alpha}+\frac{\sin \beta}{\cos \beta}$

$\sin(\alpha+\beta)=\sin \alpha \cos \beta +\sin \beta \cos \alpha$

$\mathbb{QED.}$


$\cos (A \pm B)=\cos A \cos B \mp \sin A \sin B$

$\tan (A \pm B)=\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$

The following identities can be easily derived by plugging $A=B$ into the above:

$\sin2A=2\sin A \cos A$

$\cos2A=\cos^2 A - \sin^2 A$ or $\cos2A=2\cos^2 A -1$ or $\cos2A=1- 2 \sin^2 A$

$\tan2A=\frac{2\tan A}{1-\tan^2 A}$

Pythagorean identities

$\sin^2 A+\cos^2 A=1$

$1 + \tan^2 A = \sec^2 A$

$1 + \cot^2 A = \csc^2 A$

for all $A$.

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 $a$, $b$, and $c$ opposite angles $A$, $B$, and $C$, respectively,

$c^2=a^2+b^2-2ab\cos C$

and:

Law of Sines

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

where $R$ is the radius of the circumcircle of $\triangle ABC$

Law of Tangents

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

Area of a Triangle

The area of a triangle can be found by

$\frac 12ab\sin C$

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