Difference between revisions of "User:Temperal/The Problem Solver's Resource1"

(Law of Tangents: from the wiki)
(Law of Tangents: hm)
Line 134: Line 134:
 
<cmath> \frac{a-b}{a+b}=\frac{\tan (A-B)/2}{\tan (A+B)/2} . </cmath>
 
<cmath> \frac{a-b}{a+b}=\frac{\tan (A-B)/2}{\tan (A+B)/2} . </cmath>
  
 +
The proof of this is less trivial than that of the law of sines and cosines, but still fairly easy:
 
Let <math>s</math> and <math>d</math> denote <math>(A+B)/2</math>, <math>(A-B)/2</math>, respectively. By the law of sines,
 
Let <math>s</math> and <math>d</math> denote <math>(A+B)/2</math>, <math>(A-B)/2</math>, respectively. By the law of sines,
 
<cmath> \frac{a-b}{a+b} = \frac{\sin A - \sin B}{\sin A + \sin B} = \frac{ \sin(s+d) - \sin (s-d)}{\sin(s+d) + \sin(s-d)} . </cmath>
 
<cmath> \frac{a-b}{a+b} = \frac{\sin A - \sin B}{\sin A + \sin B} = \frac{ \sin(s+d) - \sin (s-d)}{\sin(s+d) + \sin(s-d)} . </cmath>

Revision as of 21:25, 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

If $A$ and $B$ are angles in a triangle opposite sides $a$ and $b$ respectively, then \[\frac{a-b}{a+b}=\frac{\tan (A-B)/2}{\tan (A+B)/2} .\]

The proof of this is less trivial than that of the law of sines and cosines, but still fairly easy: Let $s$ and $d$ denote $(A+B)/2$, $(A-B)/2$, respectively. By the law of sines, \[\frac{a-b}{a+b} = \frac{\sin A - \sin B}{\sin A + \sin B} = \frac{ \sin(s+d) - \sin (s-d)}{\sin(s+d) + \sin(s-d)} .\] By the angle addition identities, \[\frac{\sin(s+d) - \sin(s-d)}{\sin(s+d) + \sin(s-d)} = \frac{2\cos s \sin d}{2\sin s \cos d} = \frac{\tan d}{\tan s} = \frac{\tan (A-B)/2}{\tan (A+B)/2}\] as desired.

Area of a Triangle

The area of a triangle can be found by

$\frac 12ab\sin C$

This can be easily proven by the well-known formula $\frac{1}{2}ah_a$ - considering one of the triangles which altitude $h_a$ divides $\triangle ABC$ into, we see that $h_a=b\sin C$ and hence $[ABC]=\frac 12ab\sin C$ as desired.

Back to intro | Continue to page 2

Invalid username
Login to AoPS