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

Latest revision as of 17:18, 21 January 2016

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 $\cot{A} \ne \tan^{-1} A$ , the former being the reciprocal and the latter the inverse.

$\csc A=\frac{1}{\sin A}$ , but $\csc{A} \ne \sin^{-1} A$ .

$\sec A=\frac{1}{\sin A}$ , but $\sec{A} \ne \cos^{-1} A$ .

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$",(6,0),S); draw((6,0)--(5.5,0)--(5.5,0.5)--(6,0.5)--cycle); [/asy]$

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,

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

The proof is left as an exercise for the reader. (Hint: Draw a circle with one of the sides as a radius and use the power of a point theorem)

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 $\triangle ABC$

Proof: In the diagram below, circle $O$ circumscribes triangle $ABC$ . $OD$ is perpendicular to $BC$ . Since $\triangle ODB \cong \triangle ODC$ , $BD = CD = \frac a2$ and $\angle BOD = \angle COD$ . But $\angle BAC = 2\angle BOC$ making $\angle BOD = \angle COD = \theta$ . Therefore, we can use simple trig in right triangle $BOD$ to find that

$\sin \theta = \frac{\frac a2}R \Leftrightarrow \frac a{\sin\theta} = 2R.$

The same holds for $b$ and $c$ , thus establishing the identity.

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.

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@@ Line 35: / Line 35: @@
 ===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>\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>\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>.
+<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>\tan^{-1} A=\text{atan } A=\arctan{A}</math>
-<math>\cos^{-1} A=\text{acos } A=\arccos A</math>
+<math>\cos^{-1} A=\text{acos } A=\arccos {A}</math>
-<math>\sin^{-1} A=\text{asin } A=\arcsin A</math>
+<math>\sin^{-1} A=\text{asin } A=\arcsin{A}</math>
 ===Sum of Angle Formulas===
@@ Line 122: / Line 122: @@
 <math>c^2=a^2+b^2-2ab\cos C</math>
-and:
+The proof is left as an exercise for the reader. (Hint: Draw a circle with one of the sides as a radius and use the power of a point theorem)
 ====Law of Sines====
@@ Line 128: / Line 129: @@
 where <math>R</math> is the radius of the circumcircle of <math>\triangle ABC</math>
+Proof: In the diagram below, circle <math> O </math> [[circumscribe]]s triangle <math> ABC </math>.  <math> OD </math> is [[perpendicular]] to <math> BC </math>.  Since <math> \triangle ODB \cong \triangle ODC </math>, <math> BD = CD = \frac a2 </math> and <math> \angle BOD = \angle COD </math>.  But <math> \angle BAC = 2\angle BOC </math> making <math> \angle BOD = \angle COD = \theta </math>.  Therefore, we can use simple trig in [[right triangle]] <math> BOD </math> to find that
+<center><math> \sin \theta = \frac{\frac a2}R \Leftrightarrow \frac a{\sin\theta} = 2R. </math> </center>
+The same holds for <math>b</math> and <math>c</math>, thus establishing the identity.
+<center>[[Image:Lawofsines.PNG]]</center>
 ====Law of Tangents====
@@ Line 134: / Line 143: @@
 <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,
 <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>

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