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.

Back to intro | Continue to page 2

@@ Line 14: / Line 14: @@
 <math>\cos (\pi-A) = -\cos A</math>
-<math>\cos (2\pi-A) = \cos A</math>
+<math>\sin (-A) = -\sin A</math>
+<math>\cos (-A) = \cos A</math>
 <math>\tan (\pi+A) = \tan A</math>
@@ Line 22: / Line 24: @@
 <math>\tan (\pi/2-A)=\cot A</math>
-<math>\sec{\pi/2-A}=\csc A</math>
+<math>\sec (\pi/2-A)=\csc A</math>
 <math>\cos (\pi/2-A) = \sin A</math>
@@ Line 33: / Line 35: @@
 ===Terminology and Notation===
-<math>\cot A=\frac{1}{\tan A}</math>, but <math>\cot A\ne\tan^{-1} A}</math>.
+<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>.
-Also:
+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===
 <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>
@@ Line 73: / Line 113: @@
 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===
@@ Line 78: / Line 120: @@
 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>
+<math>c^2=a^2+b^2-2ab\cos C</math>
+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)
-and:
 ====Law of Sines====
 <math>\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R</math>
+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====
-For any <math>a</math> and <math>b</math> such that <math>\tan a,\tan b \subset \mathbb{R}</math>,
+If <math>A</math> and <math>B</math> are angles in a triangle opposite sides <math>a</math> and <math>b</math> respectively, then
-<math>\frac{a-b}{a+b}=\frac{\tan(a-b)}{\tan(a+b)}</math>
+<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>
+By the angle addition identities,
+<cmath> \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} </cmath>
+as desired.
 ====Area of a Triangle====
@@ Line 94: / Line 154: @@
 <math>\frac 12ab\sin C</math>
+This can be easily proven by the well-known formula <math>\frac{1}{2}ah_a</math> - considering one of the triangles which altitude <math>h_a</math> divides <math>\triangle ABC</math> into, we see that <math>h_a=b\sin C</math> and hence <math>[ABC]=\frac 12ab\sin C</math> as desired.
 [[User:Temperal/The Problem Solver's Resource|Back to intro]] | [[User:Temperal/The Problem Solver's Resource2|Continue to page 2]]

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