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

 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: $[asy] pair A,B,C; C=(0,0); B=(10,0); A=(6,4); draw(A--B--C--cycle); label("A",A,N); label("B",B,E); label("C",C,W); draw(A--(6,0)); label("\beta",A,(-1,-2)); label("\alpha",A,(1,-2.5)); label("H",(6,0),S); draw((6,0)--(5.5,0)--(5.5,0.5)--(6,0.5)--cycle); [/asy]$ 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.