Difference between revisions of "User:Temperal/The Problem Solver's Resource1"
m (→Basic Facts: again) 
(→Sum of Angle Formulas: proof) 

Line 52:  Line 52:  
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.  +  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 image1=<asy> 
+  pair A,B,C;  
+  C=(0,0);  
+  B=(10,0);  
+  A=(6,4);  
+  draw(ABCcycle);  
+  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=right3=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>\cos (A \pm B)=\cos A \cos B \mp \sin A \sin B</math> 
Revision as of 20:03, 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
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
, but $\cot A\ne\tan^{1} A}$ (Error compiling LaTeX. ! Extra }, or forgotten $.), the former being the reciprocal and the latter the inverse.
, but $\csc A\ne\sin^{1} A}$ (Error compiling LaTeX. ! Extra }, or forgotten $.).
, but $\sec A\ne\cos^{1} A}$ (Error compiling LaTeX. ! Extra }, or forgotten $.).
Speaking of inverses:
Sum of Angle Formulas
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 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:

Figure 1 
We'll find in two different ways: and . We let . We have:
The following identities can be easily derived by plugging into the above:
or or
Pythagorean identities
for all .
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 , , and opposite angles , , and , respectively,
and:
Law of Sines
Law of Tangents
For any and such that ,
Area of a Triangle
The area of a triangle can be found by