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
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
where is the radius of the circumcircle of
Law of Tangents
If and are angles in a triangle opposite sides and respectively, then
The proof of this is less trivial than that of the law of sines and cosines, but still fairly easy: Let and denote , , respectively. By the law of sines, By the angle addition identities, as desired.
Area of a Triangle
The area of a triangle can be found by
This can be easily proven by the wellknown formula  considering one of the triangles which altitude divides into, we see that and hence as desired.