Trigonometry to Memorize, and Trigonometry to Derive

Aug 2014
6
0
unitedstates
I don't suppose anyone wants to take upon themselves the challenge/effort of writing up how each identity on the 2nd page can be derived from the stuff on the 1st page? (I'd try, but doubt I could get them all).
 
Feb 2014
1,748
651
United States
I don't suppose anyone wants to take upon themselves the challenge/effort of writing up how each identity on the 2nd page can be derived from the stuff on the 1st page? (I'd try, but doubt I could get them all).
You are both missing the point and selling yourself short.

The items on the second page involve virtually no work to derive from the items on the first page. Let's take one of the harder ones.

$tan(x + y) = \dfrac{sin(x + y)}{cos(x + y}.$

That comes from $tan( \theta ) = \dfrac{sin ( \theta )}{cos( \theta )}$ on page 1.

$So\ tan(x + y) = \dfrac{sin(x)cos(y) + cos(x)sin(y)}{cos(x + y)}$

That comes from $sin(x + y) = sin(x)cos(y) + cos(x)sin(y)$ on page 1.

$So\ tan(x + y) = \dfrac{sin(x)cos(y) + cos(x)sin(y)}{cos(x)cos(y) - sin(x)sin(y)}.$

That comes from $cos(x + y) = cos(x)cos(y) - sin(x)sin(y)$ on page 1.

Now translate into tangents using $tan( \theta ) = \dfrac{sin( \theta )}{cos( \theta )}$ from page 1.

$So\ tan(x + y) = \dfrac{cos(x)cos(y)\left \{\dfrac{sin(x)}{cos(x)} + \dfrac{sin(y)}{cos(y)}\right \}}{cos(x)cos(y)\left \{1 - \dfrac{sin(x)}{cos(x)} * \dfrac{sin(y)}{cos(y)}\right \}} \implies$

$tan(x + y) = \dfrac{tan(x) + tan(y)}{1 - tan(x)tan(y)}.$

Now this is just algebra applied to a few basic memorized formulas. Each of the formulas on the second page involve a few simple manipulations of what is on the first page. You can memorize the second page, but you do not need to. Furthermore, deriving them on your own will give you confidence in your ability to handle more complicated transformations among trigonometric functions.