Deriving+Sine+&+Cosine

= = = Deriving ** the Law **** of Sine **=

//***If you are at school and can not watch this, please download videos using the link providing:**// //**http://us.onlinevideoconverter.com/free-video-converter.aspx**//

http://www.youtube.com/watch?v=sK6J7z-cXh0&feature=youtu.be

Provided above is triangle ABC and triangle ABD. Given that this, the Sine of angle ADB can be defined as the opposite side over the hypotenuse of the triangle ADB. The opposite side is represented by lowercase c and the hypotenuse is represented by segment AD. Segment AD is also equivalent to two times the radius or diameter.

*Note: The side opposite an angle with an uppercase letter is named with the lowercase version of that letter.*



Angles ADB and ACB both intercept arc AB. Thus, these angles are congruent and the Sine of angle ACB is equal to side c divided by the diameter (2 times the radius). The above expression can be rearranged by multiplying both sides by 2r and then dividing both sides by SinC. The expression then looks like this,



The measurement of the length of side a over the Sine of angle A is equal to 2 times the radius. The Measurement of the length of side b over the Sine of angle B is also equal to 2r. Since all three of these expressions are equal to 2r, all of the expressions are also equivalent to each other.



When this formula is used the reciprocal, which also shows that the expressions are equivalent, is more common.




 * (Roberts) **

=**Deriving the Law of Cosine**=

http://www.youtube.com/watch?v=Hf6Ex9Oz89g&feature=youtu.be


 * ( Bogomolny)**

Triangle ABC has no right angles however a perpendicular is extended from angle B to side b to create two right triangles. By the definition of Sine it can be observed that Both of these expressions can be rewritten by multiplying both sides by c, like so… Focusing on triangle BDC we can use the Pythagorean theorum. The hypotenuse of triangle BDC is represented by the variable a and the two sides can be represented by variable h and b-r for side DC. Substituting this into the Pythagorean theorum we are left with We also know from previous work what h and r are equal to so lets subsitute this into the expression. To further solve this expression, we can expand using FOIL. Start with the first portion . Now focus on the second portion. Now combine these two for the full expression: Grouping Sin squared A and Cos squared A, we can recall the identity Cos squared theta plus sin squared theta is equal to 1. This same method can be used to produce other versions of this Law using the other letters:
 * ( Bogomolny)**