Page 39 - modul trigonometri_Neat
P. 39

Comparative Identity

                                      sin   
                                            = tan   
                                      cos   
                                      cos   
                                             = cot   
                                      sin   


                           3.  Pythagoras Identity

                               In the right triangle applies the following Pythagorean theorem.
                               The sum of the squares of the sides of the elbows is equal to the

                               square of the oblique side.
                                    2
                                          2
                                                 2
                               ↔    +    =   

                                                                                                 2
                               a.  The two segments of the equation (1) are divided by   
                                     2
                                      +   2       2
                                            =
                                         2        2
                                            2        2
                                   ↔ ( ) + ( ) = 1
                                                   
                                                     2
                                          2
                                   ↔           +           = 1

                                                                                                2
                               b.  The two segments of the equation (1) are divided by  
                                     2
                                      +   2       2
                                            =
                                         2        2
                                            2              2
                                   ↔ ( ) + 1 = ( )
                                                         
                                           2
                                                          2
                                   ↔           + 1 =          
                                                          2
                                                2
                                   ↔ 1 +           =          

                                                                                                2
                               c.  The two segments of the equation (1) are divided by  
                                     2
                                      +   2       2
                                            =
                                         2       
                                                 2         2
                                   ↔  1 + ( ) = ( )
                                                         
                                                          2
                                               2
                                   ↔ 1 +           =          










                                                                                                     35
   34   35   36   37   38   39   40   41   42   43   44