Periodicity of trigonometric functions

Asked by  | 12th Jul, 2008, 08:10: AM

Expert Answer:

two formula will be useful here sinC+sinD = 2 sin (C+D)/2 cos(C-D)/2

and cosC-cosD = -2 sin (C+D)/2 sin(C-D)/2

first we can write b+2c = pi-a+c = pi-(a-c)

so sin (b+2c) =sin ( pi-(a-c) ) = sin (a-c)

similarly sin (c+2a) = sin (b-a) and sin (a+2b) = sin (c-b)

sin (a-c) + sin (b-a) = 2 sin (b-c)/2   cos (2a-b-c)/2

sin (c-b) = -2 sin (b-c)/2  cos (b-c)/2

sin (a-c) + sin (b-a) + sin (b-c) = 2 sin (b-c)/2  ( cos (2a-b-c)/2 - cos (b-c)/2 )

= 2 sin (b-c)/2 . (-)2 sin (2a-2c)/4  sin ( 2a-2b)/4

= -4 sin (b-c)/2 . sin (a-c)/2  sin ( a-b)/2

=-4 sin (b-c)/2 . (-)sin (c-a)/2  sin ( a-b)/2

=4 sin (b-c)/2 . sin (c-a)/2  sin ( a-b)/2

 hence proved

Answered by  | 13th Sep, 2008, 12:29: AM

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