CBSE Class 11-science Answered
Periodicity of trigonometric functions
Asked by | 12 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 | 13 Sep, 2008, 12:29: AM
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