Grade 12th passTrigonometry

sin2 ,sin4 ,sin6, --------sin180 ka avverage how solve

rakesh Kushwaha

8 Years agoGrade 12th pass

1 Answer

Arun

8 Years ago

2 sin(2) + 178 sin(178) = 180 sin 2
4 sin (4) + 176 sin(176) = 180 sin 4
6 sin (6) + 174 sin (176) = 180 sin 6
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90 sin (90) + 90 sin (180 - 90 ) = 180 sin 90

so avg. = [ sin 2 + sin 4 + sin 6 + sin 8 + .......+ sin 90180 ] x 180

Now . sin a + sin(a+d) + sin (a+2d ) + ...... + sin (a+nd ) = sin (na /2 ) sin ( a + {n-1} d/2 )sin (d/2)

Compaaring with the given series , a = 2 , d =2 , n = 2

so avg. = sin 90 sin ( 2 + 89 )sin 1 = cot 1

Regards

Arun (askIITians forum expert)