sin2pi/7+sin4pi/7+sin8pi/7=
udaybhansingh , 10 Years ago
Grade 12th pass
Trigonometry> sin2pi/7+sin4pi/7+sin8pi/7=
1 Answers
Nishant Vora
Last Activity: 10 Years ago
Hello student, Please find the answer to your question below
sin2pi/7+sin4pi/7+sin8pi/7=
Let 2pi/7 = x
so equation becomes
sin x + sin 2x + sin 4x
[use the following formula sin x + sin 2x + sin 4x + sin 8x + ….. n terms= sin (2^n x ) / 2* sin x ]
= sin 8x / 8 * sin x
= sin (16pi/7) / 8*sin(2pi/7)
= 1/8 [because sin (16pi/7) = sin (2pi /7 )]
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