Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

prove that: sin2pi/7 + sin4pi/7 + sin8pi/7 = (root7)/2

prove that: sin2pi/7 + sin4pi/7 + sin8pi/7 = (root7)/2

Grade:11

1 Answers

Ramesh V
70 Points
11 years ago

Lets take a = 2pi/7

7a =2pi
sin4a = sin(2pi-3a)
sin4a  = -sin3a
2sin2a.cos2a = 4sin3(a) -3sina
4sin a.cos a (1-2sin2(a)) = sin a(4sin2(a) - 3)
4cos a(1-2sin2(a)) = 4sin2(a) -3

On squaring both sides


16(1-sin2(a)) [1-2sin2(a))]^2 = (4sin2(a) -3)2
64sin6(a) - 112sin4(a) + 56sin2(a) -7 =0
it is cubic in sin2(a)
its roots are sin2(2pi/7) ,sin4(pi/7) ,sin2(8pi/7)
sum of roots =7/4
sin2pi/7*sin4pi/7 +sin4pi/7*sin8pi/7 +sin8pi/7*sin2pi/7 = 0
we can simply prove it by using 2sin a.sin b= cos(a-b) - cos(a+b)
& cos(2pi-theta) = cos theta

(sin2pi/7+sin4pi/7 +sin8pi/7)2=7/4
sin2pi/7+sin4pi/7 +sin8pi/7=[(7)1/2]/2

--

Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and
we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best.

Regards,
Naga Ramesh
IIT Kgp - 2005 batch

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free