Question

Prove that 4cos(2pi/7).cos(pi/7) - 1 = 2cos(2pi/7). Urgent!!!!

Answer 1

Trigo Identities Used :
1) 2cos(A)cos(B) = cos( A+B) + cos( A-B)
2) cos(π-a) = -cos(a)
3) 2sin(A) cos(B) = sin(A+B) + sin(A-B)





Let a =π/7 ,
then we have to prove,
4cos(2a)cos(a) -1 -2cos(2a) = 0

So,
1) LHS =
$4 \cos(2a) \cos(a) - 1 - 2 \cos(2a) \\ = > 2( \cos(3a) + \cos(a) ) - 1 - 2 \cos(2a) \\ = > - 1 + 2( \cos(3a) + \cos(a) - \cos(2a) ) \\ = > - 1 - 2( \cos(\pi - 3a) + \cos(\pi - a) + \cos(2a) )$

2) Now,
Substitute a =π/7
LHS =
$- 1 - 2( \cos( \frac{4\pi}{7} ) + \cos( \frac{6\pi}{7}) + \cos( \frac{2\pi}{7} )$

Now,
Write the trigonometric function as P,
P= 2(cos(4π/7) + cos(6π/7) + cos(2π/7)) :

Multiply both sides by sin(π/7)

P sin(π/7) =
$2 \sin( \frac{\pi}{7} ) \cos( \frac{4\pi}{7} ) + 2\sin( \frac{\pi}{7} ) \cos( \frac{6\pi}{7} ) \\ + 2\sin( \frac{\pi}{7} ) \cos( \frac{2\pi}{7} )$

$= \sin( \frac{5\pi}{7} ) - \sin( \frac{3\pi}{7} ) + \sin( \frac{7\pi}{7} ) - \sin( \frac{5\pi}{7} ) \\ + \sin( \frac{3\pi}{7} ) - \sin( \frac{\pi}{7} ) \\ = \sin( \frac{7\pi}{7} ) - \sin( \frac{\pi}{7} ) \\ = 0 - \sin( \frac{\pi}{7} ) \\ = - \sin( \frac{\pi}{7} )$

We obtained
P sin(π/7) = - sin(π/7)
=> P = -1 .

3)
Now,
LHS = -1 -P
= -1 -(-1)
= -1 +1 = 0

Hence Proved,

Answer 2

{2[2sin(P/7)Cos(P/7)]Cos(2P/7)}/Sin(P/7)-1 (By dividing numerator and denominator with Sin(P/7) =[2sin(2P/7)Cos(2P/7)]/Sin(P/7)-1 =Sin(4P/7) -1 Sin(P/7) =Sin(P-4P/7) -1 Sin(P/7) =Sin(3P/7) -1 Sin(p/7) =Sin(3P/7)-Sin(P/7) Sin(P/7) =2Cos[(3P+P)/2*7]Sin[(3P-P)/2*7] {SinC-SinD=2Cos[(C+D)/2]Sin[(C-D)/2]} Sin(P/7) =2Cos(2P/7)Sin(P/7) (by cancelling Sin(P/7) in Numerator and denominator) Sin(P/7) =2Cos(2P/7)