Question

Show that cos ( 3π/ 17) cos ( 9π/17) + 1 2 [cos ( 11π/17 ) + cos ( 29π/17 ) ] = 0​

Answer 1

Given : cos ( 3π/ 17) cos ( 9π/17) + (1 /2) [cos ( 11π/17 ) + cos ( 29π/17 ) ] = 0​

To Find : Show that  

Solution:

cos ( 3π/ 17) cos ( 9π/17) + (1 /2) [cos ( 11π/17 ) + cos ( 29π/17 ) ] = 0​

LHS = cos ( 3π/ 17) cos ( 9π/17) + (1 /2) [cos ( 11π/17 ) + cos ( 29π/17 ) ]

Cos C + Cos D  = 2 Cos (C + D)/2  Cos ( C - D)/2

C = 29π/17  , D =  11π/17

=  cos ( 3π/ 17) cos ( 9π/17) + (1 /2) [2cos ( 20π/17 ) cos (9π/17 ) ]

=  cos ( 3π/ 17) cos ( 9π/17) +  cos ( π + 3π/17 ) cos (9π/17 )  

Cos ( π + α) = - Cosα

=  cos ( 3π/ 17) cos ( 9π/17) +  (-   cos ( 3π/ 17) )cos (9π/17 )  

= cos ( 3π/ 17) cos ( 9π/17) -  cos ( 3π/ 17) )cos (9π/17 )  

= 0

= RHS

QED

Hence proved

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Answer 2

Step-by-step explanation:

cos (3π/17) cos (9/17) + (1/2) [cos (11π/ 17) + cos (29π/17)] = 0

LHS = cos (3π/17) сos (9/17) + (1/2) [cos (111/17) + cos (291/17)]

Cos C + Cos D = 2 Cos (C + D)/2 Cos (C -D)/2

C = 29/17, D = 11/17

= cos (3π/17) cos (9/17) + (1/2) [2cos (201/17) cos (9π/17)]

= cos (3π/17) cos (9/17) + cos (π + 3π/ 17) cos (9π/17)

Сos (π + a) = - Cosa

= COS (3π/17) cos (9π/17) + (- сos (3π/ 17) )cos (9π/17)

= cos (3π/17) cos (9/17) - cos (3π/ 17) )cos (9π/17)

= 0

= RHS

QED

Hence proved