Question

cos pi/8 + cos 3pi/8 + cos 5pi/8 +cos 7pi/8 = 0 ... Prove. ​

Answer 1

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We need to prove that

cos π/8 + cos 3π/8 + cos 5π/8 + cos 7π/8 = 0

Let us consider L.H.S

cos π/8 + cos 7π/8 + cos 3π/8 + cos 5π/8

No we will apply here the formula of

cos C + cos D = 2cos (C+D)/2 . cos(C-D)/2

= cos π/8 + cos 7π/8 + cos 3π/8 + cos 5π/8

= 2 cos (π/8+7π/8)/2 . cos (π/8-7π/8)/2 + 2 cos (3π/8+5π/8) . cos (3π/8-5π/8)

= 2 cos π/2 . cos (-6π/16) + 2 cos π/2 . cos (-2π/16)

We know that

We know that cos π/2 = 0

= 2. 0 . cos (-6π/16) + 2. 0 . cos (-2π/16)

= 0 + 0

= 0

= R.H.S

Answer 2

Answer:

No we will apply here the formula of

cos C + cos D = 2cos (C+D)/2 . cos(C-D)/2

= cos π/8 + cos 7π/8 + cos 3π/8 + cos 5π/8

= 2 cos (π/8+7π/8)/2 . cos (π/8-7π/8)/2 + 2 cos (3π/8+5π/8) . cos (3π/8-5π/8)

= 2 cos π/2 . cos (-6π/16) + 2 cos π/2 . cos (-2π/16)

We know that

We know that cos π/2 = 0

= 2. 0 . cos (-6π/16) + 2. 0 . cos (-2π/16)

= 0 + 0

= 0

= R.H.S

No we will apply here the formula of

cos C + cos D = 2cos (C+D)/2 . cos(C-D)/2

= cos π/8 + cos 7π/8 + cos 3π/8 + cos 5π/8

= 2 cos (π/8+7π/8)/2 . cos (π/8-7π/8)/2 + 2 cos (3π/8+5π/8) . cos (3π/8-5π/8)

= 2 cos π/2 . cos (-6π/16) + 2 cos π/2 . cos (-2π/16)

We know that

We know that cos π/2 = 0

= 2. 0 . cos (-6π/16) + 2. 0 . cos (-2π/16)

= 0 + 0

= 0

= R.H.S