the expression [(1+sinπ/8+icosπ/8)/(1+sinπ/8-icosπ/8)]^8 is..
Priyanka1912:
where
ya lol but that was mistake again u can see in comments that i had told the user that its wrong
ya! ..
hmm.. i have to take care of it now.. i m sorry
have u solved that one too?
ya i have solved
no need for sorry , just take care next time
okay
ok thank u so much for ur concern dear
Ur welcome ... :)
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1 + cos π/8)(1 + cos 3π/8)(1 + cos 5π/8)(1 + cos 7π/8)
= (1 + cos π/8)(1 + cos 3π/8)(1 - cos 3π/8)(1 - cos π/8), since cos(π - t) = -cos t
= (1 - cos² π/8)(1 - cos² 3π/8)
= (1 - cos² π/8)(1 - sin² π/8), via cos(π/2 - t) = sin t
= (1 - cos² π/8)(1 - sin² π/8)
= sin ^2 π/8 cos ^2 π/8
= (1/4) (2 sin π/8 cos π/8)²
= (1/4) (sin 2π/8)², by double angle formula
= (1/4) (sin π/4)²
= (1/4) (1/√2)²
= (1/4)(1/2)
= 1/8.
= (1 + cos π/8)(1 + cos 3π/8)(1 - cos 3π/8)(1 - cos π/8), since cos(π - t) = -cos t
= (1 - cos² π/8)(1 - cos² 3π/8)
= (1 - cos² π/8)(1 - sin² π/8), via cos(π/2 - t) = sin t
= (1 - cos² π/8)(1 - sin² π/8)
= sin ^2 π/8 cos ^2 π/8
= (1/4) (2 sin π/8 cos π/8)²
= (1/4) (sin 2π/8)², by double angle formula
= (1/4) (sin π/4)²
= (1/4) (1/√2)²
= (1/4)(1/2)
= 1/8.
hey thanks for the answer but it's wrong. i mean to say in question there is iota and u have not used that... plz check it
oh I am sorry
it's ok
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