Cos 10π/13 + Cos 8π/13 + Cos 3π/13 + Cos 5π/13 = 0 ......Prove it
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cos10pi/13 + cos8pi/13 +cos3pi/13 +cos5pi/13
=> cos (pi-3pi/13) +cos (pi-5pi/13)+cos3pi/13+cos5pi/13
=>-cos3pi/13 -cos5pi/13 +cos3pi/13 +cos5pi/13
=> 0
=> cos (pi-3pi/13) +cos (pi-5pi/13)+cos3pi/13+cos5pi/13
=>-cos3pi/13 -cos5pi/13 +cos3pi/13 +cos5pi/13
=> 0
mysticd:
Nice work
Start with lhs=
Remove implies and put = symbol
Finally end with = rhs
Answered by
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Proving...: - has
=>cos (10π/13) + cos (8π/13) + cos (3π/13) + cos (5π/13) [Using 2 cosA cosB = cos(A+B) + cos(A-B) ]
Now take 1st and 3rd term and 2nd and 4th term and apply cosA + cosB = 2 cos[(A+B)/2] cos[(A-B)/2]
=> 2cos(π/2)cos(7π/26)+ 2cos(π/2)cos(3π/26) = 0 as cos(π/2) = 0
=>cos (10π/13) + cos (8π/13) + cos (3π/13) + cos (5π/13) [Using 2 cosA cosB = cos(A+B) + cos(A-B) ]
Now take 1st and 3rd term and 2nd and 4th term and apply cosA + cosB = 2 cos[(A+B)/2] cos[(A-B)/2]
=> 2cos(π/2)cos(7π/26)+ 2cos(π/2)cos(3π/26) = 0 as cos(π/2) = 0
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