The value of cos 0°+ cos pi/7+cos2pi/7+cos 3pi/7+..........+cos6pi/7 is equals to
Answers
Answer:
Recall that
cos(x) = -cos(π - x)
So
cos(π/7) = -cos(6π/7) ==> cos(π/7) + cos(6π/7) = 0
cos(2π/7) = -cos(5π/7) ==> cos(2π/7) + cos(5π/7) = 0
cos(3π/7) = -cos(4π/7) ==> cos(3π/7) + cos(4π/7) = 0
Thus
=> cos(π/7) + cos(2π/7) + cos(3π/7) + cos(4π/7) + cos(5π/7) + cos(6π/7) + cos(7π/7)
=> cos(π/7) + cos(2π/7) + cos(3π/7) + cos(4π/7) + cos(5π/7) + cos(6π/7) + cos(π)
=> cos(π/7) + cos(2π/7) + cos(3π/7) + cos(4π/7) + cos(5π/7) + cos(6π/7) - 1
=> [ cos(π/7) + cos(6π/7) ] + [ cos(2π/7) + cos(5π/7) ] + [ cos(3π/7) + cos(4π/7) ] - 1
=> 0 + 0 + 0 - 1
=> -1
Answer:
if you're talking about radian measure than the answer would be -1 and in radian measure you cannot factor out cos pie/7 cause you're adding the x-values of each given distance in the unit circle
Step-by-step explanation:
cos(x) = -cos(π - x)
So
cos(π/7) = -cos(6π/7) ==> cos(π/7) + cos(6π/7) = 0
cos(2π/7) = -cos(5π/7) ==> cos(2π/7) + cos(5π/7) = 0
cos(3π/7) = -cos(4π/7) ==> cos(3π/7) + cos(4π/7) = 0
Thus
cos(π/7) + cos(2π/7) + cos(3π/7) + cos(4π/7) + cos(5π/7) + cos(6π/7) + cos(7π/7)
= cos(π/7) + cos(2π/7) + cos(3π/7) + cos(4π/7) + cos(5π/7) + cos(6π/7) + cos(π)
= cos(π/7) + cos(2π/7) + cos(3π/7) + cos(4π/7) + cos(5π/7) + cos(6π/7) - 1
= [ cos(π/7) + cos(6π/7) ] + [ cos(2π/7) + cos(5π/7) ] + [ cos(3π/7) + cos(4π/7) ] - 1
= 0 + 0 + 0 - 1
= -1