Question : Let A1 A2 A3 A4 A5 A6 A1 be a regular hexagon. Write the x-components of the vectors represented by the six sides taken in order. Use the fact that the resultant of these six vectors is zero. Use the fact that the resultant of these six vectors is zero, to prove that cos0 + cosπ/3 + cos2π/3 + cos3π/3 + cos4π/3 + cos5π/3 = 0.
Use the known cosine values to verify the result.
Doubt by Muskan
Solution :
We know, all the sides as well as interior angles (each equals to 120°) of a regular hexagon are same.
Let
A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A1 = A (Magnitude)
We also know that
Horizontal component (X-component) of a vector is given by Acosθ and
Horizontal component (X-component) of a vector is given by Acosθ and
Vertical component (Y-component) of a vector is given by ASinθ
where θ is the angle between the given vector and x-axis.
X-component of A1A2 vector = Acos0
Y-component of A1A2 Vector = Asin0
Y-component of A1A2 Vector = Asin0
X-component of A2A3 vector = Acos60° = Acosπ/3
Y-component of A2A3 Vector = Asin60°=Asinπ/3
Y-component of A2A3 Vector = Asin60°=Asinπ/3
X-component of A3A4 vector = Acos120° = Acos2π/3
Y-component of A3A4 Vector = Asin120°=Asin2π/3
Y-component of A3A4 Vector = Asin120°=Asin2π/3
X-component of A4A5 vector = Acos180° = Acosπ = Acos3π/3
Y-component of A4A5 Vector = Asin180°=Asinπ = Asin3π/3
Y-component of A4A5 Vector = Asin180°=Asinπ = Asin3π/3
X-component of A5A6 vector = Acos240° = Acos4π/3
Y-component of A5A6 Vector = Asin240°= Asin4π/3
Y-component of A5A6 Vector = Asin240°= Asin4π/3
X-component of A6A1 vector = Acos300° = Acos5π/3
Y-component of A5A6 Vector = Asin300°= Asin5π/3
Y-component of A5A6 Vector = Asin300°= Asin5π/3
R = (Resultant of X-Component)i+(Resultant of Y-Component)j
R =(Acosπ/3+Acos2π/3+Acos3π/3+Acos4π/3+Acos5π/3)i + (Asinπ/3+Asin2π/3+Asin3π/3+Asin4π/3+Asin5π/3)j
ATQ
R = 0
R = 0i+0j
R = 0
R = 0i+0j
So
0i+0j=(Acos0+Acosπ/3+Acos2π/3+Acos3π/3+Acos4π/3+Acos5π/3)i + (Asin0+Asinπ/3+Asin2π/3+Asin3π/3+Asin4π/3+Asin5π/3)j
0i+0j=(Acos0+Acosπ/3+Acos2π/3+Acos3π/3+Acos4π/3+Acos5π/3)i + (Asin0+Asinπ/3+Asin2π/3+Asin3π/3+Asin4π/3+Asin5π/3)j
Equating the coefficient of i both sides
0=Acos0+Acosπ/3+Acos2π/3+Acos3π/3+Acos4π/3+Acos5π/3
0=A(Acos0+cosπ/3+cos2π/3+cos3π/3+cos4π/3+cos5π/3)
0=cos0+cosπ/3+cos2π/3+cos3π/3+cos4π/3+cos5π/3
cos0+cosπ/3+cos2π/3+cos3π/3+cos4π/3+cos5π/3 = 0
Hence Proved.
Hence Proved.
Verification :
cos0+cosπ/3+cos2π/3+cos3π/3+cos4π/3+cos5π/3
= cos0+cosπ/3+cos(π-π/3)+cosπ+cos(π+π/3)+cos(2π+π/3)
= cos0+cosπ/3+[-cos(π/3)]+cosπ+[-cos(π/3)]+cos(π/3)
[∵
cos(π-θ) =- cosθ
cos(π+θ) =- cosθ
[∵
cos(π-θ) =- cosθ
cos(π+θ) =- cosθ
cos(2π+θ) = cosθ
]
= cos0+cosπ/3-cos(π/3)+cosπ-cos(π/3)+cos(π/3)
= cos0+0 + cosπ + 0
= cos0+cosπ
= 1+(-1) [∵ cos0 =1, cosπ=-1]
=0
= cos0+0 + cosπ + 0
= cos0+cosπ
= 1+(-1) [∵ cos0 =1, cosπ=-1]
=0
Hence verified.