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if cos x+cosy+cosz=0 and sinx+siny+sinz=0 then find the value of cos2x +cos2y+cos2z

if cos x+cosy+cosz=0 and sinx+siny+sinz=0 then find the value of cos2x +cos2y+cos2z

Grade:11

2 Answers

Vikas TU
14149 Points
4 years ago
cos(x) + cos(y) = -cos(z) [Since given cos(x) + cos(y) + cos(z) = 0]
Squaring both sides, cos²x + cos²y + 2cos(x)*cos(y) = cos²z ----------- (1)
Similarly from the given second data, we have
sin²x + sin²y + 2sin(x)*sin(y) = sin²z ------------ (2)
Adding (1) & (2):
1 + 1 + 2{cos(x)*cos(y) + sin(x)*sin(y)} = 1 [Since by identity cos²x + sin²x = 1]
==> 2*cos(x - y) = -1
==> cos(x - y) = -1/2 --------- (3)
I am in confusion what you need? Whether {cos(x - y)}/2 (or) cos{(x - y)/2}
If it is: {cos(x - y)}/2, then it is = -1/4
If it is: cos{(x - y)/2}, then it is computed as below:
Let (x - y)/2 = u; so cos{(x - y)/2} = cos(u) = (1/2)√{1 + cos(2u)},
[Since by Multiple angle identity 2*cos²u = 1 + cos(2u)]
So, 2cos²{(x - y)/2} = [1+ cos{(x - y)}] = [1 + (-1/2)] = 1/2
==> cos²{(x - y)/2} = 1/4
So cos{(x - y)/2} = ±(1/2)
Aditya Gupta
2081 Points
4 years ago
hello anshuman, clearly vikas ans is wrong as he hasn’t answered the ques you asked, i.e, the value of cos2x +cos2y+cos2z.
lemme tell u an interesting approach:
given cos x+cosy+cosz=0 …..(1)
and sinx+siny+sinz=0 
or i*sinx+ i*siny+ i*sinz= 0 …....(2) where we have multiplied both sides by iota (i= sqrt(-1)).
add (1) and (2)
cosx+isinx + cosy+isiny + cosz+isinz= 0
or e^ix + e^iy + e^iz= 0
similarly subtract (2) from (1)
cosx – isinx + cosy – isiny + cosz – isinz= 0
or e^-ix + e^-iy + e^-iz= 0
let e^ix= a, e^iy= b, e^iz= c
so a+b+c= 0....(3)
and a^-1+b^-1+c^-1= 0 or ab+bc+ca= 0
so, square (3) both sides
a^2 + b^2 + c^2 + 2(ab+bc+ca)= 0
or a^2 + b^2 + c^2 + 2*0= 0
or a^2 + b^2 + c^2= 0
or e^i2x + e^i2y + e^i2z= 0
or cos2x+cos2y+cos2z + i(sin2x+sin2y+sin2z)= 0
so that cos2x +cos2y+cos2z= 0
kindly approve :)

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