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`        if cos x+cosy+cosz=0 and sinx+siny+sinz=0 then find the value of cos2x +cos2y+cos2z`
3 months 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 havesin²x + 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 = -1/4If 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} = 1/4So cos{(x - y)/2} = ±(1/2)
```
3 months 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= 0or e^ix + e^iy + e^iz= 0similarly subtract (2) from (1)cosx – isinx + cosy – isiny + cosz – isinz= 0or e^-ix + e^-iy + e^-iz= 0let e^ix= a, e^iy= b, e^iz= cso a+b+c= 0....(3)and a^-1+b^-1+c^-1= 0 or ab+bc+ca= 0so, square (3) both sidesa^2 + b^2 + c^2 + 2(ab+bc+ca)= 0or a^2 + b^2 + c^2 + 2*0= 0or a^2 + b^2 + c^2= 0or e^i2x + e^i2y + e^i2z= 0or cos2x+cos2y+cos2z + i(sin2x+sin2y+sin2z)= 0so that cos2x +cos2y+cos2z= 0kindly approve :)
```
3 months ago
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