if arccos x+ arccos y + arccos z= ╥ then prove x^2 +y^2 +z^2+ 2*x*y*z = 1
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Put arccosx=m, arccosy=n and arccos y=p => `m+n+p=pi`
Subtract p both sides =>`m+n=pi-p`
calculate cosine value of both sides:
`cos(m+n)=cos(pi-p)`
use the formula: `cos(m+n)=cos m*cos n-sin m*sin n`
`cos(pi-p)=cos pi*cos p+sin pi*sin p (cos pi=-1, sin pi=0)`
`cos m*cos n-sin m*sin n=-cos p`
cos m=cos(arccos x)=x
cos n=cos(arccos y)=y
cos p=cos(arccosz)=z
`sin m=sin(arccos x)=sqrt(1-x^2)`
`` `sin n=sqrt(1-y^2)`
`cos m*cos n-sin m*sin n=-cosp ` <=> `xy-sqrt((1-x^2)(1-y^2))=-z`
`sqrt((1-x^2)(1-y^2))=z+xy`
Raise to square both sides to remove the square root:
`(1-x^2)(1-y^2)=(z+xy)^2`
Remove the brackets:
`1-x^2-y^2+x^2*y^2=z^2+2xyz+x^2*y^2`
Reduce the same quantity`x^2*y^2` both sides
`1-x^2-y^2=z^2+2xyz`
Add `x^2+y^2 ` both sides
`x^2+y^2+z^2+2xyz=1`
Answer: `x^2+y^2+z^2+2xyz=1` proves the identity `arccos x+arccos y+arccos z=pi` .
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