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ghanami
hello : 
note : the polar coordinates is : (r , θ)
Another representation is  (x, y)   : x=r cos(θ)   and y = sin(θ) and : r =√(x²+y²)
in this exercice : r = 1    and  θ = 5π/6 
so : x= 1.cos(5π/6)   and y = 1. sin(5π/6)
but : cos(5π/6)  = cos (π - π/6) = - cos(π/6 ) = (- √3)/2
        sin(5π/6)  = sin (π - π/6) =  sin(π/6 ) =1/2
Another representation  is : ( (- √3)/2 , 1/2)

Answer:  The representation of the point in Cartesian co-ordinate system is

[tex]\left(\dfrac{\sqrt3}{2},-\dfrac{1}{2}\right),~\left(-\dfrac{\sqrt3}{2},\dfrac{1}{2}\right).[/tex]

Step-by-step explanation:  We are given to find the other representation of the point [tex]\left(1,\dfrac{5\pi}{6}\right)[/tex] in Cartesian system

We know that

if (x, y) are the co-ordinates of a point in two dimensional XY-plane and (r, Θ) are the co-ordinates of the point in polar co-ordinate system, then we have the following relations:

[tex]x^2+y^2=r^2,~~~~~\tan \theta=\dfrac{y}{x}.[/tex]

Given that

[tex](r,\theta)=\left(1,\dfrac{5\pi}{6}\right).[/tex]

Therefore, we have

[tex]x^2+y^2=1^2\\\\\Rightarrow x^2+y^2=1,[/tex]

and

[tex]\tan \theta=\dfrac{y}{x}\\\\\\\Rightarrow \tan \dfrac{5\pi}{6}=\dfrac{y}{x}\\\\\\\Rightarrow \tan(\pi-\dfrac{\pi}{6})=\dfrac{y}{x}\\\\\\\Rightarrow -\tan\dfrac{\pi}{6}=\dfrac{y}{x}\\\\\\\Rightarrow -\dfrac{1}{\sqrt3}=\dfrac{y}{x}\\\\\\\Rightarrow x=-\sqrt3y.[/tex]

So,

[tex]x^2+y^2=1\\\\\Rightarrow (-\sqrt3y)^2+y^2=1\\\\\Rightarrow 4y^2=1\\\\\Rightarrow y^2=\dfrac{1}{4}\\\\\\\Rightarrow y=\pm\dfrac{1}{2}.[/tex]

And,

When [tex]x=\dfrac{1}{2},[/tex],

[tex]-\dfrac{1}{\sqrt3}=\dfrac{1}{2x}\\\\\Rightarrow x=-\dfrac{\sqrt3}{2}.[/tex]

When [tex]x=-\dfrac{1}{2},[/tex],

[tex]-\dfrac{1}{\sqrt3}=-\dfrac{1}{2x}\\\\\Rightarrow x=\dfrac{\sqrt3}{2}.[/tex]

Thus, the representation of the point in Cartesian co-ordinate system is

[tex]\left(\dfrac{\sqrt3}{2},-\dfrac{1}{2}\right),~\left(-\dfrac{\sqrt3}{2},\dfrac{1}{2}\right).[/tex]

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