# What is the value of the expression `sin((5pi)/3)`?

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We wish to find the value of the expression ` ``sin((5pi)/3)` .

The domain of `x` over one whole period of the sine function `sin(x)` is ` `0 to `2pi` . If we look at a graph of a single period of the sine function we see that it has '*odd symmetry*' about `x=pi`. That is, the function has rotational symmetry of 180 degrees about `x=pi` so that `sin(x) = -sin(2pi-x)` :

Now, `(5pi)/3 = ((6-1)pi)/3 = 2pi - pi/3`

If we let `x=pi/3` then the odd/rotational symmetry of `sin(x)` shown above gives us that

`sin(pi/3) = -sin(2pi-pi/3) = -sin((5pi)/3)`

Therefore, `sin((5pi)/3) = -sin(pi/3)`

Reading off the graph, `sin(pi/3)` is approximately `0.85`. If you put it into your calculator (as 'sin' 'bracket' 'pi' / 3 'close bracket' ) you will get `sin(pi/3) = 0.866` to 3 significant figures. To be very precise,

`sin(pi/3) = sqrt(3)/2` so that `-sin(pi/3) = -sqrt(3)/2`

[ Think of an equilateral triangle with sides length 1- all three angles are 60 degrees = `pi/3`

If we cut the triangle in two we have a triangle with angles 30 degrees, 60 degrees and 90 degrees and lengths H = 1, A = 1/2 and O = ?

`cos(30) = cos(pi/3) = A/H = 1/2` and `sin(pi/3) = O /H = O` .

We can get the length O by using pythagoras: `O^2 + A^2 = H^2 = 1`

Therefore `O^2 = 1 - (1/2)^2 = 1 - 1/4 = 3/4`

and `O = sqrt(3/4) = sqrt(3)/2` . Therefore we have shown using geometry that `sin(60) = sin(pi/3) = sqrt(3)/2` ]

**Answer ` ``sin((5pi)/3) = -sin(pi/3) = -sqrt(3)/2` or 0.866 to 3 significant figures**

The value of `sin((5*pi)/3) ` has to be determined.

`sin((5*pi)/3)`

= `sin((6*pi)/3 - pi/3)`

= `sin(2*pi - pi/3)`

= `sin -pi/3`

= `-sqrt 3/2`

**The value of **`sin((5*pi)/3) = -sqrt 3/2`