You may transform the sum of sines in a product using the principle of logarithms such that:

`sin(pi/9) + sin(2pi/9) = 2sin ((pi/9 + 2pi/9)/2)*cos((pi/9- 2pi/9)/2)`

`sin(pi/9) + sin(2pi/9) = 2 sin ((3pi/9)/2)*cos((-pi/9)/2)`

You need to reduce by 3 and you need to remember that the cosine function is eve,...

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You may transform the sum of sines in a product using the principle of logarithms such that:

`sin(pi/9) + sin(2pi/9) = 2sin ((pi/9 + 2pi/9)/2)*cos((pi/9- 2pi/9)/2)`

`sin(pi/9) + sin(2pi/9) = 2 sin ((3pi/9)/2)*cos((-pi/9)/2)`

You need to reduce by 3 and you need to remember that the cosine function is eve, hence `cos((-pi/9)/2) =cos((pi/9)/2)` such that:

`sin(pi/9) + sin(2pi/9) = 2 sin ((pi/3)/2)*cos((pi/9)/2)`

`sin(pi/9) + sin(2pi/9) = 2 sin (pi/6)*cos(pi/18)`

You need to remember that `sin pi/6 = 1/2` , hence:

`sin(pi/9) + sin(2pi/9)= 2*(1/2)*cos(pi/18)`

`sin(pi/9) + sin(2pi/9) = cos pi/18 ` (in radians) or `cos 10^o` (degrees)

`sin(pi/9) + sin(2pi/9) ~~ 0.984`

**Hence, evaluating the sum of sines yields `sin(pi/9) + sin(2pi/9) = cos pi/18~~ 0.984` .**