# Solve: Sin(π/9) + Sin(2π/9) Show complete solution and explain the answer.

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,...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

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` .

Approved by eNotes Editorial Team