Asked by Kebun Baru
                find the set of values of theta lying in the interval -1/2 pi < theta < 1/2pi such that the sum to infinity of the geometric series 1+sintheta+sin^2theta +... is greater than 2
            
            
        Answers
                    Answered by
            oobleck
            
    since you have a GP with
a = 1
r = sinθ
we want
1/(1-sinθ) > 2
sinθ > 1/2
so π/6 < θ < π/2
    
a = 1
r = sinθ
we want
1/(1-sinθ) > 2
sinθ > 1/2
so π/6 < θ < π/2
                    Answered by
            mathhelper
            
    sum(all terms) = a/(1-r)
1/(1 - sinθ) > 2
1 > 2 - 2sinθ
2sinθ > 1
sinθ > 1/2
Now look at the graph of y = sinx
and knowing that sin π/6 = 1/2
www.wolframalpha.com/input/?i=graph+y+%3D+sinx+from+-%CF%80%2F2+to+%CF%80%2F2
π/6 < θ < π/2
    
1/(1 - sinθ) > 2
1 > 2 - 2sinθ
2sinθ > 1
sinθ > 1/2
Now look at the graph of y = sinx
and knowing that sin π/6 = 1/2
www.wolframalpha.com/input/?i=graph+y+%3D+sinx+from+-%CF%80%2F2+to+%CF%80%2F2
π/6 < θ < π/2
                    Answered by
            Kebun Baru
            
    thank you so much!!
    
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