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
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
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
thank you so much!!
To find the set of values of theta that satisfies the given condition, we need to determine the range of theta for which the sum of the geometric series converges to a value greater than 2.
The geometric series 1 + sin(theta) + sin^2(theta) + ... has a common ratio of sin(theta). For a geometric series to converge, the absolute value of the common ratio must be less than 1.
So, we have |sin(theta)| < 1.
Since -1/2π < theta < 1/2π, we know that the sine function is positive in this range. Therefore, we can simplify the inequality to:
sin(theta) < 1.
To find the values of theta that satisfy this inequality, we need to consider the values of sine function that are less than 1.
The sine function has a maximum value of 1 at theta = π/2 and a minimum value of -1 at theta = -π/2. In the range -1/2π < theta < 1/2π, sin(theta) has values between -1 and 1, but not including 1.
Hence, the set of values of theta that satisfies the given condition is -1/2π < theta < 1/2π.