Firstly we can use the fact that cosx is the real part of e^ix. This means that the series above is the real part of the infinite series 1 + 1/2e^ix + 1/4e^2ix + ...... This means that we have an infinite geometric series with starting term 1, and common ratio 1/2e^ix. since the modulus of the common ratio is 1/2 which is <1 , we can use the formula for an infinite convergent geometric series, Thus leaving us with the term of 1/(1 - 1/2e^ix). We can then use a useful property of the multiplication of exponentials to give us some insight, shown in a moment. We multiply the top and bottom of the fraction by (1 - 1/2e^-ix), the denominator expand out to give: 1 - 1/2(e^ix + e^-ix) + (1/4e^ix)(e^-ix), using euler's formula and the laws of the multiplications of exponentials we get: 1 - 1/2(2cosx) + 1/4, simplifying to: 5/4 - cosx. The real part of the numerator (1 - 1/2e^-ix) is 1 - 1/2cosx, therefore the sum of the trigonometric series is (1 - 1/2cosx)/(5/4 - cosx), simplifying to (4-2cosx)/(5-4cosx)
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