find an expression for the sum of the series of 1 + 1/2cosx + 1/4cos2x +1/8cos3x + ......

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)

Related Further Mathematics A Level answers

All answers ▸

Show that cosh^2(x)-sinh^2(x)=1


A=[5k,3k-1;-3,k+1] where k is a real constant. Given that A is singular, find all the possible values of k.


Find the equation of the tangent to the curve y = exp(x) at the point ( a, exp(a) ). Deduce the equation of the tangent to the curve which passes through the point (0,1) .


Use De Moivre's Theorem to show that if z = cos(q)+isin(q), then (z^n)+(z^-n) = 2cos(nq) and (z^n)-(z^-n)=2isin(nq).


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences