Question

1/2+cosx+cos2x+...+coskx=?​

Answer 1

Ответ:

$\dfrac{\sin{\frac{(2k+1)x}{2}}}{2\sin{\frac{x}{2}}}$

Пошаговое объяснение:

Домножим и разделим на $\sin{\dfrac{x}{2}}$:

$\dfrac{\sin{\frac{x}{2}}(\frac{1}{2}+\cos{x}+\cos{2x}+\ldots +\cos{kx})}{\sin{\frac{x}{2}}}=\dfrac{\dfrac{\sin{\frac{x}{2}}}{2}+\sin{\frac{x}{2}}\cos{x}+\sin{\frac{x}{2}}\cos{x}+\ldots}{\sin{\frac{x}{2}}}=\\=\dfrac{\dfrac{\sin{\frac{x}{2}}}{2}+\dfrac{\sin{\frac{3x}{2}}}{2}-\dfrac{\sin{\frac{x}{2}}}{2}+\dfrac{\sin{\frac{5x}{2}}}{2}-\dfrac{\sin{\frac{3x}{2}}}{2}+\ldots +\dfrac{\sin{\frac{(2k+1)x}{2}}}{2}-\dfrac{\sin{\frac{(2k-1)x}{2}}}{2}}{\sin{\frac{x}{2}}}=$

$=\dfrac{\dfrac{\sin{\frac{(2k+1)x}{2}}}{2}}{\sin{\frac{x}{2}}}=\dfrac{\sin{\frac{(2k+1)x}{2}}}{2\sin{\frac{x}{2}}}$

Answer 2

The expression you've provided seems to represent a series of terms involving trigonometric functions. Let's break down the expression:

1/2 + cos(x) + cos(2x) + ... + cos(kx)

It looks like you want to find the sum of this series up to the kth term. This type of series involving trigonometric functions can be challenging to simplify in a general form due to the complex nature of trigonometric identities and their interactions. However, if you're looking for a closed-form expression for the sum, it's unlikely to have a simple form.

If you're interested in a specific value for a given value of k, you could compute the sum numerically using a calculator or software. Just plug in the value of x and k, and then sum up the terms.