Question

Question #7bdf4

Answer 1

`cos(a+h) = sum_(n=0)^oo cos(a+(npi)/2) h^n/(n!)`

Explanation:

Use the trigonometric formula for the cosine of a sum of angles:

`cos(a+h) = cosacosh-sinasinh`

Now expand `cos h` and `sin h` in their McLaurin series. We know that the series expansion of `cos h` only has terms of even degree and the expansion of `sin h` only has terms of odd degree:

`cos(a+h) = cosa sum_(n=0)^oo (-1)^n h^(2n)/(2n!) -sina sum_(n=0)^oo (-1)^n h^(2n+1)/((2n+1)!)`

That is:

`cos(a+h) = cosa -sina h -cos a h^2/2 +sina h^3/6 +...`

To obtain a more compact expression, we can use the properties of trigonometric functions.

Write the series as:

`cos(a+h) = sum_(n=0)^oo c_n h^n/(n!)`

and analyze the coefficients `c_n`:

`c_0 = cosa`
`c_1 = -sina = cos(a+pi/2)`
`c_2 = -cosa = cos(a+pi)`
`c_3 = sina = cos(a+(3pi)/2)`
`...`

and we can see that the series can be written as:

`cos(a+h) = sum_(n=0)^oo cos(a+(npi)/2) h^n/(n!)`