Question #7bdf4

1 Answer
Feb 6, 2017

cos(a+h) = sum_(n=0)^oo cos(a+(npi)/2) h^n/(n!)cos(a+h)=n=0cos(a+nπ2)hnn!

Explanation:

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

cos(a+h) = cosacosh-sinasinhcos(a+h)=cosacoshsinasinh

Now expand cos hcosh and sin hsinh in their McLaurin series. We know that the series expansion of cos hcosh only has terms of even degree and the expansion of sin hsinh 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)!)cos(a+h)=cosan=0(1)nh2n2n!sinan=0(1)nh2n+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!)