# Question 7bdf4

Feb 6, 2017

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:

$\cos \left(a + h\right) = \cos a \cosh - \sin a \sinh$

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:

$\cos \left(a + h\right) = \cos a - \sin a h - \cos a {h}^{2} / 2 + \sin a {h}^{3} / 6 + \ldots$

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 ${c}_{n}$:

${c}_{0} = \cos a$
${c}_{1} = - \sin a = \cos \left(a + \frac{\pi}{2}\right)$
${c}_{2} = - \cos a = \cos \left(a + \pi\right)$
${c}_{3} = \sin a = \cos \left(a + \frac{3 \pi}{2}\right)$
$\ldots$

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!)#