# Question 0d7b8

Oct 24, 2016

i) see below
ii) In the graph of an even function the left side relative to the Y-axis is a reflection of the right side (relative to the Y-axis),
$\sin \left(x\right)$ is not an even function.

#### Explanation:

i) using the identity: $\cos \left(A - B\right) = \cos \left(A\right) \cdot \cos \left(B\right) - \sin \left(A\right) \cdot \sin \left(B\right)$

$\cos \left(- x\right) = \cos \left(0 - x\right)$

$\textcolor{w h i t e}{\text{XXX}} = \cos \left(0\right) \cdot \cos \left(x\right) - \sin \left(0\right) \cdot \sin \left(x\right)$

$\textcolor{w h i t e}{\text{XXX}} = 1 \cdot \cos \left(x\right) - 0 \cdot \sin \left(x\right)$

$\textcolor{w h i t e}{\text{XXX}} = \cos \left(x\right)$

ii) Note that the reflection of the right side of $\sin \left(x\right)$ is not equal to $\sin \left(x\right)$

Therefore $\sin \left(x\right) \ne \sin \left(- x\right)$
and $\sin \left(x\right)$ is not an even function.

Oct 24, 2016

When x is expressed in radian, the Maclaurin series for cos x and sin

x are

,cos x = 1-x^2/(2!)+x^4/(4!) + ... +(-1)^nx^(2n)/((2n)!)+... and

sin x = x - x^3/(3!)+x^5/(5!)+...(-1)^nx^(2n+1)/((2n+1)!)+...#

It follows that $\cos \left(- x\right) = \cos x \mathmr{and} \sin \left(- x\right) - - \sin x$

If $f \left(- x\right) = f \left(x\right)$, f is even and if $f \left(- x\right) = - f \left(x\right)$, f is odd.

So, sine is even and cosine is odd.

For even functions y = f(x), like cos x,

if (x, y) is on the graph, then (-x, y) is on

it. So, the graph is symmetrical about the y-axis.

For odd f like sin x,

if (x, y) is on the graph, so is (x, -y). And so, the graph is