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Question #0d7b8

2 Answers

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),
$sin(x)$ is not an even function.

Explanation:

i) using the identity: $cos(A-B)=cos(A) * cos(B) - sin(A) * sin(B)$

$cos(-x)=cos(0-x)$

$color(white)("XXX")=cos(0) * cos(x) - sin(0) * sin(x)$

$color(white)("XXX")= 1 * cos(x) - 0 * sin(x)$

$color(white)("XXX")= cos(x)$

ii) Note that the reflection of the right side of $sin(x)$ is not equal to $sin(x)$
enter image source here
Therefore $sin(x)!=sin(-x)$
and $sin(x)$ is not an even function.

A. S. Adikesavan

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 $cos(-x)=cos x and sin (-x)--sin x$

If $f(-x)=f(x)$ , f is even and if $f(-x)=-f(x)$ , 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

symmetrical about x-axis.

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