Question #0d7b8

2 Answers
Alan P.
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.

Related questions
See all questions in Proving Identities
Impact of this question
1045 views around the world
You can reuse this answer
Creative Commons License