simplify the following, where k represents any integer: cos(2kpi-x)(-sin(2kpi-x))
a) -cosx*sinx
b) (1/2)sin(2x)
c) cos((1/2)x)^2
d) cos(x)^2*sin(x)^2
e) undefined

Question

simplify the following, where k represents any integer: cos(2kpi-x)(-sin(2kpi-x))
a) -cosx*sinx
b) (1/2)sin(2x)
c) cos((1/2)x)^2
d) cos(x)^2*sin(x)^2
e) undefined

clara1223 · Answer

@Loser66 could you explain how you got that answer please?

dumbcow · Answer

https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Symmetry.2C_shifts.2C_and_periodicity https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Double-angle.2C_triple-angle.2C_and_half-angle_formulae

mathstudent55 · Answer

Here is the explanation:
Since the sine and cosine are periodic functions with a period of 2pi, 
sin (2kpi - x) = sin (-x) and
cos(2kpi - x) = cos (-x)
Ok so far?

mathstudent55 · Answer

Then because of even and odd functions,
sin (-x) = - sin x and
cos (-x) = cos x

mathstudent55 · Answer

Putting it all together, you get:

$(\cos (2k \pi-x)) ((-\sin(2k \pi-x)) $

$= \cos (-x) \sin (-x)$

$= - \sin x \cos x$

clara1223 · Answer

@mathstudent55 thanks so much!

dumbcow · Answer

wait ...... -sin(-x) = sin(x)

it equals positive sin(x) cos(x)

then use double angle identity for sin
----> sin(2x) = 2sincos
----> sincos = 1/2 sin(2x)

mathstudent55 · Answer

@dumbcow You're correct. I missed the negative with the sin in the original problem.

mathstudent55 · Answer

This is how it should read:

$(\cos (2k \pi-x)) ((-\sin(2k \pi-x))$

$= \cos (-x)(- \sin (-x))$

$= \cos x (-(-\sin x))$\)

$= \sin x \cos x$

$=\dfrac{1}{2} \sin 2x$