Question

Assuming K is a whole number, evaluate the following

Answer 1

\[\sin \frac{ \pi }{ 2 } + 2kpi\]

Answer 2

what is sin(pi/2) equal to

Answer 3

3pi/2

Answer 4

sin(pi/2) should equal to 1

Answer 5

because its on (0,1)?

Answer 6

sin ( pi/2 + 2*k * pi ) ?

Answer 7

ill re-type it hold on please

Answer 8

yes it equals to 1, since sin ( x + 2pi*k ) = sin x

Answer 9

\[\sin(\frac{ \pi }{ 2 } + 2kpi)\]

Answer 10

you can add a revolution of 2pi radians or multiples of a revolution to sin
sin (Θ + 2pi * K ) = sin (Θ)

Answer 11

if you refresh your browser that will render properly

Answer 12

i dont understand

Answer 13

This is a trig identity:
sin (theta + 2pi*k) = sin(theta)

Hence:
sin ( pi/2 + 2*k * pi ) = sin(pi/2)

Answer 14

HI!!

Answer 15

you got this or not?

Answer 16

no

Answer 17

ok lets go slow because it is really not hard at all

Answer 18

\[\sin(\frac{\pi}{2})=1\] a fact that you might should memorize, but you can see it from the unit circle

Answer 19

ok

Answer 20

the other thing you see from the unit circle is if you go around again, in other words if you add \(2\pi\) to any angle, you are right back where you started from

Answer 21

this is usually described by saying "sine is periodic with period \(2\pi\)" which just means if you add \(2\pi\) to any number the sine remains the same

Answer 22

that means
\[\sin(\frac{\pi}{2})=1=\sin(\frac{\pi}{2}+2\pi)=\sin(\frac{\pi}{2}+4\pi)=...\]

Answer 23

they are all one because they are all the same place on the unit circle
this is more succinctly written as \(\frac{\pi}{2}+2k\pi\)

Answer 24

ok

Answer 25

all these angles : Pi/2, pi/2 + 2pi, pi/2 + 4pi , ...
terminate at the same point on the unit circle