Question

Sin(3x-pi/2) ALGEBRAICALLY FIND POINTS

Answer 1

so where did we end at....

i think it was

\[\frac{\pi}{2}+2\pi n\]

Answer 2

for when Sin(u) = 1

Answer 3

yes and u told me N is any number

Answer 4

yep, alright so if we know that

\[u=\frac{\pi}{2}+2\pi n\]

to make sin(u)=0

\[sin(\frac{\pi}{2}+2\pi n)=1\]

Answer 5

just to make sure lets test it out... for n=1 we get

\[sin(\frac{\pi}{2}+2\pi)=sin(\frac{5\pi}{2})=\]

since \[\frac{4\pi}{2}=2\pi\]

starting from zero, we go around the circle once and then go another \[\frac{\pi}{2}\]

which is the same angle as \[\frac{\pi}{2}\]

sin at that point = 1! so we know that it's correct!

Answer 6

now remember how we made

\[u=3x-\frac{pi}{2}\]

Answer 7

?

Answer 8

hmmm... hold on let me see again

Answer 9

or we can look at it this way

we know that

\[sin(\frac{\pi}{2}+2\pi n)=1\]

if we want

\[sin(3x-\frac{\pi}{2})\]

to equal 1 also we should equal them to each other
if we do that we get

\[sin(\frac{\pi}{2}+2\pi n)=sin(3x-\frac{\pi}{2})\]

Answer 10

im trying to see how u got 5pi/2
hold on im trying to write it out

Answer 11

alright =]

Answer 12

let me kwow when you're done with that =]

Answer 13

okay i got it, i see how u got the 5pi/2

Answer 14

alright does what i said after make sense?

Answer 15

ya, the main problem was y=3x - pi/2


and you said that sin(pi/2 +2pi n)
and that one is from---^ is that to get a point? i forgot already.


WHY DOES THE INTERNET KEEP DOING THAT?!

Answer 16

sin(pi/2+2pi n) is for when sin(u)=1

Answer 17

okay i got that now then

Answer 18

so sin(pi/2+2pi n) will always be 1 no matter what integer you let n be

Answer 19

okay, i wrote that down and circled it

Answer 20

no you want that to be the same for your problem

\[sin(3x-\frac{\pi}{2})\]

in other words if you let

\[sin(3x-\frac{\pi}{2})=1\]

and
\[sin(\frac{\pi}{2}+2\pi n)=1

\]
they'd be equal to each other correct?

Answer 21

i mean't now* not no lol =]

Answer 22

cause \[1=1\]

Answer 23

haha ya i understand that

Answer 24

so now you can make them equal to each other

\[sin(3x-\frac{\pi}{2})=sin(\frac{\pi}{2}+\pi n)\]

Answer 25

the sins will cancel correct?