Question

HELP MEEE !!
How does the graph of y = -3cos(2θ + 45°) + 3 differ from the graph of y = cos(θ)?

Answer 1

@satellite73

Answer 2

help please

Answer 3

@gypsy1274 can you help me?

Answer 4

Sorry, I haven't gotten to trig yet. Can you graph these equations? Maybe a graphing calculator or some websites may be able to do it for you? A visual representation might give you a hint about where to start.

Answer 5

@Dreweed can you help me?

Answer 6

@satellite73 can you help??

Answer 7

Sorry. I'm still learning calculus as of right now. =( However, I agree with @gypsy1274 . A graphing calculator and some online sources will definitely help.

Answer 8

ok Thanks

Answer 9

Wait a sec.
Is it 45 degrees or just 45?

Answer 10

45 degrees

Answer 11

@mhmdrz91 can you help?

Answer 12

Ah, I see. Degrees or not, the graph is still quite similar. Your answer will be something like: reflection over x-axis, shift up vertically by 3 units, vertical stretch by a factor of 3, and shifting 45 units horizontally to the left.

Answer 13

how are they different though @Dreweed ?

Answer 14

1 -> y = -3cos(2θ + 45°) + 3
2 -> y = cos(θ)
we know that\[-1<\cos(x)<1\]then with multiplying -3 to this we have \[-3<-3\cos(x)<3\]and with add +3 to this\[0<-3\cos(x)+3<6\]so we know now the first equation is between 0 and 6! and the second is between -1 and 1...

Answer 15

ok so how are they different in transformations?

Answer 16

@mhmdrz91

Answer 17

\[y = 3-3 \cos(2 \theta+\pi/4)\]the period of this equation is \[\pi\]but in\[y=sin(\theta)\]the period is\[2\pi\]

Answer 18

ok thanks

Answer 19

yw