Question

f(x) = 2sin(3x)

Answer 1

\[2cos(3x-\phi); \phi = arctan(2/0)\]

Answer 2

what can i do with the arctangent of 2/0

Answer 3

pi/2 of course

Answer 4

because cos(x - pi/2) = sin(x)

Answer 5

is this a simple geometry relation (how do you remember that)

Answer 6

sorry, that's not right. I mean cos(pi/2 - x) = sin(x)

Answer 7

that comes looking at right triangles and see that if I swap the non-right angles, the cos and sin flip

Answer 8

now, cos x = cos(-x) so
cos(pi/2-x) = cos(x-pi/2), so it still works

Answer 9

I can also see that cos(x-pi/2) = sin(x) by thinking about the graphs of cos and sin

Answer 10

Anyway, the upshot of all this is it still hangs together:
cos(3x - pi/2) = sin(3x)

Answer 11

technically, you shouldn't use arctan(2/0) because it's not defined. But the angle is also arccot(0/2) which is defined and its principle value is pi/2.

Answer 12

\[\tan(\theta) = {\sin(\theta)\over \cos(\theta)}\] so when ever cos(θ)= 0 tan(θ) is uundefined

Answer 13

ok what you are saying makes sense,

\[f(x)=2sin3x\]\[=√2^2cos(3x- \phi); \phi =acrtan(2/0)\]\[=2cos(3x-\phi); \phi = π/2\]\[=2cos(3x-π/2)\]

Answer 14

yes, although it's a slightly torturous way to arrive at the result, it is consistent and correct.

Answer 15

\[\phi={\pi \over 2},\ {3\pi\over2}\]

Answer 16

is there a simpler way to arrive at the result?

\[f(x) = =2cos(3x−π/2+nπ)\]

Answer 17

I have a problem with this result. Are you sure there isn't a 2 missing? as in
2cos( stuff .... + *2*n.pi) ?

Answer 18

isn't 3pi/2 also a valid solution for arccot (0)

Answer 19

yes, it is. But you have to look at what we're trying to do in the context of the problem and that isn't a solution that makes sense here.

Answer 20

The big problem I have with what Rh has written down is that this "function"
f(x)=2cos(3x−π/2+nπ)
changes sign depending on whether n is even or odd, so it can't be equal to
2sin(3x)

Answer 21

that is cos(y+2pi) = cos(y), but cos(y+pi) = -cos(y)

Answer 22

so\[f(x)=2cos(3x−π/2+2nπ)\] is the best answer ?
or do i leave out the +2nπ as it is implied by the cosine function

Answer 23

Yes, I'd leave it out unless there's something you want it for later.

Answer 24

on the other hand, the original 2sin(3x) is also a perfectly respectable function. I'm guessing you're being asked to write all of these things though as cos

Answer 25

The question is just asking; to express f(x) as single sinusoid, to find amplitude and phase

Answer 26

so i guess i dident really need to rearrange the equation in this example at all