Question

Domain and range of 2sinpietheta?
and -5costheta/2?
I know the domain is all real # but the range?

Answer 1

Do you know the range of sin and cos normally? Like if it were just sinx cosx?

Answer 2

I figured out the -5costheta/2 ( I believe)
Wouldn't the range of sinx just be (-1,1)

Answer 3

Right, and the range is the same with cosx, [-1,1]. Now the absolute value of whatever number you have multiplying cosx or sinx also multiplies the end points of the range. For example, if I have 2sinx, the range becomes [-1*2, 1*2] = [-2,2]. Same thing with your problem (-5/2)cosx [-1*(5/2), 1(5/2)] = [-5/2, 5/2]

Answer 4

It was actually -5(costheta/2)

Answer 5

I figured out the range was -5,5 though. Thanks! Wouldn't it be the same for 2sin pie x? (-2,2)?

Answer 6

Ah, lol. Well, same idea though except itd just be [-1*5, 1*5]= [-5,5]

And yes, exactly. Also, the number inside of the angle does nothing to its range. You can have sin(4235734657x), its still a range of [-1.1]. Oh, and those are supposed to be brackets, not parenthesis.

Answer 7

Ah yes.

Answer 8

So yeah, you got your answer then it seems :3

Answer 9

Is there any way you can explain brackets and parenthesis more to me?

Answer 10

Well, the idea is parenthesis means the number is NOT included in the domain and brackets means the number is included. -1 and 1 still are defined, so you have to use brackets.

Answer 11

So lets say if it never hit -1 or 1. It hit just -.99 and .99. Would it be (-1,1)?

Answer 12

Right. The -1, 1 are the boundaries of the range, its just whether or not those boundaries are defined or not. Like if we had something like 1/x. Well, the domain would be (-inf, 0) U (0, inf). In this case, I have to have parenthesis around the 0. Because the boundary of the domain is 0, but 0 doesnt count, itd make the fraction undefined. On the other hand if we had something like sqrt(x), the domain would be [0,inf) The boundary is 0 because we cannot have a negative square root, but the number 0 itself can exist, so thats why I need brackets in this case. So basically it depends on whether or not those boundary values actually can exist or not.

Answer 13

Alrighty. Thank you!

Answer 14

Yep. np :3