1. ## sine function

How do you enter a complete sine function on a ti-83.

Thanks

2. ## Re: sine function

Goto 'Y=' type in 'SIN' 'X' and push 'Graph' , adjust your window accordingly.

3. ## Re: sine function

Thanks but I mean how do you enter asin(bx + c)+d.

Where abcd can be any values.

4. ## Re: sine function

Originally Posted by anthonye
Thanks but I mean how do you enter asin(bx + c)+d.

Where abcd can be any values.
You can't, you need to fix a,b,c & d to graph the function. The TI-83 only allows for one unknown, that is x.

5. ## Re: sine function

I mean how do you enter a complete sine function its where do the brackets go in a complete sine function

6. ## Re: sine function

You can enter then into the 'Y=' part just as you have shown in post #3, although you will need fix values for a,b,c, & d.

7. ## Re: sine function

Yes I know where to enter them its how ie 3sin(2pix + 2) + 4 are the brackets right.

8. ## Re: sine function

If your period is $2\pi$ then maybe its $3\sin2\pi (x + 2) + 4$ ? Hard to know without seeing the actual question.

9. ## Re: sine function

Ok so theres no standard way then also a bracket comes up as soon as you hit sin
it comes up like sin(

10. ## Re: sine function

so you cant enter sin 2pi(

2pi would be in the bracket.

11. ## Re: sine function

Originally Posted by anthonye
Ok so theres no standard way then also a bracket comes up as soon as you hit sin
it comes up like sin(
Yep, then insert an extra pair, i.e sin(b(x+c))

12. ## Re: sine function

ok thanks think i'm there got to go

thank you pickslides.

13. ## Re: sine function

I don't know the complete value of sine function.

14. ## Re: sine function

Originally Posted by kraj8995
I don't know the complete value of sine function.
If you mean you don't know what anthonye means by "complete sine function", then look at post #3 where he explains what he means.

15. ## Re: sine function

Originally Posted by anthonye
Thanks but I mean how do you enter asin(bx + c)+d.

Where abcd can be any values.
$y = 4\sin[3(x + 2)] + 1$

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