Sine

**ADY** · January 21st 2009, 05:56 AM

If you have

$Y = sin (x) - 0. 6$

How could i show the sine curve on my calculator for

X = 0 to 360 $\circ$

Thanks

**Mush** · January 21st 2009, 05:59 AM

Originally Posted by ADY

If you have

$Y = sin (x) - 0. 6$

How could i show the sine curve on my calculator for

X = 0 to 360 $\circ$

Thanks

This is just a regular sine function that has been shifted down the y axis by 0.6 units.

So it peaks at 0.4, troughs at -1.6, and it's mid point is -0.6.

**ADY** · January 21st 2009, 06:10 AM

is that with X set to 0 to 360

**Mush** · January 21st 2009, 06:25 AM

Originally Posted by ADY

is that with X set to 0 to 360

Throughout the entire interval.

**ADY** · January 21st 2009, 06:36 AM

so how can i find the two solutions to the equation? in the peaks?

**ADY** · January 21st 2009, 06:40 AM

$<br /> x= - 4.712<br />$

$x= 1.5707$

?

**Mush** · January 21st 2009, 07:05 AM

Originally Posted by ADY

so how can i find the two solutions to the equation? in the peaks?

What do you mean by SOLVE it?

You have one equation and two unknowns $x \text{ and } y$ . It is not solveable. What is it you are trying to find? I thought you only wanted to graph it.

**ADY** · January 21st 2009, 07:48 AM

well now ive graphed it - how can i find the 2 solutions of that equation that are between 0 - 360 degrees?

**mollymcf2009** · January 21st 2009, 08:24 AM

What are you trying to solve for? Is this the way the question was worded? Can you give me the question exactly as it appears in your book? I don't know how to help you because I can't figure out what it is you solving for

**ADY** · January 21st 2009, 08:54 AM

2 solutions between

$sin(x) - 0.6 = 0$

between 0 and 360degree

**chabmgph** · January 21st 2009, 09:55 AM

Originally Posted by ADY

2 solutions between

$sin(x) - 0.6 = 0$

between 0 and 360degree

That looks much better.

Graphically the solutions should be where the curve crosses the x-axis (That's where your output equals to zero).
Algebraically, you will need to solve for x by using the $sin^{-1}$ key on your calculator somewhere.

**mollymcf2009** · January 21st 2009, 11:08 AM

Ok, now I see what you need to do. The sin(x) curve has an infinite domain , but comes back to the x axis (where y equals zero)at x = 0 and at x= pi. In the case of this problem, the sine curve is shifted down .06. So it won't = 0 at x=0 or pi, it will be 0 somewhere just larger than 0 and pi. ( I'm guessing probably at .06). As far as from 0 to 2pi ( or in your case 360 deg) you will actually have 3 answers. The solutions are the x values on the curve when y = 0
Does that make sense? Good luck!

**ADY** · January 22nd 2009, 08:58 AM

Originally Posted by chabmgph

That looks much better.

Graphically the solutions should be where the curve crosses the x-axis (That's where your output equals to zero).
Algebraically, you will need to solve for x by using the $sin^{-1}$ key on your calculator somewhere.

Could you explain how to do that please chap

**chabmgph** · January 22nd 2009, 11:31 AM

Originally Posted by chabmgph

That looks much better.

Graphically the solutions should be where the curve crosses the x-axis (That's where your output equals to zero).
Algebraically, you will need to solve for x by using the $sin^{-1}$ key on your calculator somewhere.

Originally Posted by ADY

Could you explain how to do that please chap

You can do this graphically if you own a graphing calculator. However, depending on what model you have, the procedure would be different. And I believe there is a section in this forum devoted to calculator use.

To do this algebraically,
$\sin (x) - 0.6 =0$
$\sin (x) =0.6$
$x=\sin^{-1}(0.6)$
Then to find a approximation of x, find the $sin^{-1}$ key on your calculator and calculate what $sin^{-1}(0.6)$ is.

**ADY** · January 23rd 2009, 01:08 AM

How can there be two answers then? because of where the curve crosses the x axis?

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