# Sine

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• Jan 21st 2009, 05:56 AM
Sine
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
• Jan 21st 2009, 05:59 AM
Mush
Quote:

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.
• Jan 21st 2009, 06:10 AM
is that with X set to 0 to 360
• Jan 21st 2009, 06:25 AM
Mush
Quote:

is that with X set to 0 to 360

Throughout the entire interval.
• Jan 21st 2009, 06:36 AM
so how can i find the two solutions to the equation? in the peaks?
• Jan 21st 2009, 06:40 AM
$
x= - 4.712
$

$x= 1.5707$

?
• Jan 21st 2009, 07:05 AM
Mush
Quote:

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.
• Jan 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?
• Jan 21st 2009, 08:24 AM
mollymcf2009
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
• Jan 21st 2009, 08:54 AM
2 solutions between

$sin(x) - 0.6 = 0$

between 0 and 360degree

(Rock)
• Jan 21st 2009, 09:55 AM
chabmgph
Quote:

2 solutions between

$sin(x) - 0.6 = 0$

between 0 and 360degree

(Rock)

That looks much better. (Itwasntme)

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.
• Jan 21st 2009, 11:08 AM
mollymcf2009
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!
• Jan 22nd 2009, 08:58 AM
Quote:

Originally Posted by chabmgph
That looks much better. (Itwasntme)

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
• Jan 22nd 2009, 11:31 AM
chabmgph
Quote:

Originally Posted by chabmgph
That looks much better. (Itwasntme)

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.

Quote:

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.
• Jan 23rd 2009, 01:08 AM