A trig question

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June 9th 2009, 01:39 AM
Redeemer_Pie

A trig question

Hi guys

Just have a question about the sin function

If the function is given as sin2(x).

How would i know where it will cut the x-axis (assuming the x axis is denoted as using 'pi')

Cause in the original function: sin(x) if we put in 'pi' into x then it will equal 0.

So now, is there an approach that given the function: sin2(x), that i could know where it cuts the x-axis ie when y = 0?

thanks in advance guys :)

Forgot to mention:

How can i also find the trig. function values.

Like

sin when (-pi)/4 the value is -squareroot x divded by 2.

How can i also use that to figure out a value like pi/4 given the function sin2x
June 9th 2009, 01:47 AM
yeongil

I'm assuming you mean $y = \sin 2x$ and not $y = {\sin}^2 x$ .

If you know the graph of the basic sine function, you would know that the x-intercepts are $x = n\pi$ , for any integer n (in other words, multiples of pi). Putting a number greater than 1 in front of x transforms the graph -- you have a horizontal shrink by a factor of that number. So in our case, $y = \sin 2x$ is a horizontal shrink of $y = {\sin}^2 x$ by a factor of 2.

This effectively means that the x-intercepts occur twice as often, ie. $x = \frac{n\pi}{2}$ , for any integer n.

Another example: say it's $y = \sin 4x$ . Then its x-intercepts would be $x = \frac{n\pi}{4}$ , for any integer n.

01
June 9th 2009, 02:01 AM
Redeemer_Pie

Thanks a lot!!!! :)

Yes sin2x is not sin to the power of 2 x :)

So if i graphed the function sin2x, the first point it will cut the x-axis would be 'pi'/2?

As well if it was like sin4x then the first point it will cut will be 'pi'/4?

Thanks again mate :)