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.
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
Jun 9th 2009, 01:47 AM
I'm assuming you mean and not .
If you know the graph of the basic sine function, you would know that the x-intercepts are , 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, is a horizontal shrink of by a factor of 2.
This effectively means that the x-intercepts occur twice as often, ie. , for any integer n.
Another example: say it's . Then its x-intercepts would be , for any integer n.
Jun 9th 2009, 02:01 AM
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?