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# Trig and Graph Transformations watch

1. Probably being really stupid but i came across this question and am a bit confused. Wondering if you could help me out?

If y=sin(x) and I transform it so that now y=sin(x^2) why is the first root not 1?

My understanding is that sin(x^2) = sin(x*x)
I thought when you transform a graph by directly multiplying x by a number youre actually stretching the graph by the reciprocal of that number. So here i should be stretching the graph by 1/x?
The first root for sin(x) is pi so i stretched by 1/pi forming pi/pi making the new root 1, why doesnt that work?
2. (Original post by RipGeometry)
Probably being really stupid but i came across this question and am a bit confused. Wondering if you could help me out?

If y=sin(x) and I transform it so that now y=sin(x^2) why is the first root not 1?

My understanding is that sin(x^2) = sin(x*x)
I thought when you transform a graph by directly multiplying x by a number youre actually stretching the graph by the reciprocal of that number. So here i should be stretching the graph by 1/x?
The first root for sin(x) is pi so i stretched by 1/pi forming pi/pi making the new root 1, why doesnt that work?
You can't perform a stretch on a variable by using that very same variable, that's not how it works. You need to use a constant variable, unrelated to , otherwise I'm surprised you haven't said anything about that fact that since is a root, then that's stretch by s.f. on it.... which makes no sense.
3. (Original post by RipGeometry)
Probably being really stupid but i came across this question and am a bit confused. Wondering if you could help me out?

If y=sin(x) and I transform it so that now y=sin(x^2) why is the first root not 1?

My understanding is that sin(x^2) = sin(x*x)
I thought when you transform a graph by directly multiplying x by a number youre actually stretching the graph by the reciprocal of that number. So here i should be stretching the graph by 1/x?
The first root for sin(x) is pi so i stretched by 1/pi forming pi/pi making the new root 1, why doesnt that work?
Nonono xD . If a function f(x) is stretched parallel to the x axis by s.f. 1/a then the new transformed function will be f(ax). You are confusing a multiplicative constant with a variable- you can't stretch a graph by multiplying it with the exact same variable as which the function is explicitly expressed as. In other words the "a" value must be some number and cannot be x as x is a variable which can take infinitely many values.

You can work out the first positive root by simply considering the behavior of the normal sine graph. You should be familiar with this graph and should be able to figure out what the first positive root is. Now with the transformed graph you have sin(x^2) instead of sin(x) so the "new" x value is x^2. You can set x^2 equal to the old x value which is the first positive root and solve for the "new x"
4. (Original post by Anonymouspsych)
Nonono xD . If a function f(x) is stretched parallel to the x axis by s.f. 1/a then the new transformed function will be f(ax). You are confusing a multiplicative constant with a variable- you can't stretch a graph by multiplying it with the exact same variable as which the function is explicitly expressed as. In other words the "a" value must be some number and cannot be x as x is a variable which can take infinitely many values.

You can work out the first positive root by simply considering the behavior of the normal sine graph. You should be familiar with this graph and should be able to figure out what the first positive root is. Now with the transformed graph you have sin(x^2) instead of sin(x) so the "new" x value is x^2. You can set x^2 equal to the old x value which is the first positive root and solve for the "new x"
Ah i see thanks for the response. So im guessing the new roots would be rt(pi) rt(2pi) etc.?
5. to solve sin( x2 ) = 0

you want x2 to be 0, π, 2π, 3π, ....
6. (Original post by RipGeometry)
Ah i see thanks for the response. So im guessing the new roots would be rt(pi) rt(2pi) etc.?
Exactly

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