Thread: Help graphing using simple trig

I'm a little frustrated because in my class we never used calculators, and now I have to prepare for a state math test that requires extensive calculator usage. I'm using the TI-83 Plus... My review book says to graph 2sinx+cos2x in Y1 and 2sin^2x-1 in Y2 and observe their solutions (x-coordinates of intersection points) (1.57, 3.67, 5.76).

Every time I graph these I go put the cursor on where I see them intersect and they never intersect at these points. I think I am not entering the second graph in correctly... I am entering it in as 2sin(x-1)^2.

Also, any tips about graphing in general would be great since I am nearly a complete noob with graphing calculators. Help is appreciated.

As you've written it I suspect the calculator is doing - ie: it thinks it's squaring the argument rather than the sin function

Try .

I'm afraid I'd be no good at plotting it on the calculator, graphing calculators were rare and we had to do it using pencil and paper and transforming functions

You can always use Wolfram Alpha to get an idea of what it should look like

For the second graph. Is it: ? Then it's not correct to enter , enter: .

The original equation is 2sin^2x-1, make sure that it is; sin(x-1) and not sinx - 1 which is totally different. If it is 2 sin ^2 (x - 1) then enter in the calculator as 2[sin(x-1)]^2 and if it is sin^2x - 1 then enter 2 (sinx)^2 - 1 to get it right.

Originally Posted by benny92000

I'm a little frustrated because in my class we never used calculators, and now I have to prepare for a state math test that requires extensive calculator usage. I'm using the TI-83 Plus... My review book says to graph 2sinx+cos2x in Y1 and 2sin^2x-1 in Y2 and observe their solutions (x-coordinates of intersection points) (1.57, 3.67, 5.76).

Every time I graph these I go put the cursor on where I see them intersect and they never intersect at these points. I think I am not entering the second graph in correctly... I am entering it in as 2sin(x-1)^2.

Also, any tips about graphing in general would be great since I am nearly a complete noob with graphing calculators. Help is appreciated.

first (and most important) ... are you graphing in radian mode?

second ... sometimes it's a good idea to subtract one expression from the other (Y1 - Y2) when solving an equation, then one can find where Y1 - Y2 = 0 (same as Y1 = Y2, only easier to find the x-values where the equation works.)

Originally Posted by e^(i*pi)

As you've written it I suspect the calculator is doing - ie: it thinks it's squaring the argument rather than the sin function

Try .

I'm afraid I'd be no good at plotting it on the calculator, graphing calculators were rare and we had to do it using pencil and paper and transforming functions

You can always use Wolfram Alpha to get an idea of what it should look like

You're right. It was squaring the argument. The graph on the wolfalpha is the correct graph. However, when I entered it in as you initially suggested, it came up incorrect. This is because I did not communicate what the argument was effectively. My book says 2sin^2 x-1 (with no parenthesis around anything, so the -1 is separate I believe, and not part of the argument??). I'm just a tad confused because the wolfalpha graph is correct with the parenthesis around the x and not the -1. Does that mean the argument is just the x? Sorry if I am confusing you...

Originally Posted by Siron

For the second graph. Is it: ? Then it's not correct to enter , enter: .

Thanks. I think the argument is just the x since there are no parenthesis.

Originally Posted by mathjeet

The original equation is 2sin^2x-1, make sure that it is; sin(x-1) and not sinx - 1 which is totally different. If it is 2 sin ^2 (x - 1) then enter in the calculator as 2[sin(x-1)]^2 and if it is sin^2x - 1 then enter 2 (sinx)^2 - 1 to get it right.

Thank you... I think it is the latter graph since the book has no parenthesis. I initially thought that the x-1 was the argument (I don't know why.) However, when I graphed it the latter way, I came up with the correct graph.

Originally Posted by skeeter

first (and most important) ... are you graphing in radian mode?

second ... sometimes it's a good idea to subtract one expression from the other (Y1 - Y2) when solving an equation, then one can find where Y1 - Y2 = 0 (same as Y1 = Y2, only easier to find the x-values where the equation works.)

Thank you, sir. And yes, I am graphing in radians. Could you elaborate a little more on why we would subtract Y2 from Y1, and how that would give us the points of intersection?

when two expressions are equal and you subtract them, what value do you get?

0. But Y1 and Y2 are not equal to each other, are they??

Originally Posted by benny92000

0. But Y1 and Y2 are not equal to each other, are they??

at certain values of x they are ... where Y1 - Y2 = 0 ... see the last graph posted.

Ok, and I have a quick question about TI-83 Plus.. How do you make it graph immediately rather than draw it out? It saves me crucial seconds.

Originally Posted by benny92000

Ok, and I have a quick question about TI-83 Plus.. How do you make it graph immediately rather than draw it out? It saves me crucial seconds.

the TI-83 graphs at a rate dependent on the complexity of the graph ... why are seconds crucial?

I am taking a timed test and sometimes I have to enter several graphs.