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In general, the sin function is an odd function. cos and sec are even and the other 4 trig functions are odd. The y-intercept is just found by plugging in x = 0. 3sin(2*0) = 3sin(0) = 0. So you get a y-intercept of (0,0) The x-intercepts are a bit trickier. We want to solve the equation 3sin(2x) = 0. To make this easier, we can divide by 3 and just solve sin(2x) = 0. To solve this, lets first consider the equation sinx = 0. Can you tell me the x values that make sinx = 0?
I thought to find the y intercept you had to replace the x with y
I'm not actually sure what doing that would give you. Replacing x with y sounds like what you would do for finding an inverse. But think of it this way: A y-intercept is a spot where you cross the y-axis and will always be when x = 0. Any y-intercept will look like (0,y). So to find a y-intercept in any equation, its always best to just replace x with 0 and solve.
Yeah I see what you mean now, my bad
No worries. But yeah, when it comes to solving for x-intercepts, I would start off with solving sinx = 0. Do you know the solutions of that?
Pi and 2pi?
Are we considering only the interval [0,2pi), or all values?
umm I have no idea
I think it was 0,2pi
Ah, okay. Well, if its not all values, then the usual interval is [0,2pi) with 0 included, 2pi not included. In this case, the solutions would be 0 and pi. So the solutions are x = 0 and x = pi. But our problem was sin(2x) = 0. Thus we actually have 2x = 0 and 2x = pi. This implies x = 0 is a solution and x = pi/2 is a solution. Can you see thaT?
Yes I think so
Yeah. Thats usualy my strategy to solving such problems. Like, if I have sin(3x) = 1 for example, I first consider sinx = 1, which is true for \(x = \pi/2 + 2k\pi\). Then I replace x with the actual angle, in this case 3x, giving me \(3x = \pi/2 + 2k\pi\) \(x = \pi/6 + 2k\pi/3\) Kinda see the pattern and what I would be doing each time?
Yep. So everything kind of make sense then?
One more, I forgot how to figure out what it's symmetry is
Well, you kind of have to know that cosx and secx are both even functions and the other 4 are odd functions. The normal process to show a function is odd is to show that -f(x) = -f(x) and the normal process to show a function is even is to show f(x) = f(-x). Doing that method is a bit trickier with trig functions, so its simply best to just know that cosx, secx = even, the other 4 = odd.
but if x=2pi is a solution of sin(x)=0 then 2x=2pi is a solution of sin(2x)=0 and pi is also in the interval [0,2pi) So I think if you are solving sin(2x)=0 on [0,2pi) and you decide to solve sin(u)=0 well notice we replaced 2x with u so 2x=u and if 2x=u then dividing both sides by 2 gives us x=u/2 so instead of having x is between 0 and 2pi you have u/2 is between 0 and 2pi and so u is between 0 and 4pi so you actually need to solve sin(u)=0 on [0,4pi) to find all the solutions of sin(2x)=0 on [0,2pi)
Freckles is right, just overlooked that. Because sin(2x) has a period of pi, need to consider solutions in an interval twice as long, so [0,4pi) my bad, thanks, freckles :)
you are doing awesome :)
Im stil glad you caught the mistake, lol. On a random note, any good with geometry, freckles?