Thread: Tell me about trigonometric functions

You don't need to memorize all that. All you need to know is the definitions of the other trig functions, which are all in terms of sin and cos:

sec(90-x) = 1/cos(90-x) = 1/sin(x) = csc(x)

........or, just realize that coX goes with X

I pretty much just spent about 2 months in my pre-calc class extensively covering trigonometric functions, all their uses, and many side things (still doing more stuff on them).

So if you have any specific questions, just post here, PM me, or IM me. I'd be glad to help

I want to add why the trigonometric functions are what they are. Teachers pretty much teach how to use them and almost never explain what in the world they are. When teachers don't do that, it makes math look like a whole lot of random weirdness being forced onto students for no apparent reason. I believe in explaining the reasons for the rules so math will seem like the rational subject it is.

The trigonometric functions are all about how all triangles with any two given angle measures are similar (Because the third set of angles will automotically be congruent if the other two are.). When triangles are similar, their corresponding sides are proportional. Therefore, all triangles with a 90 degree angle and a (for example) 40 degree angle are going to have exactly the same shape and be similar (Since the third angle will always be 50 degrees.), which means that the ratio of the measures of the side opposite the 40 degree angle and the hypotenuse will be the same for every single triangle with a 90 degree angle and a 40 degree angle. That ratio is the sine of 40 degrees, which is also the cosine of 50 degrees. Because those ratios are always the same, they can be used to find side and angle measures if enough other information is given.

Another way to remember sine, cosine, and tangent is to learn those three words in order and then remember, "Oscar had a hit of acid." I was taught, "... hunk of apples," but the acid one sticks better.

sine: opposite/hypotenuse
cosine: adjacent/hypotenuse
tangent: opposite/adjacent

anyone have an answer for this question? More of a physics problem though heh.

A 15 N force directed to the west acts on an object for 3.0 seconds, what is the change in momentum of the object?


(while your at it...)

A car is moving in uniform circular motion. If the car's speed were to double to keep the car moving with the same radius the acceleration would what?

increase by a factor of 2

Or by a factor of 4?

Let m be the mass.

F = m*a --> a = F/m

So the change in velocity will be 3*(F/m), so the change in momentum will be 3*(F/m)*m = 3*F.

For the second question... for chrissake, there's no thinking involved. Just plug the damn thing into the equation

I won't look it up, but IIRC, the centripetal acceleration is something like v^2 / r

If the speed were to double, you'd have (2*v)^2 / r = 4 * v^2/4, i.e. factor of 4.


Now that I've shown that this is so easy someone who hasn't touched it in years can do it in his head while typing on a dreaming forum, I won't solve any more homework questions that are clearly "plug it into one equation", for that would be wrong. This stuff is for YOUR own good. Think about it before giving up. I'll happily help you along and stuff, but at the very least, apply a systematic "what do I know about this system? ok, now given the information I have, what can I figure out about it that would help me find the answer to this question?" type approach before posting it.

New problem...

sinx(1 - 2(cos^2)x + (Cos^4)x) = (sin^5)x meanwhile i'll be trying to figure it out myself heh.

Oh I figured it out, turns out all you have to do is realize that sin^4x = (sin^2x)(sin^2x) = (-cot^2x +1)(-cot^2x +1) = cos^4x - 2cos^2x + 1 which is means that since cos^4x - 2cos^2x + 1 = sin^4x, when multiplied by sin, sin(cos^4x - 2cos^2x + 1) becomese sin^5x

Please help me with these (any of these) it has been a long time since I went back and did any of these and I seriously need the help. I need it quick too =P.


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