Recent content by Wholewheat458

.. 1/2(sum)n=0, infinity (2/3)^n+1 .. ?? do you just leave it like that?.. as a sort of equation?

Homework Statement I am to find the sum of the series, but what do i do if it is infinite?? no clue. i'm also not sure how to type the symbols so i hope you can understand me:shy: : (Sum) n=0, limit = infinity: 1/2(2/3)^n Homework Equations i 'm not sure. The Attempt at a Solution...

^^ i don't know why either! ahhahaa! i think i'll just skip it for now.. thanks for your help

.. i don't think i understand the formula.. : f(x) = sinx .. then. = sinx = lim(h-->) sin(x+h) - sinx/ h ... this is so wrong wait, what is h?..

.. ? woahw.. so... how do i apply that to the sin graph?.. : no clue.. maybe i'll go over it again..

?? I'm not all that certain.. its got somethink to do with graphs, but.. idk, i been looking at a graph of y = sinx but i don't know what to do with it: oy

Homework Statement Prove using the definition of a derivative. If y=sinx, then y1=cosx Homework Equations .. ? identities maybe..? i don't really know much about derivatives.. The Attempt at a Solution y = sinx.. y1 = (1)sinx^1-1 y1 = 1.. how is that suppose to equal cosx?? ...

Homework Statement 2 cos^2 x + cos x = 0 and tan x = 3^1/2 Homework Equations .. ?? Unit circle. The Attempt at a Solution for the first, well I tried factoring to isolate x, but it really did not work i keep making up my own rules that end up messing with the entire equation...

I need to solve for x, but,.. honestly, i haven't got a clue for where to begin, or even,.. oy, not sure here are two of my problems: sin x = (3^1/2) / 2 and 2 cos 2 x + cos x = 0 I know I'm supposed to use the unit circle,.. well, maybe any help at all would be very much...

^^ WOOT! Hahahhaha! the answer is sin x! thank you so much!

oh,.. no, i forgot the carrots sorry, the sec,sec, and csc are all squared (sec^2 x csc x) / (sec^2 x + csc^2 x)

^^ oh, I'm actually suposed to simplify the expression in the most complete form

Here is my problem: (sec 2 x csc x) / (sec 2 x + csc 2 x) -first i replaced the denominator with its identities.. and i got : 2+tan^2 x +cos^2 x but the numerator.. can you just .. i really think I'm doing this wrong.. ugh, and i cannot stand it! Honestly, i would be truly gratful for...

your clue for the second one helped tons! ^^ I attempted the first one again, but i still think it's wrong heres what i did: Numerator turns out to be -sin x/cos2 x and the denominator is = to 1 (identity) so.. then i put the fractions in one layer and got: cos 2 x/ 1-sin x i replaced cos...

I do not understand these problems; I understand the trig basics (identities, inverses, and a bit more), but I have'nt got a clue of where to start with these: (sec 2 u - tan 2 u) / (cos 2 v + sin 2 v) [ (sec y - tan y)(sec y + tan y)] / [sec y] and this one: cos x - cos 3 x...