Recent content by Laceylb

im seriously mildy retarded in trig, i will kill myself if i have to take it in college

\sin ^ 2 x \cos ^ 4 x = \sin ^ 2 x \cos ^ 2 x \cos ^ 2 x = \frac{\sin ^ 2 (2x)}{4} \cos ^ 2 x = ... im not really sure how you got to the last part of it with sin^2 over 4

i actually can't understand the third problem... i never even knew you could divide the oringinal double angle identities... this is what i have on the left side = to 1-cosx: sqrt((1-cosx)(1+cosx)/(1+cosx)) sqrt(1-cosx)... i tryed to put an identity in...but i don't know if i did it right...

oh, i have them written down already

what is factor formulae?

i have all but the first and third done... for the third i have sqrt(cosx-1) but i don't know how to get rid of the sqrt, am i allowed to square it ?

the first problem is correctly written the way i had it, but i did mess up it should be cos raised to the first on the bottom i have tried these problems forever, a few days at least, we had two assignments 1-71 every other odd and 2-116 every other even... these are the only three i just...

yes for last two you are to prove the identity im not really sure what you mean about the first one? a good part of it my teacher said was right and he said from there on it was just an algebra mess

the original for that was (1 + sex(-x))/sin-x) + tan(-x) = -cscx

ok here is the first problem: sin^2 xcos^4 x and you are to simplify it, the answer is 1/32(2+cos2x-2cos4x-cos6x) i will skip a few steps just to save some time: 1/8(1+2cos2x + cos^2 2x- cos2x -2cos^2 2x -cos^3 2x) 1/8( 1+2cos2x + 1/2(1+cos4x) - cos2x- 1+cos4x - 1/2cosx(1+cos4x) )...

1-sin^2x is the conjugate... the opposite of the denominator... its useful to get a difference of squares on the bottom.. if you did stuff with radicals, you probably have seen it before...

I am sorry, i am new, but i didnt think i solved it all, he had to put the x's back into the problem

i finished it. it was due today.i asked him and someone else put it on the board..you had to replace cos^2 s with sin^x -1 and sin^2y-1 and it continued from there but i don't really remember the rest

well i think because there are two different answers for x and when you plug it back in you would have 6 total numbers, I am just taking a guess as that

ok i got it... x(x+1) = -6(x+2) x^2 + x = -6x -12 x^2+7x+12=0 (x+3)(x+4)=0 x=-3 and -4