Top Answer

(1 + tanx)/sinx

Multiply by sinx/sinx

sinx + tanxsinx

Divide by sin2x (1/sin2x) = cscx

cscx + tan(x)csc(x)

tanx = sinx/cosx and cscx = 1/sinx

cscx + (sinx/cosx)(1/sinx)

sinx cancels out

cscx + 1/cosx

1/cosx = secx

cscx + secx

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0- ln (cscx + cotx) + C You use u substitution.

You could try y = 1/sin(x) but I do not see how that helps.

pi divided by 6 is a constant and so its first derivative is 0. And since that is also a constant, the second derivative is 0. It is not clear what f(x) = csc(x) has to do with that!

2

sinx cscx = 1 is the same thing as sinx(1/sinx) = 1 which is the same as sinx/sinx = 1. This evaluates to 1=1, which is true.

d/dx csc(x) = - csc(x) tan(x)

d/dx cscx = d/dx 1/sinx = d/dx (sinx)-1= -(sinx)-2 cosx = -cosx/sin2x = -1/sinx.cosx/sinx = -cscx cotx I suggest you copy this out onto paper so it is more clear. The / signs make it harder to see what is happening compared to horizontal divide lines.

d/dx (-cscx-sinx)=cscxcotx-cosx

cscx-sinx=(cosx)(cotx) 1/sinx-sinx=(cosx)(cosx/sinx) (1/sinx)-(sin^2x/sinx)=cos^2x/sinx cos^2x/sinx=cos^2x/sinx Therefore LS=RS You have to remember some trig identities when answering these questions. In this case, you need to recall that sin^2x+cos^2x=1. Also, always switch tanx cotx cscx secx in terms of sinx and cosx.

Using the u substitution method of derivation (selecting sinx as u and cosxdx as du), you get f'(x)=cscx.

y=1/sinxy'=(sinx*d/dx(1)-1*d/dx(sinx))/(sin2x)y'=(sinx*0-1(cosx))/(sin2x)y'=(-cosx)/(sin2x)y'=-(cosx/sinx)*(1/sinx)y'=-cotx*cscx

-1

According to Wolfram Alpha, input:integral csc x it is -log[cot(x) + csc(x)] + constant You can verify this by taking the derivative of the purported integral.

Trig functions have their own special derivatives that you will have to memorize. For instance: the derivative of sinx is cosx. The derivative of cosx is -sinx The derivative of tanx is sec2x The derivative of cscx is -cscxcotx The derivative of secx is secxtanx The derivative of cotx is -csc2x

I assume the expression is cot^2 x / ( csc^2 x - csc x) express it in terms of sin x and cos x: =(cos^2 x / sin^2 x) / (1/sin^2 x - 1/sin x) =(cos^2 x / sin^2 x) / [(1 - sin x)/sin^2 x] =cos^2 x / (1 - sin x) = (1 - sin^2 x) / (1 - sin x) = (1 + sin x)(1 - sin x) / (1 - sin x) = 1 + sin x

To solve this type of problems, I found that it helps to convert everything to sines and cosines: write csc x as 1 / sin x, and cot x as cos x / sin x, for example. Because of the "2x", you'll also need to look up the double angle trigonometric identities.

It's easiest to show all of the work (explanations/identities), and x represents theta. cosxcotx + sinx = cscx cosx times cosx/sinx + sinx = csc x (Quotient Identity) cosx2 /sinx + sinx = csc x (multiplied) 1-sinx2/sinx + sinx = csc x (Pythagorean Identity) 1/sinx - sinx2/sinx + sinx = csc x (seperate fraction) 1/sinx -sinx + sinx = csc x (canceled) 1/sinx = csc x (cancelled) csc x =csc x (Reciprocal Identity)

csc(x)*{sin(x) + cos(x)} = csc(x)*sin(x) + csc(x)*cos(x) =1/sin*(x)*sin(x) + 1/sin(x)*cos(x) = 1 + cot(x)

It's difficult for me to be sure what function you are asking about, because of the limitations of answers.com. I am considering y = -1 + csc (x) csc (x) = 1/sin(x) and sin is a periodic function of x. Ignoring what happens for negative values of sin(x), csc(x) is at local minima for maximal values of sin(x), which occur at x = (2k+1/2)pi for i any integer. Putting the -1 into -1 + csc(x) simply 'lowers' the function without changing the positions of these minima. PS: Incidentally, using GeoGebra can be a big help in solving problems like these. Free. Easy to use.

what happened to Tiger is a long story. here is the short version. tiger was not named after the animal, it was named after 5 people. T. I. G. E. R. sorry i cant remember their names. if you want you can look it up under basilmarket what happened to tiger just read everything and you will get an answer. ok. so yes it is true that tiger got banned. i was logged on and can prove in this video. he was banned for 2 days. he was not banned, however, because of hacking. he leveled so fast because the 5 6th grade boys that made up Tiger took shifts. they wernt like in the Make Love not Warcraft SouthPark episode, but it was somewhat similar. When Tiger got back from his ban, it is true that he had a girlfriend named kittyz0r. look her up in maple rankings. lvl 200. but, she didn't hack him. after one of the boys moved away, that boy became greedy, and changed the password and took all the credit. this was Gavin. Gavin played the file, but gave the password to kittyz0r and she played on it for a while. School started, and he didn't play all year. summer came, and he was back. but he hated the publicity, hated all the OMFG ITS TIGER I LOVE YOU. so he changed worlds and names. we are not positive on what world or name, but what we do know is Gavin could be out there still. two possibilities are cscx. all the same clothes and same level when he disapeared from the rankings. another character that could be his is TheTigerJr. is Gavin back, or is he gone forever? did he get a life? lets hope. We will always remember you Gavin! a link that i can prove to you all that Tiger was banned for swearing at kittyz0r. hopefully you will look up other links like it to find out what you might think is the answer! http:/www.youtube.com/watch?v=MJVILnkeFXE&feature=related HAPPY MAPLING!!!! XD

First convert everything to sines and cosines:sin x + sin x cos x / sin x = 1 / sin xsin x + cos x = 1 / sin xMultiplying by sin x:sin2x + sin x cos x = 1Using the identity sin2 + cos2x = 1:sin2x + sin x cos x = sin2x + cos2xsin x cos x = cos2xDividing by cos x:sin x = cos xThe solution is therefore x = pi / 4 radians, or x = 5 pi / 4 radians.The division by cos x assumed that cos x was not equal to zero; this possibility must be explored in the original equation. When cos x = 0, sin x = 1 or -1, and the angle x = pi/2 or -pi/2. It seems both of these are solutions, too.First convert everything to sines and cosines:sin x + sin x cos x / sin x = 1 / sin xsin x + cos x = 1 / sin xMultiplying by sin x:sin2x + sin x cos x = 1Using the identity sin2 + cos2x = 1:sin2x + sin x cos x = sin2x + cos2xsin x cos x = cos2xDividing by cos x:sin x = cos xThe solution is therefore x = pi / 4 radians, or x = 5 pi / 4 radians.The division by cos x assumed that cos x was not equal to zero; this possibility must be explored in the original equation. When cos x = 0, sin x = 1 or -1, and the angle x = pi/2 or -pi/2. It seems both of these are solutions, too.First convert everything to sines and cosines:sin x + sin x cos x / sin x = 1 / sin xsin x + cos x = 1 / sin xMultiplying by sin x:sin2x + sin x cos x = 1Using the identity sin2 + cos2x = 1:sin2x + sin x cos x = sin2x + cos2xsin x cos x = cos2xDividing by cos x:sin x = cos xThe solution is therefore x = pi / 4 radians, or x = 5 pi / 4 radians.The division by cos x assumed that cos x was not equal to zero; this possibility must be explored in the original equation. When cos x = 0, sin x = 1 or -1, and the angle x = pi/2 or -pi/2. It seems both of these are solutions, too.First convert everything to sines and cosines:sin x + sin x cos x / sin x = 1 / sin xsin x + cos x = 1 / sin xMultiplying by sin x:sin2x + sin x cos x = 1Using the identity sin2 + cos2x = 1:sin2x + sin x cos x = sin2x + cos2xsin x cos x = cos2xDividing by cos x:sin x = cos xThe solution is therefore x = pi / 4 radians, or x = 5 pi / 4 radians.The division by cos x assumed that cos x was not equal to zero; this possibility must be explored in the original equation. When cos x = 0, sin x = 1 or -1, and the angle x = pi/2 or -pi/2. It seems both of these are solutions, too.

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