yes, i know. what is the problem?
yes, i know. what is the problem?
All you did was change the right side of the equation from cosec x to sinx^2. That's the problem. xDOriginally Posted by radyk05
You get marks for part one but you haven't proven 8sinx . cos x . tan x is equal to cosec x. You also haven't solved for x.
now that is the truth my friends!I hate math soooo much.
didn't i showed that 8* sin x * cos x * tan x is not equal to cosec x?All you did was change the right side of the equation from cosec x to sinx^2. That's the problem. xD
8 * sin x * (cos x * tan x) = cosec x
8 * sin x * (sin x) = cosec x
8 * (sin x)^2 = cosec x
8* (sin x)^2 = (sin x)^(-1) ---> false
I'm doing it on paper and got the EXACT same thing. But apparently there's a way to solve this.didn't i showed that 8* sin x * cos x * tan x is not equal to cosec x?All you did was change the right side of the equation from cosec x to sinx^2. That's the problem. xD
8 * sin x * (cos x * tan x) = cosec x
8 * sin x * (sin x) = cosec x
8 * (sin x)^2 = cosec x
8* (sin x)^2 = (sin x)^(-1) ---> false
Taking what's done...
8*(sin x)^2 = (sin x)^-1
8*(sin x)^3 = 1
2 sin x = 1
x = 30 degrees
i fail to recognize that as a proof.
edit: nevermind. the question is poorly writen, tho.
ahem: for which values of x is 8 * sin x * cos x * tan x = cosec x?
Last edited by radyk05; 09-24-2005 at 03:49 AM.
I say we give it to Neel even if the question isn't completely answered. I hate proving trig identities + anything to do with radians. Not that it's hard, just that I dislike it. The thing with the generators and such. Bleh.
sin<sup>2</sup>x looks better than sin^2 x
Mo never asked for a proof, simply a solution.
Here's a proof-type question for you.
Say I start walking at the Cartesian coordinate (0,0) and I need to go to (1,1). I can walk at right angles, going to (0,1) then (1,1) for a distance of 2, but it would be shorter to go at 45 degrees straight to (1,1), for a distance of 1.414. But say I move infinitesimally north, to (0,delta), then infinitesimally east, to (epsilon,delta), then repeat this in a staircase pattern. As epsilon and delta go to zero, this path should represent the 45 degree path from (0,0) to (1,1). However, as long as epsilon and delta are finite, I will have to walk a distance of 2. Is it a contradiction that my right-angle approximation of the 45 degree path still results in a distance of 2? Is the 45 degree path really shorter?
Teach me how do do a squared thingie!
he did answer the question because
8 * sin (30) * cos (30) * tan (30) = cosec (30) ---> true
if anyone cares for an easy-but-hard projectile problem i will be more than happy to bring it to the thread (even tho its a physics problem but lots of math are involved).
sin< sup >2< /sup >xTeach me how do do a squared thingie!
Wow, posts go boom. I posted a thing to solve up there.
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