can you solve this? (1 Viewer)

[tooheyz]

hmm, i have a feeling this is real easy, but i cant see it!

any help is fine!

cheers!

sin x = -cos x

OR

-sin x = cos x

 

[...]

sinx=-cosx
-1= tanx

x= - pi/4 ......and u can figure the rest

 

[tooheyz]

oh my goodness!
thanks mate, LOL

ur a legend (and so so william hung... did you know he is gonna have his single released soon?)

 

[Xayma]

sin x=-cos x
Therefore it is in the 2nd quadrant or 4th quadrant. It would be 45 degrees (or pi/4) in the second and 4th quadrant Ie 135degrees of -45 degrees or 315 degrees or the equiv radians.

 

[KeypadSDM]

Sin[x] = -Cos[x]
Tan[x] = -1
x = n(Pi) - Pi/4

(n E Z)

 

[...]

Originally posted by KeypadSDM
Sin[x] = -Cos[x]
Tan[x] = -1
x = n(Pi) - Pi/4

(n E Z)

hmm...i got lost in the x= n(pi) - pi/4

w.t.f is that?

 

[Heinz]

Originally posted by ...
hmm...i got lost in the x= n(pi) - pi/4

w.t.f is that?

thats the general solution (since there wasnt a specified domain, there are an infinite number of solutions) n is an integer

 

[hatty]

use T results or Auxillary angle method for more complex shit

 

[Grey Council]

how the hell does T results fit into this crap. Auxillary angles? :S Hatty, put on your thinking hat. (get it? hatty.. hat heheh, what a pun).

nah, seriously, T results and auxillary angles are both in the 3u course, not the 2 unit course. Anyway, this is pretty darn easy. No need to introduce all these wierd arse formulas. heheh

 

[KeypadSDM]

Umm, actually you do. You require the general solution because there is no specified domain. You're wrong unless you answer with the general solution (or some other answer which shows the infinitely many answers).

 

[tooheyz]

alright, i just got 2 more questions - i was gonna post a new thread, but umm, yeah anyway

can you guys solve this?

differeniate the following

1. a^a (a to the power of a)
(2. log (base a) x

thanks guys - asap would be great!

answers:
1. 0
2. 1/(x ln a)

 

[tooheyz]

SOMEONE?!

 

[KeypadSDM]

Actually the first one isn't 0

a^a = e^(aLn[a])
d/da a^a
= a^a * d/da(aLn[a])
= a^a * (Ln[a] + 1)

Log_a[x]
= Ln[x]/Ln[a]
d/dx Log_a[x]
= 1/Ln[a] * d/dx Ln[x]
= 1/(xLn[a])

 

[tooheyz]

Originally posted by KeypadSDM
Actually the first one isn't 0

a^a = e^(aLn[a])
d/da a^a
= a^a * d/da(aLn[a])
= a^a * (Ln[a] + 1)

yeah that was my working out, hmm the answer must be wrong then

alright thanks for that mate!

ur a legend

 

[KeypadSDM]

Originally posted by tooheyz
yeah that was my working out, hmm the answer must be wrong then

alright thanks for that mate!

ur a legend

Not necessarily. It depends what the questions said differentiate with respect to.

 

[tooheyz]

nah it doesnt say.

yeah i guess, if it is differentiate with respect to x, then it is zero.
all the questions before and after are differentiated with respect to x, but the question doesnt specify it -

hmmm - thanks for pointing that out mate

 

[~*HSC 4 life*~]

Originally posted by KeypadSDM
Umm, actually you do. You require the general solution because there is no specified domain. You're wrong unless you answer with the general solution (or some other answer which shows the infinitely many answers).

is general solutions a 3u topic?

 

[~*HSC 4 life*~]

Originally posted by ...
sinx=-cosx
-1= tanx

x= - pi/4 ......and u can figure the rest

i got that...but i just got this flashign signal in my head...i forgot..when can u never divide by something like cosx?

 

[KeypadSDM]

Originally posted by ~*HSC 4 life*~
i got that...but i just got this flashign signal in my head...i forgot..when can u never divide by something like cosx?

When Cos[x] = 0

The 11th commandment:

Thou shalt not divide by zero

 

[KeypadSDM]

Originally posted by ~*HSC 4 life*~
is general solutions a 3u topic?

Whether or not it is, the question required that answer.