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TRIG IDENTITIES
- Thread starter Uthakore
- Start date
I hope the next time you ask for help you DOUBLE CHECK your typing before submitting your problem. It is a little much to ask someone to prove an identity when it is not an identity. After spending considerable time trying to prove the identity, I determined that it doesn't check for x=30. You wasted my time.
Loren said:I hope the next time you ask for help you DOUBLE CHECK your typing before submitting your problem. It is a little much to ask someone to prove an identity when it is not an identity. After spending considerable time trying to prove the identity, I determined that it doesn't check for x=30. You wasted my time.
Ditto. I am not guessing anymore.
But, if my last guess is correct, use the sum of two cubes factorization on the numerator.
galactus, thank you for your help, the last guess did help on this, and other problems. As for Loren, I dont care if you had a bad day, there is no reason to flip out over a problem that you found to be NO SOLUTION! I checked with 5 friends who use different study helpers, and they got the same answer (without the anger it seems you exhibit). Thank you glactus, and no thank you Loren.
P.S
galactus, I'm sorry I made you guess... I am quite stressed studying for eight different finals right now, and just mixed up two problems. The first error was a computer error by the way...
P.S
galactus, I'm sorry I made you guess... I am quite stressed studying for eight different finals right now, and just mixed up two problems. The first error was a computer error by the way...
So, the last one I posted is correct?.
In that event, It is not an identity.
Use the sum of two cubes:
\(\displaystyle \frac{sin^{3}(x)+cos^{3}(x)}{sin(x)+cos(x)}=\frac{(sin(x)+cos(x))(sin^{2}(x)-sin(x)cos(x)+cos^{2}(x))}{sin(x)+cos(x)}=1-sin(x)cos(x)\)
\(\displaystyle 1-sin(x)cos(x)=1-\frac{sin(2x)}{2}\neq 2cos^{2}(x)-1\)
In that event, It is not an identity.
Use the sum of two cubes:
\(\displaystyle \frac{sin^{3}(x)+cos^{3}(x)}{sin(x)+cos(x)}=\frac{(sin(x)+cos(x))(sin^{2}(x)-sin(x)cos(x)+cos^{2}(x))}{sin(x)+cos(x)}=1-sin(x)cos(x)\)
\(\displaystyle 1-sin(x)cos(x)=1-\frac{sin(2x)}{2}\neq 2cos^{2}(x)-1\)
Uthakore said:galactus, thank you for your help, the last guess did help on this, and other problems. As for Loren, I dont care if you had a bad day, there is no reason to flip out over a problem that you found to be NO SOLUTION! I checked with 5 friends who use different study helpers, and they got the same answer (without the anger it seems you exhibit). Thank you glactus, and no thank you Loren.
P.S
galactus, I'm sorry I made you guess... I am quite stressed studying for eight different finals right now, and just mixed up two problems. The first error was a computer error by the way...
Perhaps you don't realize that we are ALL volunteers here. You may wish to consider whether it is wise to insult people you are expecting to help you FOR FREE.
We take it for granted that when you post a problem, you've taken the time to make sure you've posted it correctly. WE then spend our time trying to solve the problem you posted. When you haven't bothered to make sure your post is correct, you ARE wasting our time. Not to worry...it won't happen again, for me, at least.
Uthakore said:galactus, thank you for your help, the last guess did help on this, and other problems. As for Loren, I dont care if you had a bad day, there is no reason to flip out over a problem that you found to be NO SOLUTION! I checked with 5 friends who use different study helpers, and they got the same answer (without the anger it seems you exhibit). Thank you glactus, and no thank you Loren.
P.S
galactus, I'm sorry I made you guess... I am quite stressed studying for eight different finals right now, and just mixed up two problems. The first error was a computer error by the way...
Uthakore. I am attempting to help you. If you consider my statement as "flipping out" I guess I didn't express myself to your satisfaction. It seems to me that If you knew the teacher included some "non-identities" in the batch of "identities", you would check for that possibility after being unsuccessful for a certain amount of time. You did not think of providing us with that information. Furthermore, as you can tell by a couple of responses, I was not the only one who felt his/her time was wasted. And by double checking your typing, you will not alienate people who donate their time and effort to help students understand math.
D
Deleted member 4993
Guest
Uthakore said:Yes, you actually gave me an answer... Thank you. Our teacher said that she would put some "no solution" ones on the HW and the Test. Again, Thanks!
That statement worries me.
If the instructions were to solve for 'x' (instead of - prove the identity) - then there is a solution for that given problem - not a nice "one", but there is "one".
\(\displaystyle Let\)
\(\displaystyle \tan^{-1}(\frac{1}{2}) \, = \, \theta\)
then
\(\displaystyle x \, = \, \theta \, + \, n\pi\)
or
\(\displaystyle x \, = \, n\pi\)