Cheat Engine

pakakid1

Well this is the problem:

5cos(theta)^2-4sin(theta)-4=0.

What i have to do is find to the nearest degree the values of theta. I have no idea how to convert any of them both to sin or co. I can seem to use any function of double angles or any formulas. If your not in high school or high don't bother with this.

So anyone figure this out? Good Luck To Ya'll Razz

Cryoma · Posted: Mon Mar 31, 2008 3:40 pm Post subject:

Lol, that's easy.

pakakid1 · Posted: Mon Mar 31, 2008 3:41 pm Post subject:

whats the answer? i cant figure it out ><

Cryoma · Posted: Mon Mar 31, 2008 3:44 pm Post subject:

I said it was easy, I didn't say I knew the answer.

Eureka---renton · Posted: Mon Mar 31, 2008 3:46 pm Post subject:

Well, obviously, the answer is (4, 6), or the square root of 34.
Easy peasy lemon squeezy.

pakakid1 · Posted: Mon Mar 31, 2008 3:47 pm Post subject:

lmfao u are a good one. Btw ur gf is so fken hot Twisted Evil

Cryoma · Posted: Mon Mar 31, 2008 3:48 pm Post subject:

Ok crackdown time, let me think about this..
Tan^2 theta) / (1+ Tan^2 theta) = sin^2 theta
Not always easiest but I recommend you change everything into sine and cosine. Tan^2 theta= sin^2 theta/cos^2 theta. The left side becomes
(sin^2 theta/cos^2 theta)/(1+ sin^2 theta/cos^2 theta). Now simplify the fraction by multiplying both numerator and denominator by cos^2 theta. (And remember that sin^2 thet+ cos^2 thet= 1.)
Solve (2sin theta) (Cosine theta) = sin theta
Obvious point- any theta such that sin theta= 0 satisfies this since it makes both sides equal to 0. If sin theta is NOT 0, divide both sides by sin theta to get cos theta= 0. Can you solve sin theta= 0 and cos theta= 0?
Solve (Sin^2 theta) - (sin theta) = cosine^2 theta
Write cos^2 theta as 1- sin^2 theta. Replace sin theta by "x" and solve the quadratic equation for x. Solve sin theta= x to find theta.
Find exact value of tan15 (degrees)
15 = 30/2 so if you know either the "half angle" or "double angle" formulas for sine and cosine, this is easy.
Find exact value of Sin105 (degrees)
105= 90+ 15 degrees so use what you learned in the previous problem, together with a sum formula.
Given:
0<Beta<pi/2<alpha<pi with sin beta=24/25 and sin alpha=4/5,
FIND sin (Alpha-Beta)
You should be able to use sin^2+ cos^2= 1 to find cos beta and cos alpha and then use the fact that sin(a-b)= sin a cos b- cos a sin b.
Prove the following identities:
(Tan^2 theta) - (Sin^2 theta) = (Sin^2 theta) * (tan^2 theta)
++++2)
(Cos theta) -1 / 1- (SEC theta) = (Cos theta) +1 / 1+ (SEC theta)

pakakid1 · Posted: Mon Mar 31, 2008 3:51 pm Post subject:

Eureka---renton · Posted: Mon Mar 31, 2008 3:55 pm Post subject:

Cryoma · Posted: Mon Mar 31, 2008 3:57 pm Post subject:

It's just something I was born with, nothing much.

Eureka---renton · Posted: Mon Mar 31, 2008 4:01 pm Post subject:

Cryoma · Posted: Mon Mar 31, 2008 4:03 pm Post subject:

EDIT:
Erm, ok so maybe I didn't use a calculator.

pakakid1 · Posted: Mon Mar 31, 2008 4:07 pm Post subject:

Ok Cryoma what you said basically summed everything up that my math teacher took 2-4 weeks to teach us. However i found a way to solve my problem and i would like to share with you. Firstly the problem is 5cos(theta)-4sin(theta)-4=0. I will ignore saying theta for now.

Alriet moving on i took the 5cos2 and formed it into 5(1-sin^2)-4sin-4=0. Then i distrubuted getting:

5-5sin^2-4sin-4=0.
Then -5sin^2-4sin+1=0
Since the sqaured cannot be negative we swtich the values
5sin^2+4sin-1=0.

Now we use something called "barrow and payback"

We take the 5 divide it to 1 and dviidng it by sin^2.

This leaves us with sin^2+4sin-5=0.

Now we factor:

(5sin-1)(5sin+5)

We pay back the 5 by diving what side has the factor which elaves us with:

(5sin-1)(sin+1)

Now we sole each:

sin=1/5
sin=-1

We use all quads since we have an both positive and negative answer.

Finally answer is QI= 12 Q2=168 (this is for sin1/5) Since the other is -1 and that is an quadrentral angle we leave that at 270.

Once again your help was greatly appreciated and get me to think. Thanks alot bro, major props!

Cryoma · Posted: Mon Mar 31, 2008 4:10 pm Post subject:

No problem at all, I should do this more often.

pakakid1 · Posted: Mon Mar 31, 2008 4:13 pm Post subject:

LoL yea they can get intrestin sometimes.