Question

use a double or half angle formula to solve the equation in the interval [0,2pi)
tan(a)+cot(a)=4sin2(a)

Answer 1

Only way to solve is to set to zero. So start with that.

Answer 2

tan(a)+cot(a)-4sin2(a)=0

Answer 3

like that?

Answer 4

Yep. Now for the fun part. Seeing which double or half will help the most.

Answer 5

okay what does that mean?

Answer 6

Well, you generlly want things to factor so that you can set each term to zero. Some things are easier to factor than others.

Answer 7

Oh it's fixed now.

Answer 8

はい

Answer 9

so whats next? what?? hah

Answer 10

He picked on me about letting my japanese get overed up by th number in the lower right.... so I had to fix it so the harmony of the aviutar would be maintained.

Answer 11

hahah ight.

Answer 12

OK, using their equation editor, that is with a power already, right? \[\tan(a)+\cot(a)-4\sin^2(a)=0\]

Answer 13

man im getting tired haha its later over here on the east coast lol

Answer 14

Hmmm. what about putting tan and cot in terms of sin and cos. Then seeing what pops up that might let a half or double formula (as required) clean it up?

Answer 15

And yah, 3 hour time difference?

Answer 16

tan is sin/cos and cot is cos/tan?

Answer 17

cot = cos/sin

Answer 18

woulsnt that cancel out, and it would be -4sin2(a)=0?

Answer 19

Not quite. You can get it to 1/(sin cos) but I think I found a better way. I was playing with it a bit, then decided to dig. Man, I really need to brush up on trig. I have a calc teacher that loves trig...

Answer 20

okay whats this better way. how bout you just show me how to so it. im about to fall asleep. hah please..

Answer 21

I am going to shortcut this a bit for speed. So you see that we can get to:\[\frac{\sin}{\cos} + \frac{\cos}{\sin}=4\sin^2\]

Answer 22

okay good.

Answer 23

\[\frac{\sin^2+\cos^2}{\sin \cos}=bla\] that can be used to jump to the double angle of
\[\frac{1} {(1/2)\sin(2a))}=bla\]

Answer 24

Ge that step? Well, 2 steps. I pathagoreaned the top and double angled the bottom.

Answer 25

what does bla stand for?
is that the final answer?

Answer 26

Just repeating the same right hand side. I was not changing it, so bla.

Answer 27

Now, can you see how that left hald side would simplify to get rid of the fraction?

Answer 28

so it equals 4sin2(a)

Answer 29

\[\frac{1}{\frac{\sin(2a)}{2}}=4\sin^2a \rightarrow 2\sin(2a) =4\sin^2a \rightarrow 2\sin(2a) - 4\sin^2a = 0\]

Answer 30

Now that looks more factorable.

Answer 31

what do i factor out? -2sin(a)?

Answer 32

Hmmm.... I know we are close. Lets see. I may need to back up a step and leave part in the 2sin cos. Lets look at the 4sin^2 now.

Answer 33

With half angle, that is -4(1/2 1-cos(2a))

Answer 34

Yah, so let me put that in there. \[2\sin a \cos a - 2 - 2 \cos(2a) \; or \; 2\sin(2a)-2-2 \cos(2a)\] does anything theremake it better..

Answer 35

If you did not notice, I distributed the 4 into the half angle, which got me the 2-2cos(2a)... but I messed up the sign. It should be +2cos(2a).

Answer 36

so the answer the sin(2a) -2 +2cos(2a) ?

Answer 37

Well, that is close. But if hey want you to find values of a, it is not an answer. I was aiming for factors to set to 0 and somewhere in there I did not get what I wanted.

Answer 38

OK. I also see this in there: \[2\sin a(\cos a - 2\sin a) = 0\]which is partially factored. But I would need to get the factors of what is inside the parantisis. Which is what I thought was going to be nicer than it ended up.

Answer 39

You understand where I am going with the factoring? This is like algebra. x^2+4x+3 = 0, so you factor it to (x+1)(x+3)=0 and then I can solve each part. That is how you solve this sort of equation, unless you just needed a proof that one side equals the other.

Answer 40

Just to make sure, is that problem \(4sin^2(a)\) or \(4sin^2(2a)\)?

Answer 41

4sin(2a)

Answer 42

no squared

Answer 43

OH! Ok. That was my mistake.

Answer 44

Well that makes all the difference!

Answer 45

can you show me all the work now? cause im all confused

Answer 46

tan(a) + cot(a) = 4sin(2a) thats the problem

Answer 47

Yah, I saw up above 4sin2(a) and I thought you meant squared. Big goog there on my part for not making sure earlier.

Answer 48

The start was good then. \[\frac{\sin a}{\cos a}+\frac{\cos a}{\sin a}-4\sin(2a)=0 \rightarrow \frac{\sin^2 a + \cos^2 a}{\cos a \sin a}-4\sin(2a)=0\]
\[\rightarrow \frac{1}{\cos a \sin a}-4sin(2a)=0\]Wit me to there?

Answer 49

yeah, is that it?

Answer 50

Not quite. Then I did that step where we went to this (but corrected):\[2\sin(2a)-4\sin(2a)=0\] which simplifies as:\[-2\sin{(2a)}=0\]

Answer 51

So find every part of the interval where \(\sin{(2a)}=0\) and you have the solutions. There should be a small herd of them with that.

Answer 52

On \([0,2\pi)\) you should have something like 4 or 8 points.Lets see, on the unit circle, sine is 0 at 0 and \(\pi\). But this is of (2a) so when 2a=0 or 2a=\(\pi\) you have a zero.

Answer 53

ok, yeah i got that

Answer 54

I'm at the end of a long work day, and I know you are brain fried at this point, so write it all out and then try it to see if you can get the same thing. If you can; then I did not make any mistakes. But if you can't I may have goofed in there somewhere and you may need someone else to go over it with you.

Answer 55

And the working it out on your part should be after some sleep. Hehe.

Answer 56

ok sounds good. thank you. whats the exact answer? 0 and π?

Answer 57

Should be a={0,pi/2,pi,3pi/4}

Answer 58

Because of the 2a part.

Answer 59

thanks man. goodnight

Answer 60

Night.