Question

Analytic Trigonometry (again!)
(Sin2x + Cos2x)^2 = 1

Answer 1

Expand the binomial :)\[\Large\rm (a+b)^2=a^2+2ab+b^2\]Yes?

Answer 2

alright that gives \[\sin2x^2 + 2\sin2xcos2x + \cos2x^2 =1\]

Answer 3

Ok let's clean it up a little bit so it's easier to read,
\[\Large\rm \sin^2(2x) + 2\sin(2x)\cos(2x) + \cos^2(2x) =1\]

Answer 4

If we rearrange our addition:\[\Large\rm \color{orangered}{\sin^2(2x) +\cos^2(2x)} + 2\sin(2x)\cos(2x) =1\]Do you see anything interesting going on?

Answer 5

is that an identity I see?

Answer 6

It should be :) hehe

Answer 7

Remember your most basic Pythagorean Identity involving sine and cosine?

Answer 8

I think it's the 2x that is confusing you

Answer 9

Yes, that equals 1 correct?

Answer 10

\[\Large\rm \color{orangered}{1} + 2\sin(2x)\cos(2x) =1\]Mmmm ok good!

Answer 11

Okay. Should we subtract 1 to set 2sin(2x)cos(2x) = 0 ?

Answer 12

Yes.
And further, divide both sides by 2,
sin(2x) cos(2x) = 0

Answer 13

Alright

Answer 14

Then you apply your Zero-Factor Property again, yes? :)
OR OR OR, you can apply your Sine Double Angle if you want to simplify things a bit.

Answer 15

can we do both the sin and cos double angle properties?

Answer 16

Mmm don't need the Cosine Double Angle.\[\Large\rm \sin(4x)=2\sin(2x)\cos(2x)\]

Answer 17

Giving us,
sin(4x)=0
yah? :o

Answer 18

I am not aware of that

Answer 19

how would I go about solving a 4x then? :)

Answer 20

This is the identity you know, yes?
\[\Large\rm \sin(2u)=2\sin(u)\cos(u)\]Replace all of the u's with 2x.

Answer 21

yes

Answer 22

So then your sine function is zero at ummmm 0 and pi, yes?\[\Large\rm \sin(4x)=0\qquad\to\qquad 4x=0+k \pi\]Where k is any integer.
We solve for x by dividing both sides by 4.

Answer 23

This is very new to me xD

Answer 24

Are you not familiar with the k stuff?
Are you only looking for solutions between 0 and 2pi maybe?

Answer 25

Sine and Cosine are periodic, they will repeat their pattern over and over.
So where ever we have a solution, we'll have infinitely maybe cause of the repeating nature of sine.

Answer 26

Oh, adding from the interval [0, 2pi) would have helped, wouldn't it?

Answer 27

Yah that helps simplify things a bit :)

Answer 28

I apologize

Answer 29

Oh oh, ok so things get a little tricky here.

Answer 30

We have sine giving us zero at 0, pi, 2pi, 3pi, 4pi, 5pi, and so on.

Answer 31

So those are the angles for our 4x.
But we're interested in x.

And we want all of the angles that fall within 2pi.

Answer 32

yee

Answer 33

So if we divide each of those angles by 4 we get our values for x.\[\Large\rm 0,~\frac{\pi}{4},~\frac{2\pi}{4},~\frac{3\pi}{4},~\frac{4\pi}{4},~\frac{5\pi}{4},~\frac{6\pi}{4},~\frac{7\pi}{4},~\frac{8\pi}{4}\]

Answer 34

Ooo I'm getting a different answer set when I graph it...
hold up lemme make sure I didn't make a boo boo somewhere :(

Answer 35

haha okay

Answer 36

no no we're good :)`

Answer 37

So I'm taking all of these multiples of pi, which are equivalent to our 4x.
Dividing them all by 4 gives us the values for x.

Should we exclude any of those based on our interval that they gave us?
Any of them larger or equal to 2pi?

Answer 38

nah

Answer 39

Hmmm 8pi/4. I think that one is a problem for us.
That simplifies to 2pi, yah? :o

Answer 40

yeh

Answer 41

lol

Answer 42

o dang

Answer 43

Hmm this is a tough problem, 8 solutions D:
Your brain esplode already?

Answer 44

like 7 times already

Answer 45

Mmm yah trig will do that to ya bruh >.<

Answer 46

bbbbbrrrruuuuuhhhh

Answer 47

lol

Answer 48

Do you understand this step?\[\Large\rm \sin(\theta)=0\]Like would you be able to tell me values for \(\Large\rm \theta\) ?

Answer 49

yes

Answer 50

Because our \(\Large\rm \theta\) is our \(\Large\rm 4x\) in this problem.

Answer 51

okay

Answer 52

Ok ok chew on that for a while >.<
I think your brain needs some rest :3
mine too..

Answer 53

haha alright, I'll start on the next problem or do some comp sci