Question

Trig question

Answer 1

Show \[\sin^2\cos^2x = \frac{ 1 - \cos4x }{ 8 }\]

I used the squared identities to come up with...
\[(\frac{ 1 - \cos2x }{ 2 })(\frac{ 1+\cos2x }{ 2 })\]
but I'm not sure if that's right... and where would I go from there? Combine it into:
\[\frac{ 1-\cos^2x }{ 4 }\]
?

Answer 2

sin (2a) = 2 sina cos a, ok?

Answer 3

then \(\dfrac{sin^2(2a)}{4}= sin^2acos^2a\) right?

Answer 4

now, you have the identity \(sin^2 A =\dfrac{1-cos(2A)}{2}\)
done, right?

Answer 5

Ya I like the way Lose was doing it better.
You can certainly do it your way and then expand out all the brackets,
but I think applying Sine Double Angle Formula from the start is easier.\[\large\rm \sin^2x \cos^2x=(\sin x \cos x)^2\]Do a little something fancy here,\[\large\rm =\left(\frac122\sin x \cos x\right)^2\]And we can use the 2 for our identity,\[\large\rm =\left(\frac12\sin2x\right)^2=\frac14\sin^22x\]And then apply your Half-Angle Formula from there.

Answer 6

Wait... how did you get \[\frac{ \sin^22a }{ 4 } = \sin^2acos^2a\] ?

Answer 7

Start from the beginning... sin^2xcos^2x... what is the first step? What identity should I use?

Answer 8

0_o

Answer 9

? :3

Answer 10

We would like to apply this identity: \(\large\rm 2\sin x\cos x=\sin2x\)
But we have some squares in our way...\[\large\rm \sin^2x \cos^2x=(\sin x \cos x)^2\]Ok this is at least closer to what we want, ya?

Answer 11

Or if you prefer,
Lose was suggesting that you take your Double Angle Identity,\[\large\rm 2\sin x \cos x=\sin2x\]and square both sides,\[\large\rm 4\sin^2x \cos^2x=\sin^22x\]Dividing by 4,\[\large\rm \sin^2x \cos^2x=\frac14\sin^22x\]ya?

Answer 12

Oh my gosh :/ I'm lost..
I have sin^2xcos^2x and I need four steps to turn it into (1 - cos4x)/8

Which identities would the above be?

Answer 13

You need 4 exact steps?
That's really annoying...
They expect you to do this one specific way when there are probably 3 or 4 ways to do it...

Answer 14

Thinking :d

Answer 15

Yeah... :/

Answer 16

\(\large\rm \sin^2x\cos^2x\)

\(\large\rm (\sin x\cos x)^2\qquad\qquad~\text{1. Group the squares}\)

\(\large\rm \left(\frac12\sin2x\right)^2\qquad\qquad\quad\text{2. Sine Double Angle Formula}\)

\(\large\rm \frac14\sin^22x\qquad\qquad\qquad~\text{3. Distribute Square}\)

\(\large\rm \frac14\cdot\frac12\left(1-cos4x\right)\qquad~\text{4. Sine Half-Angle Formula}\)

This seems like 4 steps to me... I dunno :\

Answer 17

Lemme try doing it your way and see how many steps it is.

Answer 18

\(\large\rm \sin^2x\cos^2x\)

\(\large\rm \frac12(1-cos2x)\frac12(1+cos2x)\quad\qquad\text{1. Sine and Cosine Half-Angle Formulas.}\)
\(\large\rm \frac14(1-\cos^22x)\qquad\qquad\qquad\quad\qquad\text{2. Distribute.}\)

\(\large\rm \frac14\sin^22x\qquad\qquad\qquad\qquad\qquad\qquad\text{3. Pythagorean Identity.}\)

\(\large\rm \frac14\cdot\frac12(1-cos4x)\qquad\qquad\qquad\qquad\text{4. Sine Half-Angle Formula.}\)

Answer 19

Ya you can turn it into four steps either way I think......
Having 4 exact boxes is really really stupid though :P grr

Answer 20

If you prefer the write the 2 underneath when applying your Half-Angle Identities, that's fine. I think it looks cleaner to have a 1/2 in front though.

Answer 21

Thank you!
On the first one you did, where did the cosx go between step 1 and 2?

Answer 22

Which method? The first method or second? :o

Answer 23

The first one :)

Answer 24

Your Sine Double Angle looks like this,\[\large\rm 2\sin x \cos x=\sin2x\]Dividing by 2 gives us another way to write this identity,\[\large\rm \color{orangered}{\sin x \cos x=\frac12\sin2x}\]
Sooooo ya, that's what I did.
\[\large\rm (\color{orangered}{\sin x\cos x})^2=\left(\color{orangered}{\frac12\sin2x}\right)^2\]

Answer 25

Ah, okay! Thank you :) Let me go through the rest, I'll let you know if I have any questions...

Answer 26

cool beans

Answer 27

^haha. been a while since I heard that one.

The next Q says to find the equation
\[\sin^2xcos^2x = \frac{ 2 - \sqrt2 }{ }\]

.... would I rewrite it as \[\frac{ 1-\cos4x }{ 8 } = \frac{ 2 - \sqrt2 }{ 16 }\] ?

Answer 28

There should be a 16 in that first equation... on the denominator

Answer 29

Solve this equation?
Ya that's a good first step!
How bout multiplying by 16 as a second step?
And then doing some stuffs.

Answer 30

\[1 - \cos4x = \frac{ 2 - \sqrt2 }{ 2 }\]
That is what I had so far... should I multiply by 2 now?

Answer 31

Oops, actually I simplified it further... lemme know if I'm on the right track
\[\cos(4x) = \frac{ -\sqrt2 }{ 2 }\]

Answer 32

Hmm shouldn't it be positive?
Checking...

Answer 33

Ya, positive root 2 over 2.
Check your work again maybe? :O

Answer 34

You are right! I forgot about the negative sign on cos4x

Answer 35

What can I do now?

Answer 36

Come on Pebs.. you know this one..
You've been doing this type of problem all week :P

Answer 37

Sorry, I get lazy when I'm tired :(
double angle formula for cosine?

Answer 38

No, you just use your unit circle from here.

Answer 39

cosine(stuff)=sqrt2/2

so
stuff = ?

Answer 40

pi/4 ... times 4... x = pi?

Is that the only solution though? The instructions say to solve for ALL solutions to the original equation

Answer 41

Or would it be pi/16...

Answer 42

pi/4 sounds like the correct place to start\[\large\rm 4x=\frac{\pi}{4}+2k\pi\]But ya there is another one...
I think it's uhhhh 7pi/4, ya?

Answer 43

And then ya, divide by 4.

Answer 44

x = pi/16 + kpi/2
x = 7pi/16 + kpi/2

yeah?
thank you so much zep <3
My brain needs some cilantro power.

Answer 45

I've actually been out of spinach for the past couple days... it's wacking me up

Answer 46

ya that looks correct.

out of spinach? :O
oh no! that just won't do!

Answer 47

It's the worst :'(
Spinach is my staple. I'm going through withdraw

Answer 48

Most people drink coffee, I drink spinach smoothies :)

Answer 49

ok now you went too far -_-

Answer 50

Hahaha.
I'm serious though