Question

sin^4 θ − cos^4 θ = sin^2 θ − cos^2 θ

Answer 1

I know it's the power lowering thing, but I'm not very good at that

Answer 2

\[\frac{ 1-\cos2(4x)-1+\cos2(4x) }{ 4 }\]

Answer 3

is that right so far?

Answer 4

No. This is a factoring problem.
What are the factors of
\[x^2-y^2\]

Answer 5

brandon...are you there?

Answer 6

yea, I'm trying to figure it out

Answer 7

2(x-y)?

Answer 8

\[a^2-b^2=(a-b)(a+b)\]

Answer 9

Use that to factor x^2-y^2

Answer 10

ok

Answer 11

oh ok. x^2-y^2=(x-y)(x=Y)

Answer 12

*(x+y)

Answer 13

yes. (x+y)
Now by extension:
\[a^4-b^4=(a^2-b^2)(a^2+b^2)\]

Answer 14

ok

Answer 15

So use that to factor
\[x^4-y^4\]

Answer 16

\[x ^{4}+y ^{4}=(x ^{2}-y ^{2})(x ^{2}+y ^{2})\]

Answer 17

Yes. And do you see that this is the same form:
\[\sin ^4\theta-\cos ^4\theta\]

Answer 18

so it's (sin^2-cox^2)(sin^2+cos^2)

Answer 19

Yes. And do you know that sin^2 +cos^2=1?

Answer 20

yes

Answer 21

So try your problem again now.

Answer 22

ok

Answer 23

\[\sin ^{4}\theta - \cos^{4}\theta = (\sin^{2}\theta - \cos^{2})(\sin^{2}\theta +\cos^{2}\theta)\]

Answer 24

So far so good

Answer 25

ok, what am I looking for next?

Answer 26

sin^2+cos^2=1, I know, but how to I pull that out of this?

Answer 27

\[(\sin ^2\theta-\cos ^2\theta )(\sin ^2\theta+\cos ^2\theta)=(\sin ^2\theta-\cos ^2\theta)(1)=\sin ^2\theta-\cos ^2\theta\]

Answer 28

This is the way the entire problem should look. (I'm going to use x instead of theta):
\[\sin ^4x-\cos ^4x=\sin ^2x-\cos ^2x\]
\[(\sin ^2x-\cos ^2x)(\sin ^2x+\cos^2x)=\]
\[(\sin ^2x-\cos ^2x)(1)=\]
\[\sin ^2x-\cos ^2x=\sin ^2x-\cos ^x\]

Answer 29

ok, cool, I'm going to try to work through that a couple times

Answer 30

Excellent

Answer 31

cool, I see how it works. thanks so much!