Question

Factor Completely: Sin^4 - Cos ^4

Answer 1

Explain please

Answer 2

\[a^4-b^4=(a^2+b^2)(a+b)(a-b)\]

Answer 3

in this case the first term drops out because
\[\sin^2(x)+\cos^2(x)=1\]

Answer 4

what is the first term?

Answer 5

the a^2+b^2?

Answer 6

yes

Answer 7

why does it drop out? i dont really understand

Answer 8

replace \(a=\sin(x)\) and \(b=\cos(x)\) to get your answer

Answer 9

do you know what \(\sin^2(x)+\cos^2(x)\) is?

Answer 10

1

Answer 11

because of a property

Answer 12

that is why it drops out. the first term is 1, so you don't need to write it

Answer 13

ohhhh ok i didnt know what you meant by drop out but now i see

Answer 14

so then it is ( sin^2(x)+cos^2(x))* (sin^2(x)-cos^2(x))

Answer 15

IS THE ANSWER 2 or 1?

Answer 16

\[\sin ^4x-\cos ^4x=(\sin ^2x+\cos ^2x)( \sin ^2x-\cos ^2x)=1(\sin x+\cos x)(\sin x-\cos x)\]

Answer 17

The answer is (sinx+cosx)(sinx-cosx)

Answer 18

but isnt that sin^2+cos^2

Answer 19

when you foil it

Answer 20

so then 1*1=1...?

Answer 21

I'm thinking this...
\[sin^4x - cos ^4x\]\[=(sin^2x - cos ^2x)(sin^2x+cos^2x) \]\[=(sin^2x - cos ^2x) (1)\]\[=-(cos ^2x-sin^2x)\]\[=-cos2x\]

Answer 22

is cosx(-cosx)=+cos^2x?????

Answer 23

oh.. no...

Answer 24

callisto, im confused how does cos^2x-sin^2=-cos^2

Answer 25

Not ^2 ... It's 2...
\(cos2x = cos^2x-sin^2x\) It's an identity...

Answer 26

? I don't think i know that identity. Not in my book either..

Answer 27

but so far, ihave 1*sin^2-cos^2...

Answer 28

then @Mertsj 's answer is what you want ... but there is such an identity!!!!
use difference of two squares again...
a^2 - b^2 = (a+b)(a-b)
a= sinx , b= cosx

Answer 29

so the answer is the foiled version of Mertsj? sin^2-cos^2?

Answer 30

shouldn't you factor it?

Answer 31

(sinx+cosx)(sinx-cosx) is what he had

Answer 32

so foil? do you mean the same thing?

Answer 33

this is what you should have :)

Answer 34

but that is the same as

Answer 35

Sin^2-sincos+sincos-cos^2.

Sincos cancels, leaving sin^2-cos^2

Answer 36

so shouldnt it be sin^2-cos^2?

Answer 37

Yes, they are the same, sin^2x- cos^2x = (sinx+cosx)(sinx-cosx)
But!!!!
(sinx+cosx)(sinx-cosx) is the factorised form, this is what required in the question...
while sin^2x - cos^2x is not

Answer 38

oh, ok

Answer 39

can you help me with a quick one? i don't understand how this one is done...

Answer 40

Hmm... can I eat my orange first?
I'm not a good multitasker...

Answer 41

yes of course hahaha

Answer 42

i will help you practice multitask :)