Question

Prove that:
sin^8 x - cos^8x = (sin^2(x) - cos^2(x))(1- 2sin^2(x)cos^2(x))

Answer 1

@baru

Answer 2

do I take the square of sin and cos then cube the whole bracket?

Answer 3

\[[(\sin^2(x) - \cos^2(x)]^3\]

Answer 4

no,

do you know this identity\[a^2-b^2=(a+b)(a-b)\]
?

Answer 5

lets work on the left side of the equaiton
\[\sin^8(x) - \cos^8(x)\\=(\sin^4(x))^2-(\cos^4(x))^2\]

ok?

Answer 6

Okay.?

Answer 7

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

ok? ( i used the identity i mentioned in my first post)

Answer 8

okay.

Answer 9

i will use the same identity again and simplify the second term
\[(\sin^4(x)+\cos^4(x))\times \color{red}{(\sin^4(x)-\cos^4(x))}\\(\sin^4(x)+\cos^4(x))\times \color{red}{((\sin^2(x))^2-(\cos^2(x))^2)}\\(\sin^4(x)+\cos^4(x))\times \color{red}{(\sin^2(x)+\cos^2(x))(\sin^2(x)-\cos^2(x))}\]

Answer 10

followed?

Answer 11

yes..

Answer 12

sin^2 x + cos^2 x is 1..?

Answer 13

we know the identity
\[\sin^1(x)+\cos^2(x)=1\]

so the above expression becomes
\[(\sin^4(x)+\cos^4(x)) \times \color{red}{1 \times(\sin^2(x)-\cos^2(x))}\]

Answer 14

yes :)

Answer 15

what next?

Answer 16

do I square root the equation?

Answer 17

no,
now is the tricky part
do you know the property
\[(a-b)^2=a^2+b^2-2ab\]
?

Answer 18

Yes.

Answer 19

we have
\[(\sin^4(x)+\cos^4(x)) \times(\sin^2(x)-\cos^2(x))\]

add and subtract as shown
\[(\sin^4(x)+\cos^4(x) \color{red}{-2\sin^2(x)\cos^2(x)+2\sin^2(x)\cos^2(x)}) \times(\sin^2(x)-\cos^2(x))\]

ok?

Answer 20

I didn't get that. :(

Answer 21

we have this
\[(\sin^4(x)+\cos^4(x)) \times(\sin^2(x)-\cos^2(x))\]

look at the term i mark in red,\[(\color{red}{\sin^4(x)+\cos^4(x)}) \times(\sin^2(x)-\cos^2(x))\]

i will add and subtract \(\color{blue}{2sin^2(x)cos^2(x)}\)

\[(\color{red}{\sin^4(x)+\cos^4(x)} \color{blue}{-2\sin^2(x)\cos^2(x)+2\sin^2(x)\cos^2(x)}) \times(\sin^2(x)-\cos^2(x))\]

ok?

Answer 22

How'd you get 2sin^2(x)cos^2(x)?

Answer 23

if you add and subtract the same number to an equation, it remains unchanged.

ex:
2x + 3y = 8

add and subtract 11 on the left

2x + 3y +11 - 11 =8

the equation is unchanged because +11-11 =0

Answer 24

if you are wondering why specifically 2cos^2sin^2, you will see the reason in the following steps

Answer 25

ok?

Answer 26

Okay?

Answer 27

\[(\color{red}{\ \color{green}{[[}\sin^4(x)+\cos^4(x)} \color{blue}{+2\sin^2(x)\cos^2(x)\color{green}{]]}-2\sin^2(x)\cos^2(x)}) \times(\sin^2(x)-\cos^2(x))\]

look at the terms i put inside the green brackets [[ ]]

they are in the form
\[a^2+b^2+2ab\]

where
\[a=\sin^2(x)\\b=\cos^2(x)\]

so we can write it as
\[(\color{red}{\ \color{green}{[[}\sin^2(x)+\cos^2(x)} \color{green}{]]^2} \color{blue}{-2\sin^2(x)\cos^2(x)}) \times(\sin^2(x)-\cos^2(x))\]

Answer 28

ok?

Answer 29

uh,okay?

Answer 30

we know sin^2+cos^2=1

so this simplifies to
\[(\color{red}{1} \color{blue}{-2\sin^2(x)\cos^2(x)}) \times(\sin^2(x)-\cos^2(x))\]

which is the expression on the right side

so
LHS = RHS