Question

cos^6A-sin^6A=(cos^2A-sin^2A)*(1-sin^2A*cos^2A)

Answer 1

need help in breaking in simpler terms

Answer 2

power 6 ?? -_-

Answer 3

\(\cos^6A-\sin^6A=(\cos^2A)^3-(\sin^2A)^3\)
\((a^3-b^3)=(a-b)(a^2+ab+b^2)\)
here
\(a=\cos^2A\)
\(b=\sin^2A\)
also
\(\cos^2A+\sin^2A=1\)
Does that help? @koli123able

Answer 4

thanks once again!! :-)

Answer 5

Welcome:)

Answer 6

(cos^2A-sin^2A)(cos^4A+sin^4A+cos^2A*sin^2A)

Answer 7

what to do after this ^^

Answer 8

@ajprincess

Answer 9

How many of these probs you've got? Looks like you can fill the entire sunday with it :/

Answer 10

well I did almost just this one is left I did 53 questions since morning

Answer 11

Wow! Hats off to you!

Answer 12

can u help me with this last one

Answer 13

help plzz

Answer 14

????

Answer 15

OK, I'll give it a try. Give me a few secs :)

Answer 16

\((cos^2A-sin^2A)(cos^4A+sin^4A+cos^2A*sin^2A)\)
\(=(cos^2A-sin^2A)((cos^2A)^2+(sin^2A)^2+cos^2A+sin^2A)\)
\((cos^2A)^2+(sin^2A)^2=1\)

Answer 17

Shouldn't that "+" at the end of the middle line be a dot?

Answer 18

yup u r right:)@ZeHanz. really sorry abt that typo

Answer 19

We have so far: \((\cos^2A-\sin^2A)(\cos^4A+\sin^4A+\cos^2A \cdot \sin^2A)\).
By peeking at the RHS, we see that the first factor is there too. We just need to do something with the second one:

\(\cos^4A+\sin^4A+\cos^2A \cdot \sin^2A\)=
\(\cos^4A+\sin^4A+\cos^2A(1-\cos^2A)\)=
\(\cos^4A+\sin^4A+\cos^2A-\cos^4A\)=
\(\sin^4A+\cos^2A\)=
\((1-\cos^2A)^2+\cos^2A\)=
\(1-2\cos^2A+\cos^4A+\cos^2A\)=
\(1-\cos^2A+\cos^4A\)=
\(1-\cos^2A(1-\cos^2A)\)=
\(1-\cos^2A\sin^2A\)

So if we multiply this with \(\cos^2A-\sin^2A\), we're done!