1. ## prove trigonometric identities...

$sin^6x + cos^6x = 1-3sin^2x cos^2x$ how can i prove that identity...tks

2. $(\sin^2{x})^3 + (\cos^2{x})^3 =$

$(\sin^2{x} + \cos^2{x})(\sin^4{x} - \sin^2{x}\cos^2{x} + \cos^4{x}) =$

$\sin^4{x} - \sin^2{x}\cos^2{x} + \cos^4{x} =$

$\sin^4{x} + 2\sin^2{x}\cos^2{x} + \cos^4{x} - 3\sin^2{x}\cos^2{x} =$

$(\sin^2{x} + \cos^2{x})^2 - 3\sin^2{x}\cos^2{x} =$

$1 - 3\sin^2{x}\cos^2{x}$

3. $\sin^4{x} + 2\sin^2{x}\cos^2{x} + \cos^4{x} - 3\sin^2{x}\cos^2{x} =$

can u explain me what did you do in that step? i didnt get that one...

why didnt you use $\sin^4{x} + 1\sin^2{x}\cos^2{x} + \cos^4{x} - 2\sin^2{x}\cos^2{x} =$
i know that we want that 3 but can we add terms just like that?

4. $\sin^4{x} + 2\sin^2{x}\cos^2{x} + \cos^4{x} - 3\sin^2{x}\cos^2{x} =$

can u explain me what did you do in that step? i didnt get that one...

I needed the middle term $+ 2\sin^2{x}\cos^2{x}$ so that I could factor the $\sin^4{x} + \cos^4{x}$ ... to compensate adding it, I also had to also subtract it, resulting in the $-3\sin^2{x}\cos^2{x}$ term.

why didnt you use $\sin^4{x} + 1\sin^2{x}\cos^2{x} + \cos^4{x} - 2\sin^2{x}\cos^2{x} =$

why do that? it won't factor.

i know that we want that 3 but can we add terms just like that?

sure can ... isn't $a^2 - ab + b^2 = a^2 + 2ab + b^2 - 3ab$ ?

5. Originally Posted by skeeter
$\sin^4{x} + 2\sin^2{x}\cos^2{x} + \cos^4{x} - 3\sin^2{x}\cos^2{x} =$

can u explain me what did you do in that step? i didnt get that one...

I needed the middle term $+ 2\sin^2{x}\cos^2{x}$ so that I could factor the $\sin^4{x} + \cos^4{x}$ ... to compensate adding it, I also had to also subtract it, resulting in the $-3\sin^2{x}\cos^2{x}$ term.

why didnt you use $\sin^4{x} + 1\sin^2{x}\cos^2{x} + \cos^4{x} - 2\sin^2{x}\cos^2{x} =$

why do that? it won't factor.

i know that we want that 3 but can we add terms just like that?

sure can ... isn't $a^2 - ab + b^2 = a^2 + 2ab + b^2 - 3ab$ ?