Trig / Pre Cal / identities These things are killing me. Please help guide me. Thank you.

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

Am I on the right path?

Question

Trig / Pre Cal / identities  These things are killing me. Please help guide me. Thank you.

$$ (\sin^4 x-\cos^4 x) = 2\sin^2 x -1 $$
$$ (\sin^2 x+\cos^2 x)(\sin^2 x-\cos^2 x)= 2\sin^2 x -1 $$

Am I on the right path?

nnesha · Answer

apply special identity $$\huge\rm sin^2 x + \cos^2 =1 $$
solve for sin^2

phi · Answer

yes that is a good start

anonymous · Answer

so we have 

$$   (1)(\sin^2 x-\cos^2 x)= 2\sin^2 x -1  $$

phi · Answer

I would next replace the cos^2

nnesha · Answer

sin^2 x + cos ^2 = 1 solve this identity for cos^2

anonymous · Answer

$$  (1)(\sin^2 x + 1-\sin^2 x )= 2\sin^2 x -1  $$

nnesha · Answer

$$ \rm (\sin^2 x-(\color{reD}{{1-sin^2x}}))$$
write that in the parentheses and distribute by negative one

phi · Answer

close, but you lost a minus sign
-(1 - sin^2)

anonymous · Answer

$$ (1)(\sin^2 x - (1-\sin^2 x))= 2\sin^2 x -1 $$  Now what?

phi · Answer

-(a +b)  means -1*(a+b) = -1*a + -1*b

anonymous · Answer

Ah so 
$$ (1)(\sin^2 x - 1+\sin^2 x )= 2\sin^2 x -1 $$ I see it now.

anonymous · Answer

Thank you both!!!

nnesha · Answer

:-)