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?

Answer 1

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

Answer 2

yes that is a good start

Answer 3

so we have

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

Answer 4

I would next replace the cos^2

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

Thank you both!!!

Answer 13

:-)