Question

Help with verifying identities

Answer 1

\[1-\frac{ \cos^2x }{ 1+sinx }=sinx\]

Answer 2

\[1-\frac{ 1-\sin^2x }{ 1+sinx }=sinx\]

Answer 3

\[-\sin(x)=\sin(x)\]

Answer 4

does that look right?

Answer 5

@phi

Answer 6

Here's how I would go about this problem.
\[1-\cos^2x=\sin^2x\]
Then, you would have

Answer 7

\[\frac{ \sin^2x }{ 1+\sin(x) }=\sin(x)\]
Cross multiply.

Answer 8

\[\sin^2x=sinx+\sin^2x\]
So your just left with sin(x)

Answer 9

@dtan5457 recheck, it is not that

Answer 10

DarkBlue, how did you make that big step?
I would have factor 1- sin^2 (it's a difference of squares)

Answer 11

@DarkBlueChocobo let the 1 in the front aside, just calculate the second term:
\(\dfrac{cos^2x}{1+sinx}=\dfrac{cos^2x(1-sinx)}{(1+sinx)(1-sinx)}=\dfrac{cos^2(1-sinx)}{cos^2x}= 1-sinx\)
now combine with the 1 in the front you have LHS =RHS

Answer 12

So don't work out cos^2x?

Answer 13

I cancel it out, it disappears, no need to worry about it, :)

Answer 14

one thing I am confused about is where you got 1-sinx in the denominator