yes the left hand side cancels to 2, so it is clearly not an identity.

This post was edited by Rocinante on Dec 12 2013 06:09am

[cosx + sinx]/cosx + [cosx-sinx]/sinx = cscxsecx

Oops! The left side isn't just over cosx, It's over sinx and cosx I rewrote it. Although I did the method you guys recommended and got the Left = Right.

I just wrote a quiz today and got totally stumped on this one question that involved sum and differences

how do you figure out sin(3Pi/2 - x)

remember:
sin (a + b ) = sin(a)*cos(b) + sin(b)*cos(a)

so:
a = 3pi/2
b = -x

so:
sin(3Pi/2 -x) = sin(3pi/2)*cos(-x) + sin(-x)*cos(3pi/2)

remember:
cos(-x) = cos(x)
sin(-x) = -sin(x)

so:
sin(3Pi/2 - x) = sin(3Pi/2)*cos(x) + -1*sin(x)*cos(3Pi/2)

also remember/lookup:
sin(3pi/2) = -1
cos(3pi/2)=0

so
sin(3Pi/2 - x) = -1*cos(x) + -1*sin(x)*0

so:
sin(3Pi/2 - x) = -cos(x)

This post was edited by Rocinante on Dec 12 2013 01:11pm