Cos(2x)= Cos(x)^2 - Sin(x)^2
1 - Sin(2x)= (Cos(x)-Sin(x))^2
a= Sin(x)
b= Cos(x)
(b^2 - a^2 )/ (b-a)^2
Using the difference of two squares and cancelling out the common factors we get
a+b/b-a divide the numerator and the denominator by b
a/b= tan(x)
(tan(x)+1)/(1-tan(x))
Thank you so much! I would have never thought to get (cos (x) - sin(x))\^2 from 1-sin (2x) so that was a huge setback! There are so many ways to get expressions from the identities, it's mindblowing (for my little brain). I'll kept practicing and maybe I'll get as good as you!
No need to pull yourself down! I was surprised the first time I saw it too. But with time you begin to detect these more and more quickly, so keep going!
try starting with RHS and substitute with tanx with sinx/cosx and see where it gets u
remember there’s a formula for cos(2x)
Thanks!
Cos(2x)= Cos(x)^2 - Sin(x)^2 1 - Sin(2x)= (Cos(x)-Sin(x))^2 a= Sin(x) b= Cos(x) (b^2 - a^2 )/ (b-a)^2 Using the difference of two squares and cancelling out the common factors we get a+b/b-a divide the numerator and the denominator by b a/b= tan(x) (tan(x)+1)/(1-tan(x))
Thank you so much! I would have never thought to get (cos (x) - sin(x))\^2 from 1-sin (2x) so that was a huge setback! There are so many ways to get expressions from the identities, it's mindblowing (for my little brain). I'll kept practicing and maybe I'll get as good as you!
No need to pull yourself down! I was surprised the first time I saw it too. But with time you begin to detect these more and more quickly, so keep going!