Re: simplifying expressions

Quote:

Originally Posted by

**Atalante** Simplify:

1. sinx cos2x / cosx sin2x

sinx cos^{2}x - sin^{2}x / 2 sinx cosx

sinx ( cos^{2}x - 1) / 2 sinx cosx

solution: 1 - 1/2 cos^{2}x

I am not sure if what I did is right and I don't know what to do next. May you help me?? Thanks for the answers.(Clapping)

Parenthesis would be of some benefit here.

The numerator...You forgot a multiple of sin(x):

$\displaystyle sin(x) ~ cos(2x) = sin(x) \[cos^2(x) - sin^2(x) \] $

The denominator you also forgot a cos(x):

$\displaystyle cos(x)~sin(2x) = cos(x) \[ 2~sin(x)~cos(x) \] $

-Dan

Re: simplifying expressions

Quote:

Originally Posted by

**Atalante** Simplify:

1. sinx cos2x / cosx sin2x

sinx cos^{2}x - sin^{2}x / 2 sinx cosx

sinx ( cos^{2}x - 1) / 2 sinx cosx

solution: 1 - 1/2 cos^{2}x

I am not sure if what I did is right and I don't know what to do next. May you help me?? Thanks for the answers.(Clapping)

$\displaystyle \displaystyle \begin{align*} \frac{\sin{(x)}\cos{(2x)}}{\cos{(x)}\sin{(2x)}} &= \frac{\sin{(x)}\left[ \cos^2{(x)} - \sin^2{(x)} \right]}{ \cos{(x)} \left[ 2\sin{(x)}\cos{(x)} \right] } \\ &= \frac{\cos^2{(x)} - \sin^2{(x)}}{2\cos^2{(x)}} \\ &= \frac{\cos^2{(x)} - \left[ 1 - \cos^2{(x)} \right]}{2\cos^2{(x)}} \\ &= \frac{2\cos^2{(x)} - 1}{2\cos^2{(x)}} \\ &= 1 - \frac{1}{2\cos^2{(x)}} \end{align*}$