How do you simplify the expression $\frac{{cos}^{2} (x) + cos (x) - 20}{{cos}^{2} (x) - 16}$ $(cos^2(x)+cos(x)-20)/(cos^2(x)-16)$ ?

Question

1683 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-28T05:17:06

$\frac{{cos}^{2} x + cos x - 20}{{cos}^{2} x - 16} = 1 + \frac{1}{cos x + 4}$ $(cos^2x+cosx-20)/(cos^2x-16)=1+1/(cosx+4)$

$\frac{{cos}^{2} x + cos x - 20}{{cos}^{2} x - 16}$ $(cos^2x+cosx-20)/(cos^2x-16)$

= $\frac{{cos}^{2} x + 5 cos x - 4 cos x - 20}{(cos x - 4) (cos x + 4)}$ $(cos^2x+5cosx-4cosx-20)/((cosx-4)(cosx+4))$

= $\frac{cos x (cos x + 5) - 4 (cos x + 5)}{(cos x - 4) (cos x + 4)}$ $(cosx(cosx+5)-4(cosx+5))/((cosx-4)(cosx+4))$

= $\frac{(cos x - 4) (cos x + 5)}{(cos x - 4) (cos x + 4)}$ $((cosx-4)(cosx+5))/((cosx-4)(cosx+4))$

= $\frac{cos x + 5}{cos x + 4}$ $(cosx+5)/(cosx+4)$

= $1 + \frac{1}{cos x + 4}$ $1+1/(cosx+4)$

How do you simplify the expression (cos^2(x)+cos(x)-20)/(cos^2(x)-16)cos2(x)+cos(x)−20cos2(x)−16(cos^2(x)+cos(x)-20)/(cos^2(x)-16)?