Question

Prove that $y=\frac{ 4sinθ}{(2+cosθ)}-θ$is an increasing function of $θ$ in $[0,\frac π2]$.

Answer 1

We have,

y = $\frac {4sinθ}{(2+cosθ)}$ - θ

$\frac {dy}{dx}$ = $\frac {(2+cosθ)(4cosθ)-4sin(-sinθ)}{(2+cosθ)^2} - 1$

= $\frac {8cosθ+4cos^2θ+4sin^2θ}{(2+cos)^2}-1$

= $\frac {8cosθ+4}{(2+cos)^2}-1$

Now,$\frac {dy}{dx}$ = 0.

\implies $$\frac {8cosθ+4}{(2+cosθ)^2}=1

$\implies$8cosθ+4 = 4+cos2θ+4cosθ

$\implies$cos2θ-4cosθ = 0

$\implies$cosθ(cosθ-4) = 0

$\implies$cosθ=0 or cosθ=4

since cosθ≠4, cosθ=0

cosθ=0$\implies$θ=$\frac \pi2$

Now,

$\frac {dy}{dx}$ = $\frac {8cosθ+4-(4+cos^2θ+4cosθ)}{(2+cosθ)^2}$ = $\frac {4cosθ-cos^2θ}{(2+cosθ)^2 }$= $\frac {cosθ(4-cosθ)}{(2+cosθ)^2}$

In interval , we have cos θ > 0. Also, 4>cos θ ⇒ 4−cos θ>0.

$\implies$cos(4-cosθ)>0 and also (2+cosθ)2>0

\implies $$\frac {cosθ(4-cosθ)}{(2+cosθ)^2} > 0

$\frac {dy}{dx}$>0

Therefore, y is strictly increasing in interval $(0,\frac π2)$.

Also, the given function is continuous at x=0 and x=$\fracπ2$.

Hence, y is increasing in interval $[0,\frac π2]$.