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 \([0,\frac π2]\).