# How to find the maximum of the value $\frac{\sin{x}+1}{\sqrt{3+2\cos{x}+\sin{x}}}$

Find the maximun of the value $$f(x)=\dfrac{\sin{x}+1}{\sqrt{3+2\cos{x}+\sin{x}}}$$

I use wolframpha this found this maximum is $$\dfrac{4\sqrt{2}}{5}$$,But How to prove and how to find this value?(without derivative)

idea 1 let $$\tan{\dfrac{x}{2}}=t$$,then we have $$f=\dfrac{(t+1)^2}{\sqrt{t^4+2t^3+6t^2+2t+5}}$$ Therefore,it suffices to prove that $$\dfrac{(t+1)^4}{t^4+2t^3+6t^2+2t+5}\le\dfrac{32}{25}$$ or$$32(t^4+2t^3+6t^2+2t+5)-25(t+1)^4\ge 0$$ or $$(t-3)^2(7t^2+6t+15)\ge 0$$

But this method if we without derivative,we don't known the maximum is $$\dfrac{4\sqrt{2}}{5}$$.

idea 2 $$f(x)=\dfrac{\sin{x}+1}{\sqrt{(\sin{x}+1)+2(\cos{x}+1)}}$$ Let $$u=\sin{x}+1,v=\cos{x}+1$$,then $$(u-1)^2+(v-1)^2=1$$,find the maximum of the $$\dfrac{u}{\sqrt{u+2v}}$$

• have you studied the monotonicity of $f$? i.e. $f'(x),f''(x)$? May 12, 2017 at 12:45
• Why explicitly state 'without derivative' ? May 12, 2017 at 12:53
• because this problem on the book exicsie problem(before the derivative title) May 12, 2017 at 12:55
• Could you please tell the name of the book? May 12, 2017 at 12:55
• hello,@ABcDexter,this book not english book, May 12, 2017 at 12:57

Change variables so that $\sin(x)$ is $x$ and $\cos(x)$ is $\sqrt{1-x^2}$. The range of $x$ is $[0,1]$, of course. Square the function to get rid of the radical, so that now your function is: $(1+x)^2/(3+x+2\sqrt{1-x^2})$.

Try to prove by direct manipulation that $(1+x)^2/(3+x+2\sqrt{1-x^2})]\leq1$ in $[0,1]$ , i.e. get all but the radical on one side, then get rid of the radical and arrive at a polynomial inequality. You will be able to factor it and show that it is strictly negative.

We need to minimize $$\dfrac{3+2\cos x+\sin x}{(1+\sin x)^2}$$

Now WLOG let $x=\dfrac\pi2-2y$ to get $$\dfrac{3+2\sin2y+\cos2y}{(1+\cos2y)^2}$$

Using Weierstrass substitution, writing $\tan y=t$

we get $$2f(t)=(t^2+1)(t^2+2t+2)$$

Now use Second derivative test, to find the minimum value of $f(t)$ occurs at $-\dfrac12$

i.e., $$f(t)\le\dfrac{25}{32}$$

Hint:

Assume that $f(x)\leq 4\sqrt2/5$

Now cross multiply, do squares on both sides and ultimately if it's the maximum, you will get an inequality where the equality condition can hold, then write the whole sum backwards!

• How would one assume that $f(x)\leq \frac{4\sqrt{2}}{5}$, not knowing the result first? May 12, 2017 at 13:06
• @Mr.Xcoder I think he refers to Wolfram Alpha :) May 12, 2017 at 13:23