The Relationship Between Sines And Cosines

Relevant wiki: Sum and Difference Trigonometric Formulas - Problem Solving

$f(x)=\dfrac{\sin x +2}{\cos x +1}$

Use $\small{\color{teal}{\sin x=2\sin\frac x2\cos \frac x2,~1+\cos x=2\cos^2\frac x2}}$ .

$\begin{aligned}f(x)=&\dfrac{2\sin\frac x2\cos \frac x2}{2\cos^2\frac x2}+\dfrac{2}{2\cos^2\frac x2}\\=&\tan \frac x2+\underbrace{\sec^2\frac x2}_{1+\tan^2\frac x2}\\=&\tan^2\frac x2+\tan \frac x2+1\\=&\left(\tan \frac x2+\dfrac 12\right)^2+\dfrac 34\end{aligned}$

Since square of a real quantity is always non negative.

$(f(x))_{\text{min}}=\dfrac 34 \text{ when } \tan \dfrac x2=\dfrac{-1}{2}$

Let $t=\tan \frac x2$ , then $\begin{cases} \sin x = \dfrac {2t}{1+t^2} \\ \cos x = \dfrac {1-t^2}{1+t^2} \end{cases}$

Now, we have:

$\begin{aligned} \frac {\sin x +2}{\cos x +1} & = \frac {\frac {2t}{1+t^2}+2}{ \frac {1-t^2}{1+t^2}+1} \\ & = \frac {2t+2+2t^2}{1-t^2+1+t^2} \\ & = t^2 + t + 1 \\ & = \color{#3D99F6}{\left(t+\frac 12 \right)^2} + \frac 34 \quad \quad \small \color{#3D99F6}{\text{Since } \left(t+\frac 12 \right)^2 \ge 0} \\ \implies \frac {\sin x +2}{\cos x +1} & \ge \frac 34 = \boxed{0.75} \quad \quad \small \color{#3D99F6}{\text{Minimum occurs when } \tan \frac x2 = - \frac 12} \end{aligned}$

Calculus method:

Let $f(x) = \dfrac{\sin x + 2}{\cos x+1}$

For the expression to be valid, we know that $\cos x + 1 \neq 0 \implies \cos x \neq -1$

Now, at minimum points, we know that $f'(x) = 0$

$f'(x) = \dfrac{(\cos x + 1)(\cos x) - (\sin x + 2)(-\sin x)}{(\cos x + 1)^2}\\ =\dfrac{\cos^2 x + \cos x + \sin^2 x + 2 \sin x}{(\cos x + 1)^2}\\ =\dfrac{1+\cos x + 2\sin x}{(\cos x + 1)^2}$

$f'(x) = 0\\ \dfrac{1 + \cos x + 2 \sin x}{(\cos x + 1)^2}=0\\ 1+\cos x+2\sin x=0\\ 1+\cos x = -2\sin x\\ 1+\cos x = -2\sqrt{1 - \cos^2 x}\\ (1+\cos x)^2 = \left(-2\sqrt{1 - \cos^2 x}\right)^2\\ 1+2\cos x+\cos^2 x = 4(1-\cos^2 x)\\ 1+2\cos x + \cos^2 x - 4 + 4\cos^2 x = 0\\ 5 \cos^2 x + 2 \cos x - 3 = 0\\ (5\cos x -3)(\cos x+ 1)=0\\ \cos x = \dfrac{3}{5},\;\cos x = -1$

Now, we know that the second factor will result in $f(x)$ being undefined, so we ignore it. This means that

$\cos x = \dfrac{3}{5} \implies \sin x = \dfrac{1+\frac{3}{5}}{-2} =- \dfrac{4}{5}$

Therefore, the minimum value of the expression is $\dfrac{-\frac{4}{5}+2}{\frac{3}{5}+1} = \dfrac{\frac{6}{5}}{\frac{8}{5}}=\dfrac{3}{4} = \boxed{0.75}$

Note: The minimum value occurs at $x=(306.87+360n)^{\circ}$ , where $n$ is an integer

Bonus: My solution is actually incomplete. Prove that the point I found is a minimum point. If you used the second derivative test, you should get $f''(x) = \dfrac{5}{32} > 0$

Let $a=\sin x +1$ , $b=\cos x +1$ . Then $a$ and $b$ satisfy $(a-1)^2+(b-1)^2=1$ . We look to minimize $\dfrac{a+1}{b}=k$ . Through manipulation, we note that $a-1=bk-2$ . We have $(a-1)^2+(b-1)^2=(bk-2)^2+(b-1)^2=b^2(k^2+1)-b(4k+2)+5=1$ .

We know that the polynomial $b^2(k^2+1)-b(4k+2)+4=0$ has real roots, therefore its discriminant is greater than or equal to zero. $\begin{aligned}(-(4k+2))^2-4(k^2+1)(4)\geq 0\\16k-12\geq 0\\k \geq \dfrac{3}{4}=\boxed{0.75}\\ \end{aligned}$

Interesting Fact: This problem can share a similar solution to this one.