Proving ${1+\cos x\over2+\sin x} < \frac43$

Question

How to prove :- $${1+\cos x\over2+\sin x} < \frac43$$

I tried differentiating the equation but i thought there might be better and easier solution.I prefer solution using trigonometry as base.

Answer 1

We want to prove $\frac{1+\cos x}{2+\sin x}\leq \frac{4}{3}$. By multiplying by $3(2+\sin x)$ (note that $2+\sin x> 0$), we get that this is equivalent to $3(1+\cos x)\leq 4(2+\sin x)$ which is equivalent to prove that $3\cos x-4\sin x\leq 5$.

This inequality comes from Cauchy Schwartz. Note that $$(3\cos x-4\sin x)^2\leq(3^2+(-4)^2)(\cos^2x+\sin^2x)= 25$$

Answer 2

HINT

After algebraic simplification it suffices to show that $$3(1+\cos x) \le 4(2+\sin x)$$ which itself is equivalent to showing $$3\cos x - 4\sin x \le 5,$$ which should be a basic exercise in calculus.

Answer 3

If $\cos(\theta)=3/5$ and $\sin(\theta)=4/5$ (which is possible since $3^2 + 4^2 = 5^2$), the inequality (with $\le$ rather than $<$) is equivalent to $\cos(x - \theta) \le 1$.

Answer 4

$$\dfrac{2+\sin2y}{1+\cos2y}=\sec^2y+\tan y=1+\left(\tan y+\dfrac12\right)^2\ge1-\dfrac14$$

using https://en.m.wikipedia.org/wiki/Tangent_half-angle_substitution#The_substitution

More generally for $$f(y)=\dfrac{a+\cos2y}{b+\sin2y}$$

set $f(y)=u$

Now use Weierstrass substitution to form a quadratic equation in $\tan y$

As $\tan y$ is real, the discriminant must be $\ge0$

Answer 5

Let $t$ is variable and we can write $x^2+y^2=1$ as $$x=-\sin t, y=-\cos t$$

And we can rewrite $\frac{1+\cos x}{2+\sin x}$ is a slope of line which pass through two points : $(2, 1), (-\sin x, - \cos x)$

and the point $(-\sin x, -\cos x)$ is on the circle $x^2 + y^2 =1$.

Draw the graph :

We can know $\frac{1+\cos x}{2+\sin x}$ is max / min when the tangent line of the circle $x^2 + y^2 = 1$ pass $(2, 1)$ from the graph.

Thus, $$0\leq\frac{1+\cos x}{2+\sin x}\leq\frac{4}{3}$$