$\sin\alpha+\cos\alpha<2$

Question

$$\sin\alpha+\cos\alpha<2$$ I would prove this question with using Calculus. But is there any algebra or geometry method to prove it? I'm seeking a proof wihout using Calculus. Thanks in advance.

Answer 1

$$\sin\alpha+\cos\alpha$$ $$=\sqrt2\left(\cos\frac\pi4\sin\alpha+\sin\frac\pi4\cos\alpha\right)$$ $$=\sqrt2\sin\left(\alpha+\frac\pi4\right)\le\sqrt2<2$$

Answer 2

Since both $\sin\alpha$ and $\cos\alpha$ are smaller than or equal to $1$, their sum is smaller than or equal to $2$. And they can't be both equal to $1$.

Answer 3

Since $\sin(x)$ and $\cos(x)$ are both bounded above by $1$, the only way we can have $\sin(a) + \cos(a) = 2$ is if $\sin(a) = \cos(a) = 1$ for some $a \in \mathbb{R}$.

Is this possible?

Answer 4

we assume that $$\sin(x)+\cos(x)\geq 2$$ after squaring we get $$\sin(x)^2+\cos(x)^2+2\sin(x)\cos(x)\geq 4$$ and from here we get $$2\sin(x)\cos(x)\geq 3$$ or $$\sin(2x)\geq 3$$ which is impossible.

Answer 5

$f(x)=\sin(x)+\cos(x)$

$f$ has an extremum when $f'(x)=\cos(x)-\sin(x)=0\implies \tan(x)=1$

So $x_m=\frac{\pi}{4}+k\pi$.

$f(x_m)=\pm2\sin({\pi}4)=\pm\sqrt{2}$

This is a maximum for $k=0$ and a minimum for $k=1$.

Answer 6

Let's have a rectangle with dimensions $x,y$ and fixed diagonal $d$.

$\mathcal A=xy\implies \vec\nabla A=(y,x)$

Let's maximize the area under the constraint $\phi(x,y)=x^2+y^2-d^2=0\implies \vec\nabla\phi=(2x,2y)$

This happens when the two gradients are parallel: $\begin{cases}y=2\alpha x\\x=2\alpha y\end{cases}\implies \begin{cases}\alpha=\frac 12\\x=y\end{cases}$

Thus the maximum area rectangle of fixed diagonal is the square.

Now in the unit circle, let's consider the rectangle of diagonal$=Radius=1$ and sides $x=\cos(\theta)$ and $y=\sin(\theta)$, its area is maximum when $x=y$ that is $\theta=\frac{\pi}{4}$.

$\mathcal A_{max}=xy=x^2=\cos(\frac{\pi}4)^2=\frac 12$

But $x+y=\sqrt{x^2+2xy+y^2}=\sqrt{1+2xy}=\sqrt{1+2\mathcal A}$

So the sum is maximum when the area is maximum and its value is $\sqrt{1+2\times\frac 12}=\sqrt{2}$.