Showing that if $\sin x +\sin y\ge\sqrt{3}$ then $\cos x +\cos y\le 1$

Question

Let $x$ and $y$ be the measures of two acute angles, such that: $$\sin x +\sin y\ge\sqrt{3},$$ the problem is to show that: $$\cos x +\cos y\le 1$$ My approach: I've tried squaring both sides of the given inequality, and I got $\sin^2 x + \sin^2 y + 2\sin x\sin y\ge 3$, then I substituted in $1-\cos^2 x$ and $1-\cos^2 y$ to obtain: $$1-\cos^2 x +1-\cos^2 y+2\sin x\sin y\ge 3$$ But then, I got stuck and couldn't keep any further. Any hint or help is much appreciated.

Answer 1

You've made a good start. Continuing from your last statement we get

$$\begin{equation}\begin{aligned} 1 - \cos^2 x + 1 - \cos^2 y + 2\sin x\sin y & \ge 3 \\ - \cos^2 x - \cos^2 y & \ge 1 - 2\sin x\sin y \\ \cos^2 x + \cos^2 y & \le 2\sin x\sin y - 1 \\ \cos^2 x + 2\cos x\cos y + \cos^2 y & \le 2\cos x\cos y + 2\sin x\sin y - 1 \\ (\cos x + \cos y)^2 & \le 2\cos(x - y) - 1 \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

The last part is using, from the Angle sum and difference identities section of Wikipedia's "List of trigonometric identities" article, that $\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta$ to get

$$2(\cos x\cos y + \sin x\sin y) = 2\cos(x - y) \tag{2}\label{eq2A}$$

Since $\cos(x-y) \le 1 \;\to\; 2\cos(x-y) \le 2 \;\to\; 2\cos(x-y) - 1 \le 1$, then because $(\cos x + \cos y)^2 \ge 0$, we can use this in \eqref{eq1A} and take the square root of both sides to get

$$\cos x + \cos y \le 1 \tag{3}\label{eq3A}$$

Answer 2

$1-\cos^2 x +1-\cos^2 y+2\sin x\sin y\ge 3 \\ \implies -\cos^2 x -\cos^2 y+2\sin x\sin y\ge 1 \\ \implies -\cos^2 x -\cos^2y -2 \cos x \cos y + 2(\cos x \cos y + \sin x\sin y)\ge 1 \\ \implies -(\cos x + \cos y)^{2} + 2 \cos (x - y)\ge 1$.

As we know that $2\cos (x-y) \leq 2 \\ \implies 1 \leq -(\cos x + \cos y)^{2} + 2 \cos (x - y) \leq 2 - (\cos x + \cos y)^{2} \\ \implies (\cos x + \cos y)^{2} \leq 1 \\ \implies -1 \leq(\cos x + \cos y) \leq 1$.

Which proves your result.

Answer 3

Square both $\cos x+\cos y$ and $\sin x+\sin y$ and add. Plug in

$\cos^2x+\sin^2x=1$

$\cos^2y+\sin^2y=1$

$\cos x\cos y+\sin x\sin y\le1$

(What identity proves the last inequality?)

You can then prove that

$(\cos x+\cos y)^2+(\sin x+\sin y)^2\le4$

and finish.

Answer 4

Alternative approach:

Let $x$ and $y$ be the measures of two acute angles, such that: $$\sin x +\sin y\ge\sqrt{3},$$ the problem is to show that: $$\cos x +\cos y\le 1$$

$~\underline{\text{Preliminary Results}}$

From the "Sum and difference formulas" section of trigonometric functions, you have that

• $\sin(a + b) = \sin(a)\cos(b) + \sin(b)\cos(a).$
  $\sin(a - b) = \sin(a)\cos(b) - \sin(b)\cos(a).$

• $\cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b).$
  $\cos(a - b) = \cos(a)\cos(b) + \sin(a)\sin(b).$
  

From these formulas, you can deduce that

• $\sin(a+b) + \sin(a-b) = 2\sin(a)\cos(b).$

• $\cos(a+b) + \cos(a-b) = 2\cos(a)\cos(b).$

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Now set:

$$x = a+b, ~~y = a-b \implies a = \frac{x+y}{2}, ~~b = \frac{x-y}{2}. \tag1 $$

Given the specifications for $~a~$ and $~b~$ in (1) above, since you are given that $~0^\circ < x,y < 90^\circ,~$ you can deduce that

• $90^\circ > a > b > -45^\circ \implies \cos(b) > 0.$

• $a > 0^\circ.$

Further, from the preliminary results, the problem has been transformed as follows:

Premise: $~\displaystyle 2\sin(a)\cos(b) \geq \sqrt{3}.$

Conclusion : $~\displaystyle 2\cos(a)\cos(b) \leq 1.$

From the Premise, since $~\cos(b) \leq 1,~$ you have that $~\displaystyle \sin(a) \geq \frac{\sqrt{3}}{2}.$

This implies that $~\displaystyle \cos(a) \leq \sqrt{1 - \frac{3}{4}} = \frac{1}{2}.$

This result implies the Conclusion.

Answer 5

There is a nice geometric approach to this problem that also generalizes nicely. Observe the following diagram below:

Observe that $A = \cos x + \cos y$ and $B = \sin x + \sin y$, so we have to show that $A\leq 1$ when $B\geq \sqrt{3}$. We know that $A = \sqrt{C^2-B^2}$, which is maximized when $C$ is as large as possible and $B$ is as small as possible. But we know that $B \geq \sqrt{3}$ by assumption and that $C \leq 1+1=2$ by the triangle inequality, so $A = \sqrt{C^2-B^2} \leq \sqrt{2^2-\sqrt{3}^2}=1$, as was to be shown.

Answer 6

Here is a geometrical interpretation. First, to avoid confusion I'll relabel $x,y\to \alpha,\beta$. Now let $A=(\cos \alpha,\sin\alpha)$, $B=(\cos\beta,\sin\beta)$. Since both angles are acute, both points lie on the unit quarter-circle in the first quadrant. If $C=(A+B)/2$, i.e., $C$ is the midpoint of $AB$, then the constraint specifies $$y(C)=\frac12[y(A)+y(B)]\geq \frac{\sqrt{3}}{2}.$$ This yields the diagram

whence it is "evident" that $$x(C)=\frac12[x(A)+x(B)]\leq \frac12$$ which is the desired conclusion. To this into a proof, one should verify that $A,B$ on the unit circle implies $C$ lies within the unit circle: $$x(A)^2+y(A)^2=1,\; x(B)^2+y(B)^2 =1\implies x(C)^2+y(C)^2\leq 1$$ Since $A,B$ are in the first quadrant, it follows that $$x(C)\leq \sqrt{1-y(C)^2}\leq \sqrt{1-3/4}=\frac12.$$

Answer 7

Yet another way, using $\color{green}{\text{Cauchy Schwarz inequality}}$: $$2(2-\cos^2x-\cos^2y)=(1+1)\cdot(\sin^2x+\sin^2y)\color{green}{\geqslant} 3$$ $$\implies 1\geqslant 2(\cos^2x+\cos^2y) \color{green}{\geqslant} (\cos x+\cos y)^2$$