Choose the correct statement.

Question

Suppose $x,y\in(0,\frac{\pi}{2})$ and $x\neq y$ then which of the following statements are true?

A)$2sin(x+y)<sin2x + sin2y$ for all $x,y$

B)$2sin(x+y)>sin2x + sin2y$ for all $x,y$

C)There exist $x,y$ such that $2sin(x+y)=sin2x + sin2y$

D)None of these.

Based on the given problem I can only derive that expanding any of the equation we get option C as correct for $x=y$ but we cannot use it as $x\neq y$. So how should I solve the given problem? Thank you.

Answer 1

Consider the auxiliar function $$f(x,y)=2\sin(x+y)-\sin(2x)-\sin(2y).$$

Using the properties $$ \sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y), $$ $\sin(2x)=2\sin(x)\cos(x)$ and $\sin(2y)=2\sin(y)\cos(y)$, your auxiliar function simplifies to $$ f(x,y)=2\sin(x)(\cos(y)-\cos(x))+2\cos(y)(\sin(x)-\sin(y)). $$

Corrected: (using @peterwhy tip) So, since $f(x,y)>0$ for $0<y<x<\pi/2$ and $f(y,x)=f(x,y)$, we prove that that $f(x,y)>0$ for $x\neq y$ so that B is indeed the right statement.

Disclaimer: The condition $x,y\in (0,\pi/2)$ assures that $\cos(x)$ and $\sin(y)$ are positive quantities).

Answer 2

Expanding the LHS $\sin (x+y)$ and the RHS $\sin 2x + \sin 2y$,

$$\begin{align*} & 2\sin(x+y) - \sin 2x - \sin 2y\\ &= 2\sin x \cos y + 2\cos x \sin y - 2\sin x \cos x - 2\sin y \cos y\\ &= 2\sin x (\cos y - \cos x) + 2 \sin y (\cos x - \cos y)\\ &= 2(\sin x - \sin y) (\cos y - \cos x)\\ &> 0 \end{align*}$$

The final inequality is because either

• $0< x < y <\pi/2$, then $\sin x < \sin y$ and $\cos y < \cos x$; or
• $0< y < x <\pi/2$, then $\sin x > \sin y$ and $\cos y > \cos x$.