MHB Is this Trigonometric Expression a Constant Function of x?

[anemone]

Prove $\sin^2(x+a)+\sin^2(x+b)-2\cos (a-b)\sin (x+a)\sin (x+b)$ is a constant function of $x$.

Spoiler

It is not at all a hard challenge(Emo), but I am amazed at the beauty of this problem therefore the sharing of it with MHB's members.

 

[lfdahl]

Spoiler: Suggested solution

With $\alpha = x +a$ and $\beta = x + b$ the given expression reads:

\[\sin^2\alpha + \sin^2\beta - 2\cos (\alpha -\beta )\sin \alpha \sin \beta \\\\ =\sin^2\alpha + \sin^2\beta - 2 \left (\cos \alpha \cos \beta + \sin \alpha \sin \beta \right )\sin \alpha \sin \beta \\\\ =\sin^2\alpha + \sin^2\beta - 2 \sin^2 \alpha \sin^2 \beta - 2\cos \alpha \cos\beta \sin \alpha \sin \beta \\\\ =\sin^2\alpha - \sin^2 \alpha \sin^2 \beta+ \sin^2\beta - \sin^2 \alpha \sin^2 \beta - 2\cos \alpha \cos\beta \sin \alpha \sin \beta\\\\=\sin^2 \alpha(1-\sin^2 \beta) + \sin^2 \beta (1-\sin^2 \alpha )- 2\cos \alpha \cos\beta \sin \alpha \sin \beta \\\\ =\sin^2\alpha \cos^2\beta + \sin^2 \beta \cos^2 \alpha - 2\cos \alpha \cos\beta \sin \alpha \sin \beta \\\\ = \sin^2\alpha \cos^2\beta - \sin \alpha \cos \beta \sin \beta \cos \alpha+ \sin^2 \beta \cos^2 \alpha - \sin \beta \cos \alpha \sin \alpha \cos \beta \\\\ = \sin \alpha \cos \beta(\sin \alpha \cos \beta- \sin \beta\cos \alpha)+\sin \beta \cos \alpha(\sin \beta \cos \alpha-\sin \alpha \cos \beta)\\\\ =\sin \alpha \cos \beta(\sin \alpha \cos \beta- \sin \beta\cos \alpha)+\sin \beta \cos \alpha(\sin \beta \cos \alpha-\sin \alpha \cos \beta)\\\\ =\sin \alpha \cos \beta \sin(\alpha -\beta )- \sin(\alpha -\beta )\sin \beta \cos \alpha \\\\ =\sin (\alpha -\beta )(\sin \alpha \cos \beta - \sin \beta \cos \alpha )\\\\ = \sin^2(\alpha -\beta )\\\\ = \sin^2(a-b)\]

• which is independent of $x$.

 

[anemone]

Spoiler: Solution of other:

[TIKZ]\draw[thin] (0,0) circle (5cm);
\coordinate[label=left:A] (A) at (-5,0);
\coordinate[label=right:E] (E) at (5,0);
\coordinate[label=below:O] (O) at (0,0);
\draw (A) -- (O)-- (E);
\coordinate[label=left: D] (D) at (-4.33,2.5);
\coordinate[label=left: C] (C) at (-2.5,4.33);
\draw (A) -- (D)-- (E);
\draw (A) -- (C)-- (E);
\draw (D)-- (C);
\coordinate[label=right: B] (B) at (4.1,2.87);
\draw (A)-- (B);
\node (1) [font=\small,blue] at (-3.6,0.26) {$x$};
\node (2) [font=\small,orange] at (-4.3,0.65) {$\beta$};
\node (3) [font=\small,purple] at (-4.2,1.5) {$\alpha - \beta$};
\node (4) [font=\small,purple] at (3.0,1.1) {$\alpha - \beta$};
\node (5) [font=\small,teal] at (-3.8,1.0) {$\alpha$};
\begin{scope}
\path[clip] (E) -- (A) -- (B) -- cycle;
\draw[thick,blue] (A) circle (0.98);
\end{scope}
\begin{scope}
\path[clip] (B) -- (A) -- (C) -- cycle;
\draw[thick,orange] (A) circle (0.8);
\end{scope}
\begin{scope}
\path[clip] (B) -- (A) -- (D) -- cycle;
\draw[thick,teal] (A) circle (1.2);
\end{scope}
\begin{scope}
\path[clip] (C) -- (A) -- (D) -- cycle;
\draw[thick,purple,double] (A) circle (1.38);
\end{scope}
\begin{scope}
\path[clip] (C) -- (E) -- (D) -- cycle;
\draw[thick,purple,double] (E) circle (1.18);
\end{scope}
[/TIKZ]

Draw a circle with diameter 1 unit and let $O$ be the center of the circle and $AE$ be the diameter of the circle.

Using Sine Rule on the right-angled triangles $EAD$ and $EAC$, we get the following:

$ED=\sin (x+\alpha)\\CE=\sin (x+\beta)$

Now, consider triangle $CED$. Applying the Cosine Rule, we get

$CD^2=ED^2+CE^2-2(ED CE)\cos (\alpha - \beta)=\sin^2 (x+\alpha)+\sin^2 (x+\beta)-2\sin (x+\alpha) \sin (x+\beta)\cos(\alpha-\beta)$

This completes the proof.