When I research my discovered my generalization of the Simson line:
I found the Identity trigonometric as follows:
$$ \cos^2 \left( x \right) = 2\cos\left( x \right) \cos\left( y\right) \cos\left( x+y+\frac{\pi}{3} \right) + \sin^2\left( x+y \right) + 2\sin\left( x+y \right) \cos\left(y \right) \sin\left( \frac{\pi}{3} - x \right) $$
Let $y=0$ the identity become nice identity as follows:
$$\cos^2(x)-\sin^2(x)=2\cos(x)\cos\left(x+\frac{\pi}{3}\right)+2\sin(x)\sin\left(\frac{\pi}{3}-x\right)$$
My question: Is the identity new, and how can it prove?
To prove the identity which only involves $x$, we can note that \begin{align*} \cos \left( x + \frac\pi3 \right) &= \cos x \cos \frac{\pi}3 - \sin x \sin \frac{\pi}3 = \frac12 \cos x - \frac{\sqrt3}2 \sin x \\ \sin \left( \frac\pi3 - x \right) &= \sin \frac\pi3 \cos x - \cos \frac\pi3 \sin x = \frac{\sqrt3}2 \cos x - \frac12 \sin x \end{align*} This gives us \begin{align*} 2 \cos x \cos \left( x + \frac\pi3 \right) &= \cos^2 x - \sqrt{3} \cos x \sin x \\ 2 \sin x \sin \left( \frac \pi3 - x \right) &= \sqrt3 \cos x \sin x - \sin^2 x \end{align*} Adding these two together yields $\cos^2 x - \sin^2 x$, which is the left hand side of the equation.
For another proof, we can observe that \begin{align*} 1 &= 2 \cos \frac\pi 3 = 2 \cos \left( x + \frac\pi3 -x \right) = 2 \cos x \cos \left( \frac\pi3 - x\right) - 2 \sin x \sin \left( \frac\pi3 - x \right) \\ 1 &= 2 \cos \left( -\frac\pi3 \right) = 2 \cos \left( x + -\frac\pi3 - x \right) = 2 \cos x \cos \left( -\frac\pi3 - x \right) - 2 \sin x \sin \left( - \frac\pi3 - x \right) \\ &= 2 \cos x \cos \left( x + \frac\pi3 \right) + 2 \sin x \sin \left( x + \frac\pi3 \right) \end{align*} Subtracting the first equation from the second and rearranging gives us $$2 \cos x \cos \left( x + \frac\pi3 \right)+2 \sin x \sin \left(\frac\pi3 - x \right) = 2 \cos x \cos \left( \frac\pi3 - x\right) - 2 \sin x \sin \left( x + \frac\pi3 \right)$$ Adding up the two sides of this equation gives \begin{align*} &2 \cos \left( 2x + \frac\pi3 \right) + 2 \cos \left( 2x - \frac\pi 3 \right) \\&= 2 \cos 2x \cdot \frac12 - 2 \sin 2x \cdot \frac{\sqrt3}2 + 2 \cos 2x \cdot \frac12 + 2 \sin 2x \cdot \frac{\sqrt3}2 \\ &= 2 \cos 2x \end{align*} Since the two sides of the equation above are equal, they must each be equal to half of their sum, whence $$2 \cos x \cos \left( x + \frac\pi3 \right) + 2 \sin x \sin \left( \frac\pi3 - x \right) = \cos 2x = \cos^2 x - \sin^2 x$$