# Show that $5\cos^2x - 2\sqrt{3}\sin x\cos x + 3\sin^2x = \cos 2x - \sqrt{3}\sin2x + 4$

Show that $5\cos^2x - 2\sqrt{3}\sin x\cos x + 3\sin^2x = \cos 2x - \sqrt{3}\sin2x + 4$

Attempt.

I'm preparing for a trigonometry examination tomorrow. So I'm solving as many questions as possible. This is confusing. I've tried solving it but its not proofing. I've tried replacing with trigonometric identities and simplifying. Please someone should help even if it is how to start?

$$\cos^2A+\sin^2 A=1$$ $$2\sin A\cos A=\sin 2A$$ $$\cos^2 A-\sin^2 A=\cos 2A$$
$$LHS=5\cos^2 x-2\sqrt 3\sin x\cdot \cos x+3\sin^2x$$ $$=4\cos^2 x+\cos^2 x-\sqrt 3(2\sin x\cdot \cos x)+4\sin^2x-\sin^2 x$$ $$=4(\cos^2 x+\sin^2x)-\sqrt 3\sin 2x+(\cos^2 x-\sin^2 x)$$ $$=4(1)-\sqrt 3\sin 2x+(\cos 2x)$$ $$=\cos 2x-\sqrt 3\sin 2x+4=RHS$$
The first thing to remember is $$\sin2x=2\sin x\cos x$$ that settles the $-\sqrt{3}\sin2x$ part. Thus you just need to massage $$5\cos^2x+3\sin^2x$$ recalling $\cos2x=\cos^2x-\sin^2x$; so rewrite it as $$\cos^2x+4\cos^2x-\sin^2x+4\sin^2x=\cos2x+4(\cos^2x+\sin^2x)$$ You can also equivalently prove that $$5\cos^2x - 2\sqrt{3}\sin x\cos x + 3\sin^2x - \cos 2x + \sqrt{3}\sin2x - 4=0$$ that goes like \begin{multline} 5\cos^2x+3\sin^2x-\cos^2x+\sin^2x-4-\sqrt{3}(2\sin x\cos x-\sin2x)\\ =4\cos^2x+4\sin^2x-4-\sqrt{3}(\sin2x-\sin2x)=0 \end{multline}