# Prove that $\cos2\theta−\sqrt{3}\sin2\theta \equiv2 \cos (2\theta+\pi/3 )\equiv−2\sin(2\theta−\pi/6)$

Going from $2\cos(2\theta+\pi/3)$ to $\cos2\theta−\sqrt{3}\sin2\theta$ is simple enough, however I'm stuck on going from $2\cos(2\theta+\pi/3)$ to $−2\sin(2\theta−\pi /6)$. How do i do this?

• $\cos w=\sin(\pi/2-w)$ Mar 14, 2013 at 18:22
• We usually just use $=$ for identities, by the way. $\equiv$ is used for something else entirely. Mar 14, 2013 at 18:25

Use the identities for the sine and cosine of the sum or difference of two angles:

\begin{align*} \sin\left(2\theta-\frac{\pi}6\right)&=\sin2\theta\cos\frac{\pi}6-\cos2\theta\sin\frac{\pi}6\\ &=\frac{\sqrt3}2\sin2\theta-\frac12\cos2\theta\;, \end{align*}

and

\begin{align*} \cos\left(2\theta+\frac{\pi}3\right)&=\cos2\theta\cos\frac{\pi}3-\sin2\theta\sin\frac{\pi}3\\&=\frac12\cos2\theta-\frac{\sqrt3}2\sin2\theta\;. \end{align*}

First, let's recap:
$$\sin{\pi\over3}=\frac{\sqrt{3}}{2}$$
$$\cos{\pi\over3}=\frac12$$
$$\sin{\pi\over6}=\frac12$$
$$\cos{\pi\over6}=\frac{\sqrt{3}}{2}$$

Let's rock 'n roll!
Solution 1:
Going from $$\cos2\theta−\sqrt{3}\sin2\theta$$ to $$2\cos\left(2\theta+\frac\pi3\right)$$. \require{cancel} \begin{align} \cos2\theta−\sqrt{3}\sin2\theta&=\cos^2\theta-\sin^2\theta-\sqrt{3}\cdot2\sin\theta\cos\theta\\ &=2\cdot\frac12\left(\cos^2\theta-\sin^2\theta-\sqrt{3}\cdot2\sin\theta\cos\theta\right)\\ &=2\left(\frac12\left(\cos^2\theta-\sin^2\theta\right)-\frac12\sqrt{3}\cdot2\sin\theta\cos\theta\right)\\ &=2\left(\left(\cos^2\theta-\sin^2\theta\right)\cos\frac{\pi}{3}-2\sin\theta\cos\theta\sin\frac\pi3\right)\\ &=2\left(\cos2\theta\cos\frac\pi3-\sin2\theta\sin\frac\pi3\right)\\ &=2\cos\left(2\theta+\frac\pi3\right) \end{align} Solution 2:
Going from $$\cos2\theta−\sqrt{3}\sin2\theta$$ to $$\sin(2\theta−\frac\pi6)$$. \begin{align} \cos2\theta−\sqrt{3}\sin2\theta&=\cos^2\theta-\sin^2\theta-\sqrt{3}\cdot2\sin\theta\cos\theta\\ &=-2\cdot\left(-\frac12\right)\left(\cos^2\theta-\sin^2\theta-\sqrt{3}\cdot2\sin\theta\cos\theta\right)\\ &=-2\left(\left(-\frac12\right)\left(\cos^2\theta-\sin^2\theta\right)-\left(-\frac12\right)\left(\sqrt{3}\cdot2\sin\theta\cos\theta\right)\right)\\ &=-2\left(\left(-\frac12\right)\left(\cos^2\theta-\sin^2\theta\right)-\left(-\frac12\sqrt{3}\right)\left(2\sin\theta\cos\theta\right)\right)\\ &=-2\left(-\sin\frac\pi6\left(\cos^2\theta-\sin^2\theta\right)-\left(-\cos\frac\pi6\right)\left(\sin\theta\cos\theta\right)\right)\\ &=-2\left(-\sin\frac\pi6\left(\cos^2\theta-\sin^2\theta\right)+\cos\frac\pi6\left(2\sin\theta\cos\theta\right)\right)\\ &=-2\left(\cos\frac\pi6\left(2\sin\theta\cos\theta\right)-\sin\frac\pi6\left(\cos^2\theta-\sin^2\theta\right)\right)\\ &=-2\left(\cos\frac\pi6\sin2\theta-\sin\frac\pi6\cos2\theta\right)\\ &=-2\left(\sin2\theta\cos\frac\pi6-\cos2\theta\sin\frac\pi6\right)\\ &=-2\sin\left(2\theta-\frac\pi6\right) \end{align} I hope this helps.