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In $\Delta ABC$, $C = \frac{\pi}{3}$, how to prove that $\cos^2A + \cos^2B = \frac{1}{2} \cos(2A + \frac{\pi}{3}) + 1$?

My attempt: \begin{align} L.H.S &= \cos^2A + \cos^2B \\ &= \cos^2A + \cos^2(A + \frac{\pi}{3}) \\ &= \frac{1 + \cos 2A}{2} + \frac{1 + \cos(2A + \frac{2}{3}\pi)}{2} \\ &= 1 + \frac{1}{2} \cos 2A + \frac{1}{2}(\cos 2A \cos\frac{2\pi}{3} - \sin \frac{2\pi}{3}\sin 2A) \\ &= 1 + \frac{1}{4} \cos 2A - \frac{\sqrt 3}{4} \sin 2A \\ R.H.S &= \frac{1}{2}\cos (2A + \frac{\pi}{3}) + 1 \\ &= \frac{1}{2}(\cos 2A \cos\frac{\pi}{3} - \sin \frac{\pi}3 \sin 2A) + 1 \\ &= 1 + \frac{1}{4}\cos 2A - \frac{\sqrt{3}}{4} \sin 2A \\ &= L.H.S \end{align} I wonder if there is a way to show it more efficiently.

Y.T.
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2 Answers2

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At the step $$\frac {1+\cos 2A}2+\frac{1+\cos(2A+\frac23\pi)}2$$

we can use the identity

$$\cos \alpha+\cos\beta=2\cos\left(\frac{\alpha+\beta}2\right)\cos\left(\frac{\alpha-\beta}2\right)$$

to obtain

$$\begin{align}\frac {1+\cos 2A}2+\frac{1+\cos(2A+\frac23\pi)}2 &= 1+\frac12(\cos2A+\cos(2A+\frac23\pi)) \\&=1+\cos\left(\frac{2A+(2A+\frac23\pi)}2\right)\cos\left(\frac{2A-(2A+\frac23\pi)}2\right) \\&=1+\cos\left(2A+\frac\pi3\right)\cos\left(-\frac\pi3\right) \\&=1+\frac12\cos\left(2A+\frac\pi3\right) \end{align}$$

player3236
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Use Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$

$$\cos^2A+\cos^2B=\cos(A+B)\cos(A-B)+1$$

Now $A+B=\pi-C=?$

$A-B=A-(\pi-A-C)=A-\left(\pi-A-\dfrac\pi3\right)=?$

lab bhattacharjee
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