Question

prove the identity cos³2A+3cos2A=4(cos^6A-sin^6)

Answer 1

Prove the identity $\cos^3 2A + 3\cos 2A = 4(\cos^6 A - \sin^6 A)$

Solution.

Lets start with the right side. Use the power-reduction formula:

$$\cos^6 A = \left(\frac{1 + \cos 2A}{2}\right)^3 = \frac{1}{8} (1 + 3 \cos 2A + 3 \cos^2 2A + \cos^3 2A)$$ $$\sin^6 A = \left(\frac{1 - \cos 2A}{2}\right)^3 = \frac{1}{8} (1 - 3 \cos 2A + 3 \cos^2 2A - \cos^3 2A)$$

Then we add these expressions:

$$4 \left(\cos^6 A - \sin^6 A\right) = \frac{1}{2} (1 + 3 \cos 2A + 3 \cos^2 2A + \cos^3 2A - 1 + 3 \cos 2A - 3 \cos^2 2A + \cos^3 2A)$$

And finally simplify the expressions:

$$\frac{1}{2} (1 + 3 \cos 2A + 3 \cos^2 2A + \cos^3 2A - 1 + 3 \cos 2A - 3 \cos^2 2A + \cos^3 2A) = \frac{1}{2} (6 \cos 2A + 2 \cos^3 2A) = 3 \cos 2A + \cos^3 2A$$

Check the left and right sides:

$$\cos^3 2A + 3 \cos 2A = \cos^3 2A + 3 \cos 2A$$

We have the identity.