<s>[INST] Prove that $\cos (5A) = 16 \cos^5 (A) - 20 \cos^3 (A) + 5 \cos (A)$ 
Prove the given trigonometric identity
$$\cos (5A) = 16 \cos^5 (A) - 20 \cos^3 (A) + 5 \cos (A)$$


My attempt
L.H.S.$=\cos5A$
$$\cos(A+4A)$$
$$\cos A\cos4A-\sin A\sin4A$$
Now how should I move further?
 [/INST] $\cos{5a} = \cos{(3a+2a)}$
$=\cos 3a \cos 2a -\sin 3a \sin 2a$
$=\cos{(2a+a)}\cos 2a -\sin{(2a+a)} \sin 2a$
$=(\cos 2a\cos a-\sin2a\sin a)\cos 2a -(\sin2a \cos a + \cos 2a \sin a) \sin 2a$
$=(\cos 2a\cos a-\sin2a\sin a)\cos 2a -(\sin2a \cos a + \cos 2a \sin a) \sin 2a$
$=((\cos^2a-\sin^2a)\cos a-(2\sin a \cos a)\sin a)(\cos^2a-\sin^2a) -((2\sin a \cos a) \cos a + (\cos^2a-\sin^2a) \sin a) (2\sin a\cos a)$
$=(\cos^3a-\sin^2 a\cos a -2\sin^2a\cos a)(\cos^2a-\sin^2a) -(2\sin a \cos^2a + \cos^2a \sin a - \sin^3 a) (2\sin a \cos a)$
$=(\cos^3a - \sin^2 a \cos a - 2 \sin^2a \cos a)(\cos^2a)-(\cos^3a - \sin^2 a \cos a - 2 \sin^2a \cos a)(\sin^2a) -(2\sin a \cos^2a + \cos^2a \sin a - \sin^3 a) (2\sin a \cos a)$
$=(\cos^5a - \sin^2 a \cos^3 a - 2 \sin^2a \cos^3 a)-(\cos^3a \sin^2a - \sin^4 a \cos a - 2 \sin^4a \cos a) - (4\sin^2 a \cos^3a + 2\cos^3a \sin^2 a - 2\sin^4 a \cos a)$
$=(\cos^5a - \sin^2 a \cos^3 a - 2 \sin^2a \cos^3 a)+(-\cos^3a \sin^2a + \sin^4 a \cos a + 2 \sin^4a \cos a) + (-4 \sin^2 a \cos^3a - 2\cos^3a \sin^2 a + 2\sin^4 a \cos a)$
$\bf =\cos^5a - 10 \sin^2a \cos^3 a + 5 \sin^4a \cos a$ // Another proof
$=\cos^5a - 10 (1 - \cos^2a) \cos^3 a + 5 (1- \cos^2a)^2 \cos a$
$=\cos^5a - 10 \cos^3a + 10 \cos^5a + (5 \cos a- 10 \cos^3a + 5 \cos^5a)$
$\cos{5a}=16\cos^5a - 20 \cos^3a + 5 \cos a$
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