Taking A = 30°, verify that

(i) cos⁴ A - sin⁴ A = cos 2A

(ii) 4 cos A cos (60° - A) cos (60° + A) = cos 3A.

(i) To verify,

cos⁴ A - sin⁴ A = cos 2A

Substituting value of A in L.H.S. of the above equation, we get :

$\Rightarrow \text{cos}^4 \space A - \text{sin}^4 \space A \\[1em] \Rightarrow \text{cos}^4 \space 30° - \text{sin}^4 \space 30° \\[1em] \Rightarrow \Big(\dfrac{\sqrt{3}}{2}\Big)^4 - \Big(\dfrac{1}{2}\Big)^4 \\[1em] \Rightarrow \dfrac{9}{16} - \dfrac{1}{16} \\[1em] \Rightarrow \dfrac{8}{16} \\[1em] \Rightarrow \dfrac{1}{2}.$

Substituting value of A in R.H.S. of the above equation, we get :

⇒ cos 2A = cos 2(30°) = cos 60° = $\dfrac{1}{2}$.

Since, L.H.S. = R.H.S.

Hence, proved that cos⁴ A - sin⁴ A = cos 2A.

(ii) To verify,

4 cos A cos (60° - A) cos (60° + A) = cos 3A.

Substituting value of A in L.H.S. of the above equation, we get :

$\Rightarrow \text{4 cos 30° cos (60° - 30°) cos (60° + 30°)} \\[1em] \Rightarrow \text{4 cos 30° cos 30° cos 90°} \\[1em] \Rightarrow 4 \times \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{2} \times 0 \\[1em] \Rightarrow 0.$

Substituting value of A in R.H.S. of the above equation, we get :

$\Rightarrow \text{cos 3(30°)} = \text{cos 90°} = 0.$

Since, L.H.S. = R.H.S.

Hence, proved that 4 cos A cos (60° - A) cos (60° + A) = cos 3A