Question

Verify the following
(i) If A = 30$°$ then show that cos3A = 4cos³ A $-$ 3cosA

Answer 1

Given:
A = 30$°$

To show:
cos 3A = 4 cos³ A - 3 cos A

Here, LHS = cos 3A
= cos 3(30$°$)
= cos 90$°$
= 0 (From table)

Now, RHS = 4 cos³ A - 3 cos A
= 4 cos³ (30$°$) - 3 cos (30$°$)
$$\begin{gathered} =4{\left(\frac{\sqrt{3}}{2}\right)}^3-3\left(\frac{\sqrt{3}}{2}\right)(\mathrm{From}\mathrm{the}\mathrm{table}) \\ =\frac{12\sqrt{3}}{8}-\frac{3\sqrt{3}}{2} \\ =\frac{3\sqrt{3}}{2}-\frac{3\sqrt{3}}{2}=0 \end{gathered}$$
∴ LHS = RHS