If θ = 30°, verify that

(i) sin 2θ = 2 sin θ cos θ

(ii) cos 2θ = 2 cos² θ - 1

(iii) sin 3θ = 3 sin θ - 4 sin³ θ

(iv) cos 3θ = 4 cos³ θ - 3 cos θ

(i) To verify,

sin 2θ = 2 sin θ cos θ

Substituting value of θ in L.H.S. of the equation we get :

⇒ sin 2θ = sin 2(30°) = sin 60° = $\dfrac{\sqrt{3}}{2}$.

Substituting value of θ in R.H.S. of the equation we get :

⇒ 2 sin θ cos θ = 2 sin 30° cos 30° = $2 \times \dfrac{1}{2} \times \dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{3}}{2}$.

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

Hence, proved that sin 2θ = 2 sin θ cos θ.

(ii) To verify,

cos 2θ = 2 cos² θ - 1

Substituting value of θ in L.H.S. of the equation we get :

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

Substituting value of θ in R.H.S. of the equation we get :

⇒ 2cos² θ - 1 = 2cos² 30° - 1

= $2 \times \Big(\dfrac{\sqrt{3}}{2}\Big)^2 - 1$

= $2 \times \dfrac{3}{4} - 1 = \dfrac{3}{2} - 1 = \dfrac{1}{2}$.

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

Hence, proved that cos 2θ = 2cos² θ - 1.

(iii) To verify,

sin 3θ = 3 sin θ - 4 sin³ θ

Substituting value of θ in L.H.S. of the equation we get :

⇒ sin 3θ = sin 3(30°) = sin 90° = 1.

Substituting value of θ in R.H.S. of the equation we get :

⇒ 3 sin θ - 4 sin³ θ = 3 sin 30° - 4 sin³ 30°

= $3 \times \dfrac{1}{2} - 4 \times \Big(\dfrac{1}{2}\Big)^3$

= $\dfrac{3}{2} - \dfrac{4}{8}$

= $\dfrac{12 - 4}{8} = \dfrac{8}{8}$ = 1.

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

Hence, proved that sin 3θ = 3 sin θ - 4 sin³ θ.

(iv) Given,

Equation : cos 3θ = 4 cos³θ - 3 cos θ

Substituting θ = 30°, in L.H.S. of the given equation, we get :

⇒ cos 3θ = cos 3(30°) = cos 90° = 0.

Substituting θ = 30°, in R.H.S. of the given equation, we get :

⇒ 4 cos³θ - 3 cos θ = 4 cos³ 30° - 3 cos 30°

$= 4 \times \Big(\dfrac{\sqrt{3}}{2}\Big)^3 - 3 \times \dfrac{\sqrt{3}}{2}\\[1em] = 4 \times \dfrac{3\sqrt{3}}{8} - \dfrac{3\sqrt{3}}{2} \\[1em] = \dfrac{3\sqrt{3}}{2} - \dfrac{3\sqrt{3}}{2} \\[1em] = 0.$

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

Hence, proved that cos 3θ = 4 cos³θ - 3 cos θ.