If θ = 30°, verify that

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.

Answer :

θ = 30°

(i) sin 2θ = 2 sin θ cos θ

LHS = sin 2θ

= sin 2 × 30°

= sin 60°

= √3/2

RHS = 2 sin θ cos θ

= 2 sin 30° cos 30°

= 2 × ½ × √3/2

= 1 × √3/2

= √3/2

Hence, LHS = RHS.

(ii) cos 2θ = 2 cos²θ – 1

LHS = cos 2θ

= cos 2 × 30°

= cos 60°

= ½

RHS = 2 cos² θ – 1

= 2 cos² 30° – 1

= 2 (√3/2)² – 1

= 2 × ¾ – 1

= 3/2 – 1

= (3 – 2)/2

= ½

Hence, LHS = RHS.

(iii) sin 3θ = 3 sin θ – 4 sin³ θ

Consider

LHS = sin 3θ

= sin 3 × 30°

= sin 90°

= 1

RHS = 3 sin θ – 4 sin³ θ

= 3 sin 30° – 4 sin³ 30°

= (3×½) – 4×(1/2)³

= 3/2 – 4× 1/8

= 3/2 – ½

Taking LCM

= (3 – 1)/2

= 2/2

= 1

Hence, LHS = RHS.

(iv) cos 3θ = 4 cos³ θ – 3 cos θ

LHS = cos 3θ

= cos 3 × 30°

= cos 90°

= 0

RHS = 4 cos³ θ – 3 cos θ

= 4 cos³ 30° – 3 cos 30°

= 4×(√3/2)³ – 3×(√3/2)

= 4× 3√3/8 – 3√3/2

= 3√3/2 – 3√3/2

= 0

Hence, LHS = RHS.

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