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

Question

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

KnowledgeBoat · Accepted Answer

(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° = . Substituting value of θ in R.H.S. of the equation we get : ⇒ 2 sin θ cos θ = 2 sin 30° cos 30° = . 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° = . Substituting value of θ in R.H.S. of the equation we get : ⇒ 2cos² θ - 1 = 2cos² 30° - 1 = = . 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° = = = = 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° Since, L.H.S. = R.H.S. Hence, proved that cos 3θ = 4 cos³θ - 3 cos θ.

Mathematics

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

Trigonometrical Ratios

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Mathematics

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

Trigonometrical Ratios

Related Questions

If θ = 30°, find the ratio 2 sin θ : sin 2θ.

If x = 15°, verify that 4 sin 2x cos 4x sin 6x = 1.

By, means of an example, show that sin(A + B) ≠ sin A + sin B.

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