If 4 cos 2 A - 3 = 0 and 0° A 90°; then prove that : (i) sin 3A = 3 sin A - 4 sin 3 A (ii) cos 3A = 4 cos 3 A - 3 cos A

Given, ⇒ 4 cos 2 A - 3 = 0 ⇒ 4 cos 2 A = 3 ⇒ cos 2 A = ⇒ cos A = . ⇒ cos A = cos 30° ⇒ A = 30°. (i) To prove: sin 3A = 3 sin A - 4 sin 3 A Solving L.H.S. of the equation : ⇒ sin 3A = sin 3(30°) = sin 90° = 1. Solving R.H.S. of the equation : ⇒ 3 sin A - 4 sin 3 A ⇒ 3 sin 30° - 4 sin 3 30° ⇒ 3 ⇒ ⇒ ⇒ ⇒ 1. Since, L.H.S. = R.H.S. Hence, proved that sin 3A = 3 sin A - 4 sin 3 A. (ii) To prove: cos 3A = 4 cos 3 A - 3 cos A Solving L.H.S. ⇒ cos 3A = cos 3(30°) = cos 90° = 0. Solving R.H.S. ⇒ 4 cos 3 A - 3 cos A ⇒ 4 cos 3 30° - 3 cos 30° ⇒ 4 ⇒ 4 ⇒ ⇒ 0. Since, L.H.S. = R.H.S. Hence, proved that cos 3A = 4 cos 3 A - 3 cos A

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Find A, if 0° ≤ A ≤ 90° and :
(i) 2 cos² A - 1 = 0
(ii) sin 3A - 1 = 0
(iii) 4 sin² A - 3 = 0
(iv) cos² A - cos A = 0
(v) 2 cos² A + cos A - 1 = 0

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Mathematics

If 4 cos² A - 3 = 0 and 0° ≤ A ≤ 90°; then prove that :
(i) sin 3A = 3 sin A - 4 sin³ A
(ii) cos 3A = 4 cos³ A - 3 cos A

Trigonometric Identities

Related Questions

Find A, if 0° ≤ A ≤ 90° and :
(i) 2 cos² A - 1 = 0
(ii) sin 3A - 1 = 0
(iii) 4 sin² A - 3 = 0
(iv) cos² A - cos A = 0
(v) 2 cos² A + cos A - 1 = 0

Mathematics

If 4 cos2 A - 3 = 0 and 0° ≤ A ≤ 90°; then prove that :(i) sin 3A = 3 sin A - 4 sin3 A(ii) cos 3A = 4 cos3 A - 3 cos A

Trigonometric Identities

Related Questions

Find A, if 0° ≤ A ≤ 90° and :(i) 2 cos2 A - 1 = 0(ii) sin 3A - 1 = 0(iii) 4 sin2 A - 3 = 0(iv) cos2 A - cos A = 0(v) 2 cos2 A + cos A - 1 = 0

If 4 cos² A - 3 = 0 and 0° ≤ A ≤ 90°; then prove that :
(i) sin 3A = 3 sin A - 4 sin³ A
(ii) cos 3A = 4 cos³ A - 3 cos A

Find A, if 0° ≤ A ≤ 90° and :
(i) 2 cos² A - 1 = 0
(ii) sin 3A - 1 = 0
(iii) 4 sin² A - 3 = 0
(iv) cos² A - cos A = 0
(v) 2 cos² A + cos A - 1 = 0