We will learn how to express the multiple angle of sin 3A in terms of A or sin 3A in terms of sin A.
Trigonometric function of sin 3A in terms of sin A is also known as one of the double angle formula.
If A is a number or angle then we have, sin 3A = 3 sin A - 4 sin^3 A.
Now we will proof the above multiple angle formula step-by-step.
Proof: sin 3A
= sin (2A + A)
= sin 2A cos A + cos 2A sin A
= 2 sin A cos A ∙ cos A + (1 - 2 sin^2 A) sin A
= 2 sin A (1 - sin^2 A) + sin A - 2 sin^3 A
= 2 sin A - 2 sin^3 A + sin A - 2 sin^3 A
= 3 sin A - 4 sin^3 A
Therefore, sin 3A = 3 sin A - 4 sin^3 A Proved
Note: (i) In the above formula we should note that the angle on the R.H.S. of the formula is one-third of the angle on L.H.S. Therefore, sin 60° = 3 sin 20° - 4 sin^3 20°.
(ii) To find the formula of sin 3A in terms of sin A we have used cos 2A = 1 - 2 sin^2 A
Now, we will apply the formula of multiple angle of sin 3A in terms of A or sin 3A in terms of sin A to solve the below problems.
1. Prove that sin A ∙ sin (60 - A) sin (60 + A) = ¼ sin 3A.
Solution:
L.H.S. = sin A ∙ sin (60° - A) sin (60° + A)
= sin A (sin^2 60° - sin^2 A), [Since, sin (A + B) sin (A - B) = sin^2 A - sin^2 B]
= sin A [(√3/2)^2 - sin^2 A), [Since we know that sin 60° = ½]
= sin A (3/4 - sin^2 A)
= ¼ sin A (3 - 4 sin^2 A)
= ¼ (3 sin A - 4 sin^3 A)
Now apply the formula of sin 3A in terms of A
= ¼ sin 3A = R.H.S. Proved
2. If cos θ = 12/13 find the value of sin 3θ.
Solution:
Given, cos A = 12/13
We know that sin^2 A + cos^2 A = 1
⇒ sin^2 A = 1 - cos^2A
⇒ sin A = √(1 - cos^2A)
Therefore, sin A = √[1 - (12/13)^2]
⇒ sin A = √[1 - 144/169]
⇒ sin A = √(25/169)
⇒ sin A = 5/13
Now, sin 3A = 3 sin A - 4 sin^3 A
= 3 ∙ 5/13 - 4 ∙ (5/13)^3
= 15/13 - 500/2199
= (2535 - 500)/2199
= 2035/2199
3. Show that, sin^3 A + sin^3 (120° + A) + sin^3 (240° + A) = - ¾ sin 3A.
Solution:
L.H.S = sin^3 A + sin^3 (120° + A) + sin^3 (240° + A)
= ¼ [4 sin^3 A + 4 sin^3 (120° + A) + 4 sin^3 (240° + A)]
= ¼ [3 sin A - sin 3A + 3 sin (120° + A) - sin 3 (120° + A) + 3 sin (240° + A) - sin 3 (240° + A)]
[Since we know that, sin 3A = 3 sin 3A - 4 sin^3 A
⇒ 4 sin^3 A = 3 sin A − sin 3A]
= ¼ [3 {sin A + sin (120° + A) + sin (240° + A)} - {sin 3A + sin (360° + 3A) + sin (720° + 3A)}]
= 1/4 [3 {sin A + 2 sin (180° + A) cos 60°) - (sin 3A + sin 3A + sin 3A)}
= ¼ [3 {sin A + 2 ∙ (- sin A) ∙ 1/2} - 3 sin A]
= ¼ [3 {sin A - sin A} - 3 sin A]
= - ¾ sin 3A = R.H.S. Proved
11 and 12 Grade Math
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