sin 3A in Terms of A

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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