Question

Could u pls solve it.Trigonometry question

Answer 1

Answer:

3/2

Step-by-step explanation:

Well!!!!!!!!

LHS = Sin^2 (120-A)+Sin^2 A +Sin^2 (120+A)

Now, cos 2A = 1 - 2 sin^2 A ==> sin^2 A = (1 - cos 2A) / 2

==> LHS = [ 1 - cos 2(120-A)]/2 + [1 - cos 2A]/2 + [ 1 - cos 2(120+A)] / 2

= (1/2) * [ 3 - [ cos 2(120-A) + cos 2A + cos 2(120+A) ] ]

Now, cos A + cos B = 2 cos ½ (A + B) cos ½ (A − B)

==> LHS =(1/2) * [ 3 - [ cos 2A + [ cos 2(120-A) +  cos 2(120+A) ] ] ]

= (1/2) * [ 3 - [ cos 2A +

           2  cos (  (½) * 2(120-A+ 120+A))  cos (  (½) * 2(120-A- 120-A))

= (1/2)* [ 3 - [ cos 2A +  2  cos 240  cos (  -2A) ]

= (1/2) * [ 3 - [ cos 2A +  2  (-1/2)  cos (2A) ]

= (1/2) * [ 3 - [ cos 2A -  cos (2A) ] ]

= 3/2

SOLVED!!!

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Answer 2

Answer:

3/2

Step-by-step explanation: