sin theta - sin 2 theta + sin 3 theta ÷ cos theta - cos 2 theta + cos 3 theta = cos A
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Answer:
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Answer: α/2
Given,
sin θ + sin 2θ + sin 3θ = sin α
(sin 3θ + sin θ) + sin 2θ = sin α
2 sin(3θ + θ)/2 cos(3θ – θ)/2 = sin α
2 sin 2θ cos θ + sin 2θ = sin α
sin 2θ(2 cos θ + 1) = sin α….(i)
Also, given:
cos θ + cos 2θ + cos 3θ = cos α
(cos 3θ + cos θ) + cos 2θ = cos α
2 cos(3θ + θ)/2 cos(3θ – θ)/2 + cos 2θ = cos α
2 cos 2θ cos θ + cos 2θ = cos α
cos 2θ(2 cos θ + 1) = cos α….(ii)
Dividing (i) by (ii),
[sin 2θ(2 cos θ + 1)]/ [cos 2θ(2 cos θ + 1)] = sin α/cos α
tan 2θ = tan α
2θ = α
θ = α/2
Step-by-step explanation:
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