Question

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Answer 1

Answer:

If sin θ + sin 2θ + sin 3θ = sin α and cos θ + cos 2θ + cos 3θ = cos α, then θ is equal to

1) α/2

2) α

3) 2α

4) α/6

Answer: (1) α/2

Solution:

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

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Answer 2

Answer:

0=a/2 is write answer

Step-by-step explanation:

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