Question

find the value of sin 36°​

Answer 1

Answer:

Let A = 18°

Let A = 18° Therefore, 5A = 90°

Let A = 18° Therefore, 5A = 90° ⇒ 2A + 3A = 90˚

Let A = 18° Therefore, 5A = 90° ⇒ 2A + 3A = 90˚⇒ 2θ = 90˚ - 3A

Let A = 18° Therefore, 5A = 90° ⇒ 2A + 3A = 90˚⇒ 2θ = 90˚ - 3ATaking sine on both sides, we get

Let A = 18° Therefore, 5A = 90° ⇒ 2A + 3A = 90˚⇒ 2θ = 90˚ - 3ATaking sine on both sides, we get sin 2A = sin (90˚ - 3A) = cos 3A

Let A = 18° Therefore, 5A = 90° ⇒ 2A + 3A = 90˚⇒ 2θ = 90˚ - 3ATaking sine on both sides, we get sin 2A = sin (90˚ - 3A) = cos 3A ⇒ 2 sin A cos A = 4 cos3 A - 3 cos A

Let A = 18° Therefore, 5A = 90° ⇒ 2A + 3A = 90˚⇒ 2θ = 90˚ - 3ATaking sine on both sides, we get sin 2A = sin (90˚ - 3A) = cos 3A ⇒ 2 sin A cos A = 4 cos3 A - 3 cos A⇒ 2 sin A cos A - 4 cos3 A + 3 cos A = 0

Let A = 18° Therefore, 5A = 90° ⇒ 2A + 3A = 90˚⇒ 2θ = 90˚ - 3ATaking sine on both sides, we get sin 2A = sin (90˚ - 3A) = cos 3A ⇒ 2 sin A cos A = 4 cos3 A - 3 cos A⇒ 2 sin A cos A - 4 cos3 A + 3 cos A = 0 ⇒ cos A (2 sin A - 4 cos2 A + 3) = 0

Answer 2

Answer:

Step-by-step explanation:

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