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Sin18°=?

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Sin18°=?

    • Posted by Saurav Singh 6 years, 10 months ago

    CBSE > Class 11 > Mathematics
    • 1 answers

    Shaina Rajput 6 years, 10 months ago

    Let A = 18° Therefore, 5A = 90° ⇒ 2A + 3A = 90˚ ⇒ 2θ = 90˚ - 3A Taking sine on both sides, we get sin 2A = sin (90˚ - 3A) = cos 3A ⇒ 2 sin A cos A = 4 cos^3 A - 3 cos A ⇒ 2 sin A cos A - 4 cos^3A + 3 cos A = 0 ⇒ cos A (2 sin A - 4 cos^2 A + 3) = 0 Dividing both sides by cos A = cos 18˚ ≠ 0, we get ⇒ 2 sin θ - 4 (1 - sin^2 A) + 3 = 0 ⇒ 4 sin^2 A + 2 sin A - 1 = 0, which is a quadratic in sin A Therefore, sin θ = −2±−4(4)(−1)√2(4) ⇒ sin θ = −2±4+16√8 ⇒ sin θ = −2±25√8 ⇒ sin θ = −1±√5/4 Now sin 18° is positive, as 18° lies in first quadrant. Therefore, sin 18° = sin A = −1±√5/4

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