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Q.Value of sin 36° is

(a) (1/4)√(10-2√5)
(b) (1/4)√(10+2√5)
(c) (1/4)√(10+√5)
(d) (1/4)√(10-√5).
West Bengal WbchseWest Bengal HS First Year (WBCHSE Class XI) Annual Examination 2018MCQ· 1mImportance★★★★★est
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Using the pentagon-derived value cos⁡36∘=5+14\cos36^\circ=\dfrac{\sqrt5+1}{4}, we get sin⁡36∘=1410−25\sin36^\circ=\dfrac14\sqrt{10-2\sqrt5}.

Let θ=36∘\theta=36^\circ, so 5θ=180∘5\theta=180^\circ, giving 3θ=180∘−2θ3\theta=180^\circ-2\theta, hence sin⁡3θ=sin⁡2θ\sin3\theta=\sin2\theta. Expanding both sides in terms of sin⁡θ=x\sin\theta=x using the triple- and double-angle formulas and solving the resulting equation 4x3−2x2−3x+1=04x^3-2x^2-3x+1=0 (with x=sin⁡18∘x=\sin18^\circ as one root, or working with cos⁡36∘\cos36^\circ) gives the classical value cos⁡36∘=1+54\cos36^\circ=\dfrac{1+\sqrt5}{4}.

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