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JEE Main 2023 · Mathematics · Trigonometry and Inverse Trigonometric Functions · easy

The value of cos 20^°cos 40^°cos 80^° is

Single correct option

The value of cos⁡20∘cos⁡40∘cos⁡80∘\cos20^\circ\cos40^\circ\cos80^\circ is
  1. A. 18\dfrac18
  2. B. 14\dfrac14
  3. C. 38\dfrac{\sqrt3}{8}
  4. D. 38\dfrac{3}{8}
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Answer: (A) 18\dfrac18

Multiply and divide by sin⁡20∘\sin20^\circ and apply 2sin⁡θcos⁡θ=sin⁡2θ2\sin\theta\cos\theta=\sin2\theta repeatedly: sin⁡20∘cos⁡20∘cos⁡40∘cos⁡80∘=12sin⁡40∘cos⁡40∘cos⁡80∘=14sin⁡80∘cos⁡80∘=18sin⁡160∘=18sin⁡20∘\sin20^\circ\cos20^\circ\cos40^\circ\cos80^\circ=\tfrac12\sin40^\circ\cos40^\circ\cos80^\circ=\tfrac14\sin80^\circ\cos80^\circ=\tfrac18\sin160^\circ=\tfrac18\sin20^\circ. Hence the product is 18\dfrac18.
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