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Trigonometric Ratio and Identites question

2022 · 26 Jun · Shift 2 · Q37
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  5. /2022 · 26 Jun · Shift 2 · Q37

Trigonometric Ratio and Identites question

2022 · 26 Jun · Shift 2 · Q37

JEE MainMathematicsTrigonometric Ratio and IdentitesMCQ+4 / −1
16sin⁡(20∘)sin⁡(40∘)sin⁡(80∘)16\sin (20^\circ )\sin (40^\circ )\sin (80^\circ )16sin(20∘)sin(40∘)sin(80∘) is equal to :
  1. A
    3\sqrt 33​
  2. B
    2 3\sqrt 33​
  3. C
    3
  4. D
    4 3\sqrt 33​
View written solutionFree

Correct answer: B

  1. We need to evaluate 16sin⁡20∘sin⁡40∘sin⁡80∘.16\sin 20^\circ\sin 40^\circ\sin 80^\circ.16sin20∘sin40∘sin80∘.

  2. Use the standard identity: sin⁡3x=4sin⁡xsin⁡(60∘+x)sin⁡(60∘−x).\sin 3x = 4\sin x\sin(60^\circ+x)\sin(60^\circ-x).sin3x=4sinxsin(60∘+x)sin(60∘−x).

  3. Put x=20∘x=20^\circx=20∘. Then sin⁡60∘=4sin⁡20∘sin⁡80∘sin⁡40∘.\sin 60^\circ = 4\sin 20^\circ\sin 80^\circ\sin 40^\circ.sin60∘=4sin20∘sin80∘sin40∘. Since sin⁡60∘=32\sin 60^\circ=\frac{\sqrt{3}}{2}sin60∘=23​​, we get 4sin⁡20∘sin⁡40∘sin⁡80∘=32.4\sin 20^\circ\sin 40^\circ\sin 80^\circ = \frac{\sqrt{3}}{2}.4sin20∘sin40∘sin80∘=23​​.

  4. Multiply both sides by 444: 16sin⁡20∘sin⁡40∘sin⁡80∘=4⋅32=23.16\sin 20^\circ\sin 40^\circ\sin 80^\circ = 4\cdot \frac{\sqrt{3}}{2} = 2\sqrt{3}.16sin20∘sin40∘sin80∘=4⋅23​​=23​.

  5. Therefore the value is 23.\boxed{2\sqrt{3}}.23​​.

  6. Checking options:

  • A: 3\sqrt33​ ❌
  • B: 232\sqrt323​ ✅
  • C: 333 ❌
  • D: 434\sqrt343​ ❌

So the correct option is B.

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