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JEE Main
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JEE Main
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main · Mathematics · Trigonometric Ratios & Identities

The value of 18^ ^ ^ is

The value of sin⁡18∘\sin 18^{{}^{{}^{\circ}}} is

Options

  1. A45−1\frac{4}{\sqrt{5}-1}
  2. B5−14\frac{\sqrt{5}-1}{4}
  3. C5+14\frac{\sqrt{5}+1}{4}
  4. D45+1\frac{4}{\sqrt{5}+1}

Correct answer

B. 5−14\frac{\sqrt{5}-1}{4}

Step-by-step solution

sin⁡90∘=sin⁡5(81∘) and let 18∘=Asin⁡90∘=sin⁡5 A=sin⁡(3 A+2 A)∴90∘=3 A+2 A⇒sin⁡(90∘−3 A)=sin⁡2 A∴sin⁡2 A−cos⁡3 A⇒2sin⁡Acos⁡A=4cos⁡3 A−3cos⁡A∴cos⁡A(2sin⁡A−4cos⁡2 A+3)=0∴cos⁡A=0 or [2sin⁡A−4(1−sin⁡2 A)+3]=0 and ∴A=π2 or 4sin⁡2 A+2sin⁡A−1=0\begin{aligned}& \sin 90^{{}^{{}^{\circ}}}=\sin 5 \left(81^{{}^{{}^{\circ}}}\right)\text { and let }18^{{}^{{}^{\circ}}}=\mathrm{A}\\ & \sin 90^{{}^{{}^{\circ}}}=\sin 5 \mathrm{~A}=\sin (3 \mathrm{~A}+2 \mathrm{~A})\\ & \therefore 90^{{}^{{}^{\circ}}}=3 \mathrm{~A}+2 \mathrm{~A} \Rightarrow \sin \left(90^{{}^{{}^{\circ}}}-3 \mathrm{~A}\right)=\sin2 \mathrm{~A}\\ & \therefore \sin 2 \mathrm{~A}-\cos3 \mathrm{~A} \Rightarrow2 \sin \mathrm{A} \cos \mathrm{A}=4 \cos ^3 \mathrm{~A}-3 \cos \mathrm{A}\\ & \therefore \cos \mathrm{A}\left(2 \sin \mathrm{A}-4 \cos ^2 \mathrm{~A}+3\right)=0 \\ & \therefore \cos \mathrm{A}=0 \text { or }\left[2 \sin \mathrm{A}-4\left(1-\sin ^2 \mathrm{~A}\right)+3\right]=0 \\ \text{ and } \therefore \mathrm{A}=\frac{\pi}{2} \text { or }4 \sin ^2 \mathrm{~A}+2 \sin \mathrm{A}-1=0 \end{aligned}Since, A=18∘,A≠π2A=18^{{}^{{}^{\circ}}}, A \neq \frac{\pi}{2} and ∴4sin⁡2 A+2sin⁡A−1=0 and ∴sin⁡A=−2±4+162(4)=−2±208=2−±258=−1±54 and ∴sin⁡A=−1+54 or sin⁡A=−1+54\begin{aligned} \text{ and } \therefore 4 \sin ^2 \mathrm{~A}+2 \sin \mathrm{A}-1=0 \\ \text{ and } \therefore \sin \mathrm{A}=\frac{-2 \pm \sqrt{4+16}}{2(4)}=\frac{-2 \pm \sqrt{20}}{8}=\frac{2- \pm 2 \sqrt{5}}{8}=\frac{-1 \pm \sqrt{5}}{4}\\ \text{ and } \therefore \sin \mathrm{A}=\frac{-1+\sqrt{5}}{4} \text { or } \sin \mathrm{A}=\frac{-1+\sqrt{5}}{4} \end{aligned}$Since,Since,\sin \mathrm{A}>0, \sin \mathrm{A}=\frac{\sqrt{5}-1}{4}$
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