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Triangle and trigonometry

 
Trigonometry 1 & 2
Last Post by Staff 4 years ago
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June 15, 2021 12:46 pm
Topic starter
Staff 
(@armor)
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June 15, 2021 12:46 pm
Topic starter
Staff 
(@armor)
Admin
  Honorable Member
317 Posts
156 153 0

\small sin^2(n+1)\Theta =sin^2n\Theta +sin^2(n-1)\Theta \Rightarrow sin^2(n+1)\Theta -sin^2(n-1)\Theta =sin^2n\Theta

\small \Rightarrow [sin(n+1)\Theta -sin(n-1)\Theta ][sin(n+1)\Theta +sin(n-1)\Theta ]=sin^2n\Theta

\small \Rightarrow 2sin\Theta cosn\Theta 2sinn\Theta cos\Theta =sin^2n\Theta \Rightarrow sin2n\Theta sin2\Theta =sin^2n\Theta     (I)

Since \small (n+1)\Theta +n\Theta +(n-1)\Theta =\Pi ,   then  \small n\Theta =\frac{\Pi }{3}.   Substituting it into (I), then 

\small sin2\Theta =\frac{1}{2}tan\frac{\Pi }{3}=\frac{\sqrt{3}}{2}\Rightarrow 2\Theta =\frac{\Pi }{3}\Rightarrow \Theta =\frac{\Pi }{6}.

Since   \small n\Theta =\frac{\Pi }{3},   we have \small n=2.

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