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[Solved] Logarithm trigonometry

 
Trigonometry 1 & 2
Last Post by Staff 4 years ago
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June 15, 2021 3:06 pm
Topic starter
Staff 
(@armor)
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June 15, 2021 3:06 pm
Topic starter
Staff 
(@armor)
Admin
  Honorable Member
317 Posts
156 153 0

Changing the base number of the given equation, we have 

\small \frac{lgcos\Theta }{lgsin\Theta -lgcos\Theta }=\frac{2}{3}\Rightarrow \frac{lgcos\Theta }{lgsin\Theta }=\frac{2}{5}\Rightarrow log_{sin\Theta }cos\Theta =\frac{2}{5}.

Hence,  

\small log_{csc^2\Theta }(\frac{sin2\Theta }{2})=-log_{sin^2\Theta }(sin\Theta cos\Theta )

\small =-log_{sin\Theta }(sin\Theta cos\Theta )^\frac{1}{2}=-\frac{1}{2}log_{sin\Theta }(sin\Theta cos\Theta )=-\frac{1+log_{sin\Theta }cos\Theta }{2}=-\frac{1+\frac{2}{5}}{2}=-\frac{7}{10}.

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