Răspuns:
[tex]2 - \sqrt{3} [/tex]
Explicație pas cu pas:
55° = 60° - 5°
65° = 60° + 5°
folosim identitatea:
[tex]\tan( \alpha ) \times \tan(\frac{\pi}{3} - \alpha) \times \tan( \frac{\pi}{3} + \alpha ) = \tan( 3\alpha )[/tex]
deci:
[tex]\tan(5) \times \tan(60 - 5) \times \tan(60 + 5) = \tan(3 \times 5)[/tex]
[tex] = > \tan(5) \times \tan(55) \times \tan(65) = \tan(15)[/tex]
15° = 45° - 30°
folosim formula:
[tex]\tan( \alpha - \beta ) = \frac{ \tan( \alpha ) - \tan( \beta ) }{1 + \tan( \alpha ) \tan( \beta ) } [/tex]
=>
[tex] \tan(15) = \tan(45 - 30) = \frac{ \tan(45) - \tan(30) }{1 + \tan(45) \times \tan(30) } = \frac{1 - \frac{1}{ \sqrt{3} } }{1 + 1 \times \frac{1}{ \sqrt{3} } } = \frac{ \sqrt{3} - 1}{ \sqrt{3} + 1} = 2 - \sqrt{3} [/tex]
deci:
[tex]\tan(5) \times \tan(55) \times \tan(65) = 2 - \sqrt{3}[/tex]
demonstație pentru:
[tex]\tan(5) \times \tan(60 - 5) \times \tan(60 + 5) =\tan(5) \times \frac{ \tan(60) - \tan(5) }{1+ \tan(60) \tan(5) } \times \frac{ \tan(60) + \tan(5) }{1- \tan(60) \tan(5)} = \tan(5) \times \frac{ \sqrt{3} - \tan(5) }{1 + \sqrt{3} \tan(5)} \frac{ \sqrt{3} + \tan(5)}{1 - \sqrt{3} \tan(5) } = \tan(5) \times \frac{3 - \tan^{2} (5) }{1 - 3 \tan^{2} (5) } = \frac{3 \tan(5) - \tan^{3} (5) }{1 - 3 \tan^{2} (5) } = \tan(3 \times 5)= \tan(15)[/tex]
unde:
[tex]\tan(3 \alpha ) = \frac{3 \tan( \alpha ) - \tan^{3} ( \alpha ) }{1 - 3 \tan^{ \alpha } ( \alpha ) }[/tex]
