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Cum pot deduce tg3x si ctg3x?

Răspuns :

   
[tex]\displaystyle\\ \text{Folosim formula: } \text{tg}(x+y) \text{ unde } y = 2x\\ \Longrightarrow x+y = x+2x = 3x \\ \\ \text{tg}(3x) =\text{tg}(x+2x) = \text{tg}(x+y) = \frac{\text{tg}(x) + \text{tg}(y)}{1-\text{tg}(x)\text{tg}(y)} =\frac{\text{tg}(x) + \text{tg}(2x)}{1-\text{tg}(x)\text{tg}(2x)}\\ \\ \text{Inlocuim tg(y) din formula de sus cu formula tg(2x).} [/tex]


[tex]\displaystyle \\ \text{tg}(2x) = \frac{2\text{tg}(x)}{1-\text{tg}^2(x)}\\\\\\ \text{tg}(3x)= \text{tg}(x+2x) =\frac{\text{tg}(x) + \text{tg}(2x)}{1-\text{tg}(x)\text{tg}(2x)} =\frac{\text{tg}(x) + \frac{2\text{tg}(x)}{1-\text{tg}^2(x)} }{1-\text{tg}(x)\frac{2\text{tg}(x)}{1-\text{tg}^2(x)} } =\\\\\\ =\frac{\text{tg}(x)[1-\text{tg}^2(x)] + 2\text{tg}(x)} {1-\text{tg}^2(x)-\text{2tg}^2(x)} = \boxed{\frac{3\text{tg}(x)-\text{tg}^3(x) } {1-\text{3tg}^2(x)}} \\\\ \text{La ctg(x) calculele sunt similare.}[/tex]