Стойност на тригонометричен израз

[Гост]

Трябва да пресметна стойността, но мисля, че нещо не се получава... Може ли помощ?

[ammornil]

[tex]\alpha \in (90 ^\circ;180 ^\circ) \Rightarrow \text{ II квадрант } \Rightarrow \begin{cases} \sin{\alpha} > 0 \\ \cos{\alpha} <0 \end{cases}[/tex]

[tex]\tg{\alpha}\cdot \cotg{\alpha}=1 \Rightarrow \cotg{\alpha}=\frac{1}{\tg{\alpha}}=-\frac{1}{2}[/tex]

[tex]\tg{\alpha}=\frac{\sin{\alpha}}{\cos{\alpha}}=-2 \Rightarrow \sin{\alpha}=-2\cdot \cos{\alpha}[/tex]

[tex]\sin^{2}{\alpha} + \cos^{2}{\alpha} = 1 \cdots[/tex]

Скрит текст: покажи

[tex]\Leftrightarrow (-2\cos{\alpha})^{2} +\cos^{2}{\alpha}= 1 \Rightarrow 5\cos^{2}{\alpha}=1 \Rightarrow \cos^{2}{\alpha}=\frac{1}{5} \Rightarrow \cos{\alpha}=\pm \frac{\sqrt{5}}{5}[/tex]

[tex]\cos{\alpha}<0 \Rightarrow \cos{\alpha}=-\frac{\sqrt{5}}{5} \rightarrow \sin{\alpha}=-2\cos{\alpha}=\frac{2\sqrt{5}}{5}[/tex]

[tex]\frac{\tg{\alpha}-\cotg{\alpha}}{\sin^{4}{\alpha}-\cos^{4}{\alpha}}=\frac{-2-\left( -\frac{1}{2} \right)}{\left(\frac{2\sqrt{5}}{5} \right)^{4}-\left(-\frac{\sqrt{5}}{5} \right)^{4}}=\frac{-2+\frac{1}{2}}{\large{\frac{2^{4}\cdot 5^{2}}{5^{4}}-\frac{5^{2}}{5^{4}}}}=\frac{\frac{-2\cdot 2+1}{2}}{\frac{16}{25}-\frac{1}{25}}=\frac{-\frac{3}{2}}{\frac{15}{25}}=\frac{-3\cdot 25}{2\cdot 15}=-\frac{5}{2}[/tex]


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[tex](\sqrt{5})^{4}=(5^{\frac{1}{2}})^{4}=5^{\frac{1}{2}\cdot 4}=5^{2}[/tex]

[S.B.]

Ето и още един поглед върху задачата:

[tex]A = \displaystyle\frac{\tg \alpha - \cotg \alpha }{ \sin^{4 } \alpha - \cos^{4 } \alpha } =\displaystyle \frac{\displaystyle \frac{\sin \alpha }{\cos \alpha } - \displaystyle \frac{\cos \alpha }{\sin \alpha } }{( \sin^{2 } \alpha+ \cos^{2 } \alpha)( \sin^{2 } \alpha - \cos^{2 } \alpha) } = \displaystyle\frac{\displaystyle \frac{\ sin^{2 } \alpha - \cos^{2 } \alpha }{\\sin \alpha .\cos \alpha } }{ \sin^{2 } \alpha - \cos^{2 } \alpha } =\displaystyle \frac{1}{\sin \alpha\cos \alpha }[/tex]
$$\Rightarrow A = \frac{1}{\sin \alpha\cos \alpha } $$

От [tex]\tg \alpha = -2 \rightarrow \begin{array}{|l} \displaystyle \frac{\sin \alpha }{\cos \alpha }= -2 \\ \sin^{2 } \alpha + \cos^{2 } \alpha = 1 \end{array} \Leftrightarrow \begin{array}{|l} \sin \alpha = -2.\cos \alpha \\ 4 \cos^{2 } \alpha + \cos^{2 } \alpha = 1 \end{array} \Leftrightarrow \cos \alpha = \displaystyle \frac{ \pm \sqrt{5} }{5} , \sin \alpha = -2\cos \alpha[/tex]
[tex]\alpha \in ( 90 ^\circ , 180 ^\circ) \Rightarrow[/tex]
$$\cos \alpha = \frac{- \sqrt{5} }{5} , \sin \alpha = \frac{2 \sqrt{5} }{5} $$
[tex]A = \displaystyle\frac{1}{\sin \alpha\cos \alpha } = \displaystyle\frac{1}{\displaystyle \frac{- \sqrt{5} }{5}.\displaystyle \frac{2 \sqrt{5} }{5} }[/tex]
$$\Rightarrow A = - \frac{5}{2} $$