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Rozwiązane

tg alfa = 5
oblicz (2sin alfa - 3cos alfa)/(4sin alfa + 7cos alfa)​

Odpowiedź :

Hankago

[tex]\frac{2sin\alpha-3cos\alpha}{4sin\alpha+7cos\alpha}=\frac{\frac{2sin\alpha}{cos\alpha}-\frac{3cos\alpha}{cos\alpha}}{\frac{4sin\alpha}{cos\alpha}+\frac{7cos\alpha}{cos\alpha}}=\frac{2tg\alpha-3}{4tg\alpha+7}=\frac{2\cdot5-3}{4\cdot5+7}=\frac{10-3}{20+7}=\frac{7}{27}[/tex]

Basetlago

[tex]tg\alpha =5\\\\\frac{sin\alpha}{cos\alpha} = 5 \ \ \rightarrow \ \ sin\alpha = 5cos\alpha\\\\\\\frac{2sin\alpha-3cos\alpha}{4sin\alpha+7cos\alpha} = \frac{2\cdot5cos\alpha-3cos\alpha}{4\cdot5cos\alpha+7cos\alpha} = \frac{10cos\alpha-3cos\alpha}{20cos\alpha+7cos\alpha} =\frac{7cos\alpha}{27cos\alpha} = \frac{7}{27}[/tex]

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