Question

[PFVR EH URGENTE]

2)Conhecendo o valor de cos alfa = 3/10 com alfa no primeiro quadrante, calcule;

sen alfa
tg alfa
cotg alfa
sec alfa
cossec alfa

3)conhecendo o valor de sen alfa = 4/9 com alfa no primeiro quadrante, calcule:

tg alfa
cotg alfa
sec alfa
cossec alfa

Answer 1

Resposta:

Explicação passo-a-passo:

2) cos α = 3/10, assim:

i)

sen² α + cos² α = 1

sen² α + (3/10)² = 1

sen² α = 1 - 9/100

sen² α = (100 - 9)/100

sen² α = 91/100

sen α = ± √91/100

sen α = ± √91/10

ii)

tg α = sen α/cos α

$tg\alpha= \frac{sen\alpha}{cos\alpha}=\frac{\frac{\sqrt{91}}{10}}{\frac{3}{10}}=\frac{\sqrt{91}}{10}.\frac{10}{3}=\frac{\sqrt{91}}{3}$

iii)

$cotg\alpha \frac{cos\alpha}{sen\alpha}=\frac{\frac{3}{10}}{\frac{\sqrt{91}}{10}}=\frac{3}{10}.\frac{10}{\sqrt{91}}=\frac{3}{\sqrt{91}}=\frac{3\sqrt{91}}{\sqrt{91}\sqrt{91}}=\frac{3\sqrt{91}}{91}$

iv)

$sec\alpha=\frac{1}{cos\alpha}=\frac{1}{\frac{3}{10}}=1.\frac{10}{3}=\frac{10}{3}$

v)

$cossec\alpha=\frac{1}{sen\alpha}=\frac{1}{\frac{\sqrt{91}}{10}}=1.\frac{10}{\sqrt{91}}=\frac{10\sqrt{91}}{\sqrt{91}\sqrt{91}}=\frac{10\sqrt{91}}{91}$

3)

i)

sen² α + cos² α = 1

(4/9)² + cos² α = 1

16/81 + cos² α = 1

cos² α = 1 - 16/81

cos² α = (81 - 16)/81

cos² α = 65/81

cos α = ± √65/81

cos α = ± √65/9

ii)

$tg\alpha \frac{sen\alpha}{cos\alpha}=\frac{\frac{4}{9}}{\frac{\sqrt{65}}{9}}=\frac{4}{9}.\frac{9}{\sqrt{65}}=\frac{4\sqrt{65}}{\sqrt{65}\sqrt{65}}=\frac{4\sqrt{65}}{65}$

iii)

$cotg\alpha=\frac{cos\alpha}{sen\alpha}=\frac{\frac{\sqrt{65}}{9}}{\frac{4}{9}}=\frac{\sqrt{65}}{9}.\frac{9}{4}=\frac{\sqrt{65}}{4}$

iv)

$sec\alpha=\frac{1}{cos\alpha}=\frac{1}{\frac{\sqrt{65}}{9}}=1.\frac{9}\sqrt{65}}=\frac{9\sqrt{65}}{\sqrt{65}\sqrt{65}}=\frac{9\sqrt{65}}{65}$

v)

$cossec\alpha=\frac{1}{sen\alpha}=\frac{1}{\frac{4}{9}}=1.\frac{9}{4}=\frac{9}{4}$