Question

Sendo Sen a= 1/4 e Cos b=1 Calcule:
A) Cos a=
B) Sen b=
C) Tg a=

Answer 1

a)

Vamos usar a relação trigonométrica fundamental:

${seno}^{2} \alpha + {cosseno}^{2} \alpha = 1$

${( \frac{1}{4} )}^{2} + cosseno {}^{2} \alpha = 1 \\ \\ \frac{1}{16} + {cosseno}^{2} \alpha = 1 \\ \\ {cosseno}^{2} \alpha = 1 - \frac{1}{16} \\ \\ {cosseno}^{2} \alpha = \frac{16}{16} - \frac{1}{16} \\ \\ {cosseno}^{2} \alpha = \frac{15}{16} \\ \\ \sqrt{ {cosseno}^{2} \alpha } = ± \sqrt{ \frac{15}{16} } \\ \\ cosseno \: \alpha = ± \frac{ \sqrt{15} }{ \sqrt{16} } \\ \\ cosseno \: \alpha = ± \frac{ \sqrt{15} }{4}$

b)

${seno}^{2} \beta + {cosseno}^{2} \beta = 1 \\ \\ {seno }^{2} \beta + (1 ){}^{2} = 1 \\ \\ seno {}^{2} \beta + 1 = 1 \\ \\ seno {}^{2} \beta = 1 - 1 \\ \\ seno {}^{2} \beta = 0 \\ \\ \sqrt{seno {}^{2} \beta } = \sqrt{0} \\ \\ seno \: \beta = 0$

c)

$tangente \: \alpha = \frac{seno \: \alpha }{cosseno \: \alpha } \\ \\ \\ tangente \: \alpha = \frac{ \frac{1}{4} }{ \frac{ \sqrt{15} }{4} } \: \: ou \: \: \frac{ \frac{1}{4} }{ - \frac{ \sqrt{15}}{4} } \\ \\ tangente \: \alpha = \frac{1}{4} \times \frac{4}{ \sqrt{15} } \: \: ou \: \: \frac{1}{4} \times ( - \frac{4}{ \sqrt{15} } ) \\ \\ tangente \: \alpha = \frac{1}{ \sqrt{15} } \: \: ou \: \: - \frac{1}{ \sqrt{15} }$