Respuesta :

xpluto

Step-by-step explanation:

[tex] { \sin(x) }^{2} + { \cos(x) }^{2} = 1 \\ { \cos(theta) }^{2} = \frac{9}{16} \\ \: { \sin(theta) }^{2} = \frac{6}{16} = \frac{3}{8} \\ \: \sin(theta) = \frac{ \sqrt{3} }{ \sqrt{8} } \\ \sin(theta ) = \frac{ \sqrt{24} }{8}

= \frac{ 2\sqrt{6} }{8} = \frac{ \sqrt{6} }{4}

[/tex]

Answer:

[tex]\frac{\sqrt{7} }{4}[/tex]

Step-by-step explanation:

Using the Pythagorean identity

sin²x + cos²x = 1, then

sinx = [tex]\sqrt{1-cos^2x}[/tex]

sinΘ = [tex]\sqrt{1-(3/4)^2}[/tex]

        = [tex]\sqrt{1-\frac{9}{16} }[/tex] = [tex]\sqrt{\frac{7}{16} }[/tex] = [tex]\frac{\sqrt{7} }{4}[/tex]

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