Answer :
Answer:
sinΘ = [tex]\frac{\sqrt{55} }{8}[/tex]
Step-by-step explanation:
using the identity
sin²Θ + cos²Θ = 1 ( subtract cos²Θ from both sides )
sin²Θ = 1 - cos²Θ ( take square root of both sides )
sinΘ = ± [tex]\sqrt{1 -cos^20}[/tex]
given
cosΘ = [tex]\frac{3}{8}[/tex] , then
sinΘ = ± [tex]\sqrt{1-(\frac{3}{8})^2 }[/tex]
= ± [tex]\sqrt{1-\frac{9}{64} }[/tex]
= ± [tex]\sqrt{\frac{64}{64}-\frac{9}{64} }[/tex]
= ± [tex]\sqrt{\frac{55}{64} }[/tex]
= ± [tex]\frac{\sqrt{55} }{\sqrt{64} }[/tex]
= ± [tex]\frac{\sqrt{55} }{8}[/tex]
since Θ is in the first quadrant then sinΘ > 0 , so
sinΘ = [tex]\frac{\sqrt{55} }{8}[/tex]