Answer:
[tex] \sqrt{ \frac{13}{16} } [/tex]
Step-by-step explanation:
[tex]cos \: theta = \frac{ \sqrt{3} }{4} \\ sin \: theta = \sqrt{1 - {cos}^{2} theta} \\ = \sqrt{1 - {( \frac{ \sqrt{3} }{4} ) }^{2}} \\ = \sqrt{1 - \frac{3}{16}} \\ = \sqrt{ \frac{13}{16} } [/tex]
Answer:
see explanation
Step-by-step explanation:
Using the trigonometric identity
• sin²x + cos²x = 1 ⇒ sin x = ± [tex]\sqrt{1-cos^2x}[/tex]
Since sinΘ < 0 , then
sinΘ = - [tex]\sqrt{1-(\frac{\sqrt{3} }{4})^2 }[/tex]
= - [tex]\sqrt{1-\frac{3}{16} }[/tex]
= - [tex]\sqrt{\frac{13}{16} }[/tex] = - [tex]\frac{\sqrt{13} }{4}[/tex]
