Respuesta :

abhaysilver1
Here is your answer

[tex][I am using ☆ instead of theta][/tex]

[tex]cos☆= -8/17[/tex]

[tex]{cos}^{2}☆= {(-8/17)}^{2}= 64/289[/tex]

Now,

[tex]sin☆= \sqrt{1-{cos}^{2}☆}[/tex]

[tex]sin☆= \sqrt{1- (64/289)}[/tex]

[tex]sin☆= \sqrt{(289-64)/289}[/tex]

[tex]sin☆= \sqrt{225/289}[/tex]

[tex]sin☆= 15/17 [/tex]

Since, 180<theta<270

so, it must lie in 3rd quadrant.

In third quadrant

[tex]sin☆ is -ve[/tex]

[tex]Hence, sin☆= -15/17[/tex]

HOPE IT IS USEFUL

Answer:

sinΘ = - [tex]\frac{15}{17}[/tex]

Step-by-step explanation:

Using the trigonometric identity

sin²x + cos²x = 1

⇒ sinx = ± [tex]\sqrt{1-cos^2x}[/tex]

Since 180 < Θ < 270 then sinΘ < 0

sinΘ = - [tex]\sqrt{1-(-8/17)^2}[/tex]

        = - [tex]\sqrt{1-\frac{64}{289} }[/tex]

        = - [tex]\sqrt{\frac{225}{289} }[/tex] = - [tex]\frac{15}{17}[/tex]

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