Question

If cos ⁡θ=−8/17 and 180°θ < 270°, what is sin θ?

Answer 1

Answer:

sinΘ = - $\frac{15}{17}$

Step-by-step explanation:

Using the trigonometric identity

sin²x + cos²x = 1

⇒ sinx = ± $\sqrt{1-cos^2x}$

Since 180 < Θ < 270 then sinΘ < 0

sinΘ = - $\sqrt{1-(-8/17)^2}$

        = - $\sqrt{1-\frac{64}{289} }$

        = - $\sqrt{\frac{225}{289} }$ = - $\frac{15}{17}$

Answer 2

Here is your answer

$[I am using ☆ instead of theta]$

$cos☆= -8/17$

${cos}^{2}☆= {(-8/17)}^{2}= 64/289$

Now,

$sin☆= \sqrt{1-{cos}^{2}☆}$

$sin☆= \sqrt{1- (64/289)}$

$sin☆= \sqrt{(289-64)/289}$

$sin☆= \sqrt{225/289}$

$sin☆= 15/17$

Since, 180<theta<270

so, it must lie in 3rd quadrant.

In third quadrant

$sin☆ is -ve$

$Hence, sin☆= -15/17$

HOPE IT IS USEFUL