Question

please help me

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

Answer 1

Starting from the fundamental trigonometric equation, we have

$\cos^2(\alpha)+\sin^2(\alpha)=1 \iff \sin(\alpha)=\pm\sqrt{1-\cos^2(\alpha)}$

Since $180<\alpha<270$, we know that the angle lies in the third quadrant, where both sine and cosine are negative. So, in this specific case, we have

$\sin(\alpha)=-\sqrt{1-\cos^2(\alpha)}$

Plugging the numbers, we have

$\sin(\alpha)=-\sqrt{1-\dfrac{64}{289}}=-\sqrt{\dfrac{225}{289}}=-\dfrac{15}{17}$

Now, just recall that

$\sin(-\alpha)=-\sin(\alpha)$

to deduce

$\sin(-\alpha)=-\sin(\alpha)=-\left(-\dfrac{15}{17}\right)=\dfrac{15}{17}$

Answer 2

Answer:

15/17

Step-by-step explanation:

opposite² = 17² - 8²

opposite² = 225

opposite = 15

sin a = - 15/17

(sin is -ve in the third quadrant)

sin(-X) = -sinX

sin(-a) = -sin(a)

= -(-15/17)

= 15/17