3 years ago

Can you help me please??

Mathematics

1 answer:

max2010maxim [7]3 years ago

3 0

Answer:

Its either B or C, I'm not sure which one tho, but I'm betting on C, sorry if its wrong.

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3 years ago

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If sine ⁡theta equals negative 8/17 and the terminal side of theta lies in quadrant IV, find cosine ⁡ theta .

Lelechka [254]

$\sin^2\theta+\cos^2\theta=1\implies \cos\theta=\pm\sqrt{1-\sin^2\theta}$

Since $\sin\theta=-\dfrac8{17}$ , it follows that

$\cos\theta=\pm\sqrt{1-\dfrac{64}{289}}=\pm\sqrt{\dfrac{225}{289}}=\pm\dfrac{15}{17}$

where the sign depends on the location of the terminal side of the angle. Since the terminal side lies in the fourth quadrant, that means the cosine of the angle is positive, so

$\cos\theta=\dfrac{15}{17}$