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asambeis [7]
3 years ago
14

8. Angle α is in quadrant III, and angle β is in quadrant II. If sin α = –4∕5 and sin β = 1∕2, find cos (α + β).

Mathematics
1 answer:
katovenus [111]3 years ago
6 0

Answer:

D

Step-by-step explanation:

Using the addition formula for cosine

cos(α + β) = cosαcosβ - sinαsinβ

Given α in quadrant 3 then and sinα = - \frac{4}{5} = \frac{opposite}{hypotenuse} , then

cosα < 0 in the third quadrant

The adjacent side of the triangle = 3  since 3- 4- 5 right triangle

Thus

cosα = \frac{adjacent}{hypotenuse} = - \frac{3}{5}

---------------------------------------

Given β in second quadrant and sinβ = \frac{1}{2} = \frac{opposite}{hypotenuse} , then

cosβ < 0 in the second quadrant

The adjacent side of the triangle = \sqrt{3} since 1- \sqrt{3} - 2 right triangle

Thus

cosβ = \frac{adjacent}{hypotenuse} = - \frac{\sqrt{3} }{2}

Hence

cos(α + β )

= ( - \frac{3}{5} × - \frac{\sqrt{3} }{2} ) - (- \frac{4}{5} × \frac{1}{2} )

= \frac{3\sqrt{3} }{10} + \frac{4}{10}

= \frac{3\sqrt{3}+4 }{10} → D

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