Question

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).

Answer 1

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

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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