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

D

Explanation:

Answer 2

D

Explanation:

Using the addition formula for cosine

cos(α + β) = cosαcosβ - sinαsinβ

Given α in quadrant 3 then and sinα = -
`(4)/(5)` =
`(opposite)/(hypotenuse)` , then

cosα < 0 in the third quadrant

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

Thus

cosα =
`(adjacent)/(hypotenuse)` = -
`(3)/(5)`

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

Given β in second quadrant and sinβ =
`(1)/(2)` =
`(opposite)/(hypotenuse)` , then

cosβ < 0 in the second quadrant

The adjacent side of the triangle =
`√(3)` since 1-
`√(3)` - 2 right triangle

Thus

cosβ =
`(adjacent)/(hypotenuse)` = -
`(√(3) )/(2)`

Hence

cos(α + β )

= ( -
`(3)/(5)` × -
`(√(3) )/(2)` ) - (-
`(4)/(5)` ×
`(1)/(2)` )

=
`(3√(3) )/(10)` +
`(4)/(10)`

=
`(3√(3)+4 )/(10)` → D