Find the exact value for the expression under the given conditions. (cosα+β), sinα = −8/ 17for α in Quadrant III and cosβ = −8/15 for β in Quadrant II.

Answer 1

We have the trigonometric identity:

cos(α+β) = cos(α)cos(β) - sin(α)sin(β)

We know that sin(α) = -8/17 and that α is in quadrant 3.

We also know that cos(β) = −8/15 and that β is in quadrant 2

We need to find cos(α). This cosine must be negative, since the angle α is in quadrant 3. In what follows, sqrt is the square root.

1 = cos(α)^2+sin(α)^2

1 - sin(α)^2 = cos(α)^2

+-sqrt(1 - sin(α)^2) = cos(α)

Since the value must be negative, we only take the negative square root:

cos(α) = -sqrt(1 - sin(α)^2) = -sqrt(1 - (-8/17)^2) = -sqrt(1-64/289) = -sqrt(225/289) = -15/17

We need to also find sin(β). This sine must be positive, since β is in quadrant II.

1 = cos(β)^2+sin(β)^2

1-cos(β)^2 = sin(β)^2

+-sqrt(1-cos(β)^2) = sin(β)

Since the value must be positive, we take the positive of the square root.

sin(β) = sqrt(1-cos(β)^2) = sqrt(1 - (−8/15)^2) = sqrt(1 - 64/225) = sqrt(161/225) = sqrt(161)/15

We now have all our values and can return to the initial trigonometric identity:

cos(α+β) = cos(α)cos(β) - sin(α)sin(β) = -15/17 * -8/15 - (-8/17) * sqrt(161)/15 = 8/17 + 8sqrt(161)/255 = (120 + 8sqrt(161))/255

Answer 2

sinα = -8/17 and α lies in Quadrant 3. So, cosα < 0 and cosα = -√(1 - sin²α) = -15/17

cosβ = -8/15 and β lies in Quadrant 2. So, sinβ > 0 and sinβ = √(1 - cos²β) = √(161)/15

cos(α+β) = cosαcosβ - sinαsinβ = (-15/17)(-8/15) - (-8/17)(√161 / 15) = (120 + 8√161) / 255

Answer 3

Determine sin β and cos α

Use the formula for the cosine of the sum of two angles.