If tan α = − 21/20 and cot β = 12 /5 for a second-quadrant angle α and a third-quadrant angle β, find the following

Question

sin(α + β)

 cos(α + β)

tan(α + β)

sin(α − β)

cos(α − β)

tan(α − β)

Answer 1

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Answer 2

Since tanα = -21/20, we'll use the Pythagorean Theorem to find the length of the hypotenuse and then find sinα and cosα

a² + b² = c²

21² + 20² = c²

441 + 400 = c²

841 = c²

29 = c

α is in Quadrant II so sinα = 21/29 and cosα = -20/29

Since cotß = 12/5, we'll use the Pythagorean Theorem to find the length of the hypotenuse and then find sinß and cosß

a² + b² = c²

5² + 12² = c²

25 + 144 = c²

169 = c²

13 = c

ß is in Quadrant III so tanß= 5/12, sinß = -5/13 and cosß = -12/13

Now we can use the Sum and Difference formulas for sine, cosine, and tangent to complete the problem

sin(α + ß) = sin(α)cos(ß) + cos(α)sin(ß)

sin(α + ß) = (21/29)(-12/13) + (-20/29)(-5/13)

sin(α + ß) = -152/377

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

cos(α + ß) = (-20/29)(-12/13) - (21/29)(-5/13)

cos(α + ß) = 345/377

tan(α + ß) = [tan(α) + tan(ß)] / [1 - tan(α)tan(ß)]

tan(α + ß) = [(-21/20) + (5/12)] / [1 - (-21/20)(5/12)]

tan(α + ß) = -152/345

sin(α - ß) = sin(α)cos(ß) - cos(α)sin(ß)

sin(α - ß) = (21/29)(-12/13) - (-20/29)(-5/13)

sin(α - ß) = -352/377

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

cos(α - ß) = (-20/29)(-12/13) + (21/29)(-5/13)

cos(α - ß) = 135/377

tan(α - ß) = [tan(α) - tan(ß)] / [1 + tan(α)tan(ß)]

tan(α - ß) = [(-21/20) - (5/12)] / [1 + (-21/20)(5/12)]

tan(α - ß) = -352/135