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Given that tanα=4/3 (α is in Q1) and cosβ= -4/5 (β is in Q3), find cos(α+β)

User MSingh
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Answer:

We can use the following formula for the cosine of the sum of two angles:

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

To use this formula, we need to find the values of cos(α), sin(α), cos(β), and sin(β).

Since tan(α) = 4/3 and α is in the first quadrant, we can use the Pythagorean identity to find sin(α) and cos(α):

tan^2(α) + 1 = sec^2(α)

(4/3)^2 + 1 = sec^2(α)

16/9 + 1 = sec^2(α)

25/9 = sec^2(α)

sec(α) = 3/5

cos(α) = 1/sec(α) = 5/3

sin(α) = tan(α) * cos(α) = (4/3) * (5/3) = 20/9

Since cos(β) = -4/5 and β is in the third quadrant, we can use the Pythagorean identity again to find sin(β):

sin^2(β) + cos^2(β) = 1

sin^2(β) = 1 - cos^2(β)

sin(β) = -√(1 - cos^2(β)) (since sin(β) is negative in Q3)

sin(β) = -√(1 - (-4/5)^2) = -√(1 - 16/25) = -√(9/25) = -3/5

Now we can substitute these values into the formula for cos(α+β):

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

cos(α + β) = (5/3) * (-4/5) - (20/9) * (-3/5)

cos(α + β) = -4/3 + 4/3

cos(α + β) = 0

Therefore, cos(α+β) = 0.

User Jwehrle
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