# How do you solve #sin( alpha + beta) # given #sin alpha = 12/13 # and #cos beta = -4/5#?

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To solve sin(α + β) when sin α = 12/13 and cos β = -4/5, you can use the sum-to-product identities:

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

Given sin α = 12/13 and cos β = -4/5:

sin α = 12/13 cos β = -4/5

We need to find cos α and sin β.

Since sin α = 12/13, we can find cos α using the Pythagorean identity:

cos α = ±√(1 - sin² α)

cos α = ±√(1 - (12/13)²) cos α = ±√(1 - 144/169) cos α = ±√((169 - 144)/169) cos α = ±√(25/169) cos α = ±(5/13)

Given that α is in the first or second quadrant, where cos α > 0, we take the positive value:

cos α = 5/13

Now, we need to find sin β. Since cos β = -4/5, we can find sin β using the Pythagorean identity:

sin β = ±√(1 - cos² β)

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

Given that β is in the second or third quadrant, where sin β < 0, we take the negative value:

sin β = -3/5

Now, plug the values into the formula for sin(α + β):

sin(α + β) = (12/13)(-4/5) + (5/13)(-3/5)

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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