Question

Cos (a+b), если sin a= 4/5; cos b= -3/5П/2 < а < П, П/2 < b < П

Answer 1

Ответ: cos(α+β)=-7/25.

Пошаговое объяснение:

sinα=4/5      cosβ=-3/5     π/2<α<π     π/2<β<π      cos(α+β)=?

cosα=±√(1-sin²α)=±√(1-(4/5)²)=±√(1-(16/25))=±√(9/25)=±3/5.

Так как π/2<α<π    ⇒    cosα=-3/5.

sinβ=±√(1-(-3/5)²)=±√(1-(9/25))=±√(16/25)=±4/5.

Так как π/2<β<π    ⇒     sinβ=4/5.

cos(α+β)=cosα*cosβ-sinα*sinβ=(-3/5)*(-3/5)-(4/5)*(4/5)=

=(9/25)-(16/25)=(9-16)/25=-7/25.

Answer 2

To find the value of cos(a + b), we'll first need to determine the signs of sin(a) and cos(b) based on the given information:

Given:

1. sin(a) = 4/5
2. cos(b) = -3/5
3. π/2 < a < π (90° < a < 180°)
4. π/2 < b < π (90° < b < 180°)

Let's start by finding the value of cos(a) using the Pythagorean trigonometric identity: cos^2(a) + sin^2(a) = 1

cos^2(a) = 1 - sin^2(a) cos^2(a) = 1 - (4/5)^2 cos^2(a) = 1 - 16/25 cos^2(a) = 9/25

cos(a) = ±√(9/25) = ±3/5

Since a is in the second quadrant (π/2 < a < π), the value of cos(a) will be negative: cos(a) = -3/5.

Next, we'll find the value of sin(b) using the Pythagorean trigonometric identity: cos^2(b) + sin^2(b) = 1

sin^2(b) = 1 - cos^2(b) sin^2(b) = 1 - (-3/5)^2 sin^2(b) = 1 - 9/25 sin^2(b) = 16/25

sin(b) = ±√(16/25) = ±4/5

Since b is in the second quadrant (π/2 < b < π), the value of sin(b) will be positive: sin(b) = 4/5.

Now we have the following values: sin(a) = 4/5 cos(a) = -3/5 sin(b) = 4/5 cos(b) = -3/5

Now, we can use the sum-to-product identity for cosine to find cos(a + b): cos(a + b) = cos(a) * cos(b) - sin(a) * sin(b)

cos(a + b) = (-3/5) * (-3/5) - (4/5) * (4/5) cos(a + b) = 9/25 - 16/25 cos(a + b) = (9 - 16)/25 cos(a + b) = -7/25

So, the value of cos(a + b) is -7/25.