Question

CosA,если sinA=-15/17;П<a<3П/2​

Answer 1

Ответ:

Задачка на основное тригонометрическое тождество

Пошаговое объяснение:

Answer 2

To find cos(A) given that sin(A) = -15/17 and A lies in the interval (π, 3π/2), we can use the identity sin^2(A) + cos^2(A) = 1.

Since sin(A) = -15/17, we can square both sides to get:

sin^2(A) = (-15/17)^2 sin^2(A) = 225/289

Next, we can rearrange the identity to solve for cos^2(A):

cos^2(A) = 1 - sin^2(A) cos^2(A) = 1 - 225/289 cos^2(A) = 289/289 - 225/289 cos^2(A) = 64/289

Taking the square root of both sides, we find:

cos(A) = ±√(64/289) cos(A) = ±8/17

Since A lies in the interval (π, 3π/2), cosine is negative in that range. Therefore, we can conclude that:

cos(A) = -8/17