Question

Найти cosx,если sin x = -15/17 и П<x<3П/2

Answer 1

По основному тригонометрическому тождеству: sin^2X + cos^2X = 1, значит cos^2X = 1 - sin^2X = 1 - 225/289 = 64/289. Значит cosX = плюс/минус корень квадратный из 64/289 = плюс/минус 8/17. При условии, что П<x<3П/2, cosX = - 8/17. ^2 - вторая степень.

Answer 2

To find the value of cos(x) given sin(x) = -15/17 and π < x < 3π/2, we can use the Pythagorean trigonometric identity, which states that sin^2(x) + cos^2(x) = 1.

Given sin(x) = -15/17, we can find cos(x) using the following steps:

Step 1: Find cos^2(x): cos^2(x) = 1 - sin^2(x) cos^2(x) = 1 - (-15/17)^2 cos^2(x) = 1 - 225/289 cos^2(x) = (289 - 225)/289 cos^2(x) = 64/289

Step 2: Find cos(x): Since x is in the second quadrant (π < x < 3π/2), cos(x) is negative. cos(x) = -sqrt(64/289) cos(x) = -8/17

So, the value of cos(x) is -8/17.