Question

Sinx= 3/5; х есть ( п; п/2) Найти: а) cos x, б) ctgx

Answer 1

а) sin²x + cos²x = 1;

9/25 + cos²x = 1;

cos²x = 1- 9/25

cos²x = 16/25

cosx = +/- 4/5; Косинус на промежутке (p/2; p) - отрицателен, поэтому

cosx = - 4/5 = -0.8

б) ctg x = cosx/sinx = - 0.8 / 0.6 = -1 1/3

Answer 2

To find the values of cos(x) and ctg(x) given that sin(x) = 3/5 and x lies in the interval (π, π/2), we can use the Pythagorean identity and the definition of the cotangent function.

1. We know that sin(x) = 3/5, and x lies in the second quadrant (between π/2 and π). In the second quadrant, sin(x) is positive, and cos(x) is negative. The Pythagorean identity is:

sin^2(x) + cos^2(x) = 1

Substitute sin(x) = 3/5 into the equation:

(3/5)^2 + cos^2(x) = 1 9/25 + cos^2(x) = 1 cos^2(x) = 1 - 9/25 cos^2(x) = 16/25

Since x is in the second quadrant, cos(x) is negative, so:

cos(x) = -√(16/25) = -4/5

2. To find the cotangent (ctg(x)), we know that:

ctg(x) = 1/tan(x)

The tangent function is defined as:

tan(x) = sin(x)/cos(x)

Substitute sin(x) = 3/5 and cos(x) = -4/5:

tan(x) = (3/5) / (-4/5) tan(x) = -3/4

Now, find the cotangent:

ctg(x) = 1/tan(x) ctg(x) = 1 / (-3/4) ctg(x) = -4/3

So, the solutions are: a) cos(x) = -4/5 b) ctg(x) = -4/3