Question

Найдите Cosa, если Sina=-12/13, п<a<3п/2

Answer 1

$Cos A = - \sqrt{1 - sinA^{2} } = - 5/13$
Ответ: - 5/13

Answer 2

Соs²a=1-sin²x=1-(-12/13)²=1-144/169=
25/169
π<a<3π/2
cosa=-√(25/169)=-5/13

Answer 3

To find the value of cos(a) given that sin(a) = -12/13 and π/2 < a < 3π/2, we can use the Pythagorean trigonometric identity: sin^2(a) + cos^2(a) = 1.

We already know that sin(a) = -12/13. Let's solve for cos(a):

sin^2(a) + cos^2(a) = 1 (-12/13)^2 + cos^2(a) = 1 144/169 + cos^2(a) = 1 cos^2(a) = 1 - 144/169 cos^2(a) = (169 - 144) / 169 cos^2(a) = 25 / 169

Now, since π/2 < a < 3π/2, the cosine function is negative. Therefore:

cos(a) = -sqrt(25 / 169)

Calculating the square root:

cos(a) = -5/13

So, the value of cos(a) is -5/13.