Question

Найти sin2a, если cosa = -8/17 и п<а<3п/2 распишите подробно пожалуйста

Answer 1

Sin2x= 2 sinxcosx    Нам известен cos. для того чтобы решить нам нужен еще синус. sin²x=√(1-cos²)
sin²a=1-64/289=289/289-64/289=√225/289=15/17=-15/17    Знак определяем п<a<3п/2  3 четверть. у синуса знак -. значит  -15/17
Находим sin2x = 2 * (-15/17) * (-8/17) =240/289

Answer 2

Sin2a=2sina *cosa
sIn^2a+cos^2a=1;   sina=-√(1-cos^2a);  3-ья етверть
sina=-√(1-64/289)=-√225/289=-15/17
sin2a=2*(-15/17)*(-8/17)=(2*15*8) /(17^2)=20=240/289

Answer 3

Finding sin(2a) given cos(a) = -8/17 and π < a < 3π/2

To find sin(2a) given cos(a) = -8/17 and the range of a, we can use the double-angle identity for sine, which states that sin(2a) = 2sin(a)cos(a).

First, we need to find sin(a) using the given value of cos(a) = -8/17. Then, we can use the double-angle identity to find sin(2a).

Calculating sin(a)

We know that cos(a) = -8/17. To find sin(a), we can use the Pythagorean identity, which states that sin^2(a) + cos^2(a) = 1.

Using this identity, we can calculate sin(a): sin^2(a) + (-8/17)^2 = 1 sin^2(a) + 64/289 = 1 sin^2(a) = 1 - 64/289 sin^2(a) = 225/289 sin(a) = ±√(225/289) sin(a) = ±15/17

Since a is in the range π < a < 3π/2, sin(a) is negative. Therefore, sin(a) = -15/17.

Calculating sin(2a)

Now that we have sin(a) = -15/17, we can use the double-angle identity for sine to find sin(2a): sin(2a) = 2sin(a)cos(a) sin(2a) = 2 * (-15/17) * (-8/17) sin(2a) = 240/289

So, sin(2a) = 240/289.

This calculation is based on the provided information and the trigonometric identities.