Question

Найти sin a, cos a, если cos a/2= -8/17 и sin a/2 >0​

Answer 1

Решение задания приложено

Answer 2

Ответ:

Объяснение: sinα=-240/289       cosα=-161/289.

cos(α/2)=-8/17     sin(α/2)>0      sinα=?       cosα=?

1-cos²(α/2)=sin²(α/2)

sin²(α/2)=1-(-8/17)²=1-(64/289)=(289-64)/289=225/289.

sin(α/2)=±√(225/289)=±15/17.

Так как sin(α/2)>0    ⇒    sin(α/2)=15/17

sinα=2*sin(α/2)*cos(α/2)=2*(15/17)*(-8/17)=-2**15*18/(17*17)=-240/289.

cosα=cos²(α/2)-sin²(α/2)=(-8/17)²-(15/17)²=(64/289)-(225/289)=-161/289.

Answer 3

To find the values of sin(a) and cos(a), we can use the half-angle trigonometric identities. Given that cos(a/2) = -8/17 and sin(a/2) > 0, we can follow these steps:

Step 1: Find sin(a/2) using the Pythagorean identity: sin^2(a/2) + cos^2(a/2) = 1

sin^2(a/2) + (-8/17)^2 = 1

sin^2(a/2) = 1 - (-8/17)^2

sin^2(a/2) = 1 - 64/289

sin^2(a/2) = (289 - 64) / 289

sin^2(a/2) = 225 / 289

sin(a/2) = √(225 / 289)

sin(a/2) = 15 / 17 (we took the positive value because we are given that sin(a/2) > 0)

Step 2: Use the half-angle identity for sin(a): sin(a) = 2 * sin(a/2) * cos(a/2)

sin(a) = 2 * (15/17) * (-8/17)

sin(a) = -240/289

Step 3: Use the half-angle identity for cos(a): cos(a) = cos^2(a/2) - sin^2(a/2)

cos(a) = (-8/17)^2 - (15/17)^2

cos(a) = (64/289) - (225/289)

cos(a) = (64 - 225) / 289

cos(a) = -161/289

So, the values of sin(a) and cos(a) are:

sin(a) = -240/289

cos(a) = -161/289