Question

Вычислите sin2α и cos2α, если sinα = - 5/13 и π < α < 3π/2

Answer 1

Дано: sinα = - 5/13 и π < α < 3π/2 (это третья четверть).

cos α = √(1 - (- 5/13)²) =  √(1 - (25/169)) = √(144/169) = -12/13 (для 3 четверти).

Переходим к двойному углу 2α .

sin2α = 2sin α*cosα = 2*(- 5/13)*(-12/13) = 120/169 (это 1 четверть).

cos 2α = cos²α - sin²α = (-12/13)² - (-5/13)² = (144/169) - (25/169) = 119/169.

Answer 2

To calculate sin(2α) and cos(2α), we can use the double-angle formulas for trigonometric functions:

1. sin(2α) = 2 * sin(α) * cos(α)
2. cos(2α) = cos^2(α) - sin^2(α)

Given that sin(α) = -5/13 and π < α < 3π/2, we need to find cos(α) first. We can use the Pythagorean identity to find cos(α):

cos^2(α) + sin^2(α) = 1

Now, substitute sin(α) = -5/13 into the equation:

cos^2(α) + (-5/13)^2 = 1 cos^2(α) + 25/169 = 1 cos^2(α) = 1 - 25/169 cos^2(α) = 144/169

Taking the square root of both sides (we use the positive square root since α is in the third quadrant), we get:

cos(α) = √(144/169) = 12/13 (positive since α is in the third quadrant)

Now, we can use the double-angle formulas to find sin(2α) and cos(2α):

1. sin(2α) = 2 * sin(α) * cos(α) sin(2α) = 2 * (-5/13) * (12/13) = -120/169

2. cos(2α) = cos^2(α) - sin^2(α) cos(2α) = (12/13)^2 - (-5/13)^2 cos(2α) = 144/169 - 25/169 cos(2α) = 119/169

So, sin(2α) = -120/169 and cos(2α) = 119/169.