Question

Найдите: sin2a, cos2a,tg2a если sina= 24/25 и пи/2 < a < пи решите уравнение: cos2x = 1 - cos ( пи/2 - x )

Answer 1

Основные тригонометрические равенства

Answer 2

To find the values of sin(2a), cos(2a), and tan(2a) when sin(a) = 24/25 and π/2 < a < π, we can use trigonometric identities.

First, we know that sin(a) = 24/25. Let's find cos(a) using the Pythagorean identity:

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

(24/25)^2 + cos^2(a) = 1

cos^2(a) = 1 - (24/25)^2 cos^2(a) = 1 - 576/625 cos^2(a) = (625 - 576) / 625 cos^2(a) = 49 / 625

cos(a) = ±√(49 / 625) = ±7/25

Since we are given that a is in the second quadrant (π/2 < a < π), the cosine value should be negative:

cos(a) = -7/25

Now, let's find sin(2a):

sin(2a) = 2 * sin(a) * cos(a) sin(2a) = 2 * (24/25) * (-7/25) sin(2a) = -48/625

Next, we'll find cos(2a):

cos(2a) = cos^2(a) - sin^2(a) cos(2a) = (7/25)^2 - (24/25)^2 cos(2a) = (49/625) - (576/625) cos(2a) = (49 - 576) / 625 cos(2a) = -527/625

Finally, we'll find tan(2a):

tan(2a) = sin(2a) / cos(2a) tan(2a) = (-48/625) / (-527/625) tan(2a) = 48/527

So, the values are: sin(2a) = -48/625 cos(2a) = -527/625 tan(2a) = 48/527

Now, let's solve the equation: cos(2x) = 1 - cos(π/2 - x)

We'll use the trigonometric identity: cos(π/2 - x) = sin(x)

cos(2x) = 1 - sin(x)

Using the double-angle identity for cosine:

cos(2x) = 1 - 2 * sin^2(x)

Now, we can rewrite the equation:

1 - 2 * sin^2(x) = 1 - sin(x)

Now, move everything to one side:

2 * sin^2(x) - sin(x) = 0

Factor out sin(x):

sin(x) * (2 * sin(x) - 1) = 0

Now, set each factor to zero and solve for x:

1. sin(x) = 0

Since 0 ≤ x ≤ π, the solutions for this part are x = 0 and x = π.

2. 2 * sin(x) - 1 = 0

2 * sin(x) = 1

sin(x) = 1/2

The solution for this part is x = π/6.

So, the solutions to the equation are: x = 0, x = π, and x = π/6.