Question

Ctg(π/2+L)+tg(2π-L); 1-cos*2L+sin*2L; sin x/2= -1; cos4xcos2x-sin4xsin2x=1/2

Answer 1

$ctg( \pi /2+ \alpha ) +tg (2 \pi - \alpha ) = -tg \alpha - tg \alpha = -2tg \alpha$

$sin x/2= -1$
$x/2 = - \pi /2 + 2 \pi k$
$x = - \pi + 4 \pi k$

Answer 2

Let's solve each of these trigonometric expressions one by one:

1. Ctg(π/2+L)+tg(2π-L): We can use the following identities: ctg(x) = 1/tan(x) tan(-x) = -tan(x) Therefore, ctg(π/2+L) = 1/tan(π/2+L) = 1/-cot(L) = -tan(L) Similarly, tg(2π-L) = -tan(L) Hence, Ctg(π/2+L)+tg(2π-L) = -tan(L) + (-tan(L)) = -2tan(L)

2. 1-cos²2L+sin²2L: We can use the following identity: sin²x + cos²x = 1 Therefore, 1 - cos²2L + sin²2L = sin²2L + (1 - cos²2L) = sin²2L + sin²2L = 2sin²2L

3. sinx/2 = -1: There is no solution to this equation. The range of sine function is between -1 and 1, so it can never be equal to -1.

4. cos4x cos2x - sin4x sin2x = 1/2: We can use the following identities: cos(x+y) = cos(x)cos(y) - sin(x)sin(y) cos(x-y) = cos(x)cos(y) + sin(x)sin(y) Therefore, cos4x cos2x - sin4x sin2x = cos(4x-2x) = cos(2x) Now, we can use the following identity: cos²x + sin²x = 1 cos²(2x) + sin²(2x) = 1 cos²(2x) = 1 - sin²(2x) Therefore, cos(2x) = ±sqrt(1 - sin²(2x)) We need to determine the sign based on the given equation. cos4x cos2x - sin4x sin2x = 1/2 2cos4x cos2x - 2sin4x sin2x = 1 We can use the following identities: cos(2x) = ±sqrt(1 - sin²(2x)) sin(2x) = 2sin(x)cos(x) cos(4x) = 2cos²(2x) - 1 sin(4x) = 2sin(2x)cos(2x) Substituting these identities, we get: 2(2cos²(2x) - 1)(±sqrt(1 - sin²(2x))) - 4sin(2x)(±sqrt(1 - sin²(2x))) = 1 Let's simplify this equation: 2(2cos²(2x) - 1)(±cos(2x)) - 4sin(2x)(±sin(2x)) = 1 ±2cos(2x)(4cos²(2x) - 2) - 2sin²(2x) = 1 ±8cos³(2x) - 4cos(2x) - 2sin²(2x) = 1 ±8cos³(2x) - 4cos(2x) - (1-cos²(2x)) = 1 ±8cos³(2x) - 4cos(2x) - 1 + cos²(2x) =