Question

(sin(pi÷2+a)+sin(2pi+a))÷(2cos(-a)sin(-a)+1)

Answer 1

(sin(π/2+a)+sin(2π+a))/(2cos(-a)sin(-a)+1)

=(cosa+sina)/(1-sin2a)

Answer 2

Решение внизу на фото

Answer 3

To simplify the expression (sin(pi÷2+a) + sin(2pi+a)) ÷ (2cos(-a)sin(-a) + 1), we'll start by using some trigonometric identities to reduce it.

1. Trigonometric identities used:

   • sin(π/2 + θ) = cos(θ)
   • sin(θ + 2π) = sin(θ)
   • cos(-θ) = cos(θ)
   • sin(-θ) = -sin(θ)

2. Applying the identities: (cos(a) + sin(2π + a)) ÷ (2cos(a)(-sin(a)) + 1)

3. Further simplification: Since sin(2π + a) = sin(a), the expression becomes: (cos(a) + sin(a)) ÷ (-2cos(a)sin(a) + 1)

That's the simplest form of the given expression.