Question

Sin (П-2a) * tg (360-a) / 2cos (180+a) * ctg (П/2 +a)

Answer 1

Использую формулы приведения, формулу синуса двойного аргумента    sin2α=2sinα*cosα, получу

(sin(π-2α)*tg(360-α))/(2cos(180+α)ctg(π/2+α))=

((sin2α)tg(-α))/((-cosα)(-tgα))=-((2sinα)*(cosα)(tgα))/((cosα)*(tgα))= -2sinα

Answer 2

To simplify the expression, let's break it down step by step:

1. Recall some trigonometric identities:

   • sin(π - θ) = sin(θ)
   • tan(360° - θ) = tan(θ)
   • cos(180° + θ) = -cos(θ)
   • cot(π/2 + θ) = -tan(θ)

2. Apply these identities to the given expression: sin(π - 2a) * tan(360° - a) / (2 * cos(180° + a) * cot(π/2 + a))

   Since sin(π - 2a) = sin(2a) and tan(360° - a) = tan(a), and cos(180° + a) = -cos(a) and cot(π/2 + a) = -tan(a):

   sin(2a) * tan(a) / (2 * -cos(a) * -tan(a))

3. Simplify further: sin(2a) * tan(a) / (2 * cos(a) * tan(a))

   Since tan(a) / tan(a) = 1:

   sin(2a) / (2 * cos(a))

So, the simplified expression is sin(2a) / (2 * cos(a)).