Ctg t * sin(-t)+cos(2π-t)

Предыдущий
вопрос Следующий
вопрос

Let's simplify the given expression:

ctg(t) * sin(-t) + cos(2π - t)

First, note that ctg(t) = 1/tan(t), so ctg(t) * sin(-t) can be rewritten as sin(-t) / tan(t). Since sin(-t) = -sin(t), and tan(t) = sin(t) / cos(t), we get:

sin(-t) / tan(t) = -sin(t) / (sin(t) / cos(t)) = -cos(t)

Next, cos(2π - t) can be simplified using the difference of angles formula for cosine: cos(2π - t) = cos(2π)cos(t) + sin(2π)sin(t). Since cos(2π) = 1 and sin(2π) = 0, this simplifies to:

cos(2π - t) = cos(t)

Putting it all together, the expression simplifies to:

Therefore, ctg(t) * sin(-t) + cos(2π - t) simplifies to 0.