Question

Sin(2π+a)+cos(π+a)+sin(-a)+cos(-a)​

Answer 1

Sin a - cos a - sin a + cos a = 0

Здесь работают формулы приведения и четность, нечетность(sin(-a)=-sina; cos (-a)= cosa;tg(-a)=-tga; ctg(-a)=-ctg a)

Удачки:))

Answer 2

To simplify the expression Sin(2π + a) + Cos(π + a) + Sin(-a) + Cos(-a), we can use trigonometric identities and properties. Let's break it down step by step:

1. Sin(2π + a) = Sin(2π) * Cos(a) + Cos(2π) * Sin(a) = 0 * Cos(a) + 1 * Sin(a) = Sin(a)

2. Cos(π + a) = Cos(π) * Cos(a) - Sin(π) * Sin(a) = -1 * Cos(a) - 0 * Sin(a) = -Cos(a)

3. Sin(-a) = -Sin(a) (since the sine function is an odd function)

4. Cos(-a) = Cos(a) (since the cosine function is an even function)

Now, we can substitute these simplified expressions back into the original equation:

Sin(2π + a) + Cos(π + a) + Sin(-a) + Cos(-a) = Sin(a) + (-Cos(a)) + (-Sin(a)) + Cos(a) = Sin(a) - Cos(a) - Sin(a) + Cos(a)

Since the like terms cancel each other out, the final simplified expression is:

Sin(a) - Sin(a) + Cos(a) - Cos(a) = 0

Therefore, the value of the expression Sin(2π + a) + Cos(π + a) + Sin(-a) + Cos(-a) is 0.