Question

Sin(2П+α)+cos(П+α)+sin(-α)+cos(α)

Answer 1

$sin(2 \pi + \alpha )+cos( \pi + \alpha )+sin(- \alpha )+cos \alpha =$ $=sin \alpha -cos \alpha -sin \alpha +cos \alpha =0$

Answer 2

We can start by using some trigonometric identities to simplify the expressions:

• sin(2π + α) = sin(2π)cos(α) + cos(2π)sin(α) = 0*cos(α) + (-1)*sin(α) = -sin(α)
• cos(π + α) = -cos(α)
• sin(-α) = -sin(α)
• cos(α) = cos(α)

Substituting these values, we get:

sin(2π + α) + cos(π + α) + sin(-α) + cos(α)

= -sin(α) - cos(α) - sin(α) + cos(α)

= -2sin(α)

Therefore, the final simplified expression is -2sin(α).