(parity refers to π2? n+απ2？ Whether the integer nn in n+α is odd or even, consider α α as an acute angle when looking at the quadrant)
sin(π2？ n+α)={(？ 1)π2sinα，n？ Even number? (? 1)n+ 12cosα，n？ Odd numbers? Sin? (π2? n+α)={(？ 1)π2sin？ α，n？ Even number? (? 1)n+ 12cos？ α，n？ Odd numbers? cos(π2？ n+α)={(？ 1)n2cosα，n？ Even number? (? 1)n+ 12sinα，n？ Odd numbers?
Formula: (1)? sin(α+2kπ)= sinαsin(α+2kπ)= sinα； cos(α+2kπ)=cosαcos？ (α+2kπ)=cos？ α? ; tan(α+2kπ)=tanαtan？ (α+2kπ)=tan？ α.
It is easy to get from the definition of trigonometric function.
(2)? sin(π+α)=？ sinαsin？ (π+α)=? Sin? α; cos(π+α)=？ cosαcos？ (π+α)=? Because? α; ? tan(π+α)=tanαtan？ (π+α)=tan？ α.
It is proved that as shown in Figure 2, the terminal edge of α α intersects the unit circle at P 1(x, y)P 1(x, y), then the terminal edge of π+α π+α intersects the unit circle at P2(x, y)P2(x, y).
Obviously P2P2 and P 1P 1 are symmetrical about the origin, so P2 (? x，？ y)P2(？ x，？ y)。