The following angles are reduced to 0 to 2 π plus 2K π (K ∈ z) (1) -25/6π (2)-5π (3)-45° (4)400°
(1) -25π/6
=24π/6-π/6
=-4π-π/6
=-6π+2π-π/6
=-6π+11π/6
(2)-5π
=-6π+π
(3)-45°
=-π/4
=-2π+7π/4
(4)400°
=400/180 *π
=2π+2π/9
RELATED INFORMATIONS
- 1. The form of - 1125 ° to 2K π + α (K ∈ Z, 0 ≤ α < 2 π) is () A. −6π−π4B. −6π+7π4C. −8π−π4D. −8π+7π4
- 2. 16 π of the angle is reduced to α + 2K π (K ∈ Z 0)
- 3. What is the temperature of 4.2K
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- 7. Simple calculation by factorization (1)22.4^2-27.6^2 (2)2.03*2.5+4.83*2.5+3.14*2.5 (3)1.25*14^2-125*8.6^2 (4)102^2-2*(100^2-2^2)+98^2 (5)27*1.8^2-12*1.2^2 (6)(1-1/2^2)(1-1/3^2)(1-1/4^2)…… (1-1/2008^2)
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- 11. The - 1125 ° is reduced to 2K π + α (0 ≤ α < 2 π, K ∈ z)
- 12. The following angles are reduced to 2K π + a (0
- 13. The form of transforming - 1125 ° into α + 2K π (0 ≤ α & lt; 2 π, K ∈ z =) is ()
- 14. Change - 18 π / 7 to 2K π + α (0 ≤ α
- 15. -Square of a · (- a) · (- a) · (- a) · (- a) · (- a) · (- a) · 2K + 1
- 16. Calculation: - (a-b) of 2K + 1 power × (B-A) of 2K power × (a-b) of 2k-1 power (k is a positive integer) rt
- 17. What is the negative (2k + 1) power of 2 minus the negative (2k-1) power of 2 plus the negative 2K power of 2?
- 18. Simplification: - (a-b) to the power of 2K + 1 * (B-A) to the power of 2K * (a-b) to the power of 2k-3. (k is a positive integer) With a multiply sign, write as clearly as a question
- 19. An exciting Let a = {x | x = 2K, K belongs to Z (integer set)} Let a = {x | x = 2K, K belong to Z (integer set)}, B = {x | x = 2K + 1, K belong to Z}, C = {x | x = 2 (K + 1), K belong to Z}, d = {x | x = 2k-1, K belong to Z}, in a, B, C, D, which sets are equal, which sets are empty, which sets are union Z? Which friend is on the first floor?
- 20. Given M1 = {m | M = x square - y square, X is an integer, y is an integer}, M2 = {m | M = 2K + 1 or M = 4K, K is an integer} to prove M 1 = m 2