The following angles are reduced to 2K π + a (0
(50π)/3=8*2π+2π/3
(-50π)/3=-9*2π+4π/3
-108°=-108π/180=-3π/5=-2π+7π/5
-225°=-225π/180=-5π/4=-2π+3π/4
RELATED INFORMATIONS
- 1. The - 1125 ° is reduced to 2K π + α (0 ≤ α < 2 π, K ∈ z)
- 2. The following angles are reduced to 0 to 2 π plus 2K π (K ∈ z) (1) -25/6π (2)-5π (3)-45° (4)400°
- 3. The form of - 1125 ° to 2K π + α (K ∈ Z, 0 ≤ α < 2 π) is () A. −6π−π4B. −6π+7π4C. −8π−π4D. −8π+7π4
- 4. 16 π of the angle is reduced to α + 2K π (K ∈ Z 0)
- 5. What is the temperature of 4.2K
- 6. Given that point a (0, - 1) and point B are moving points on the curve X ^ 2-y ^ 2 + X + 2Y + 3 = 0, then the trajectory equation of the midpoint of line AB is obtained
- 7. It is known that 2x + y = 7 is a quadratic equation of X and y, and the solutions of the equations are positive integers
- 8. If we keep fixed points F1 (- 2,0) and F2 (2,0), then the locus of the moving point P satisfying the following conditions in the plane is hyperbolic A./PF1/-/PF2/=_ +3 B./PF1/-/PF2/=_ +4 C./PF1/-/PF2/=_ +5 D./PF1/^2-/PF2/^2=_ +4 Note: Pf1 / medium / is absolute value_ +3 is positive and negative
- 9. 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)
- 10. How many days are there from October 24, 2010 to today
- 11. The form of transforming - 1125 ° into α + 2K π (0 ≤ α & lt; 2 π, K ∈ z =) is ()
- 12. Change - 18 π / 7 to 2K π + α (0 ≤ α
- 13. -Square of a · (- a) · (- a) · (- a) · (- a) · (- a) · (- a) · 2K + 1
- 14. Calculation: - (a-b) of 2K + 1 power × (B-A) of 2K power × (a-b) of 2k-1 power (k is a positive integer) rt
- 15. What is the negative (2k + 1) power of 2 minus the negative (2k-1) power of 2 plus the negative 2K power of 2?
- 16. 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
- 17. 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?
- 18. 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
- 19. Given a = {x | x = (2k + 1) π, K ∈ Z},}, B = {x | x = (4k-1) π, K ∈ Z}, then the relation between a and B is
- 20. Set M = {XIX = 2K + 1, K ∈ Z}, n = {XIX = 4K ± 1, K ∈ Z} How to prove n ∈ m? It is said in the book that n ∈ M can be explained in detail because m is an odd number set