التصنيف: مادة الرياضيات في التعليم الثانوي

صيغ اعتيادية

معادلات مثلثية

Z تنتمي الى k

sin(a) = sin(b)
alors a = b + 2kn
ou a =n – b + 2kn

cos(a) = cos(b)
alors a = b + 2kn
ou a = -b + 2kn

tan(a) = tan(b)
alors a = b + kn;

2-صيغ مجموع

sin(a + b) = sin(a)cos(b) + sin(b)cos(a)
sin(a – b) = sin(a)cos(b) – sin(b)cos(a)
cos(a + b) = cos(a)cos(b) – sin(a)sin(b)
cos(a – b) = cos(a)cos(b) + sin(a)sin(b)
tan(a + b) = (tan(a) + tan(b)) / (1 – tan(a)tan(b))
tan(a – b) = (tan(a) – tan(b)) / (1 + tan(a)tan(b))

sin(p) + sin(q) = 2sin( p + q)/2)cos (p – q)/2)
sin(p) – sin(q) = 2sin((p – q)/2)cos((p + q)/2)
cos(p) + cos(q) = 2cos((p + q)/2)cos((p – q)/2)
cos(p) – cos(q) = -2sin((p + q)/2)sin((p – q)/2)
tan(p) + tan(q) = sin(p + q) / (cos(p)cos(q))
tan(p) – tan(q) = sin(p – q) / (cos(p)cos(q))

sin(a)sin(b) = (1/2)(cos(a – b) – cos(a + b))
cos(a)cos(b) = (1/2)(cos(a + b) + cos(a – b))
sin(a)cos(b) = (1/2)(sin(a + b) + sin(a – b))

3-صيغ مضاعفات

sin(2a) = 2sin(a)cos(a)
= 2tan(a) / (1 + tan²(a))

cos(2a) = cos²a – sin²a
= 2cos²a – 1
= 1 – 2sin²a

tan(2a) = 2tan(a) / (1 – tan²(a))

sin²(a) = (1 – cos(2a)) / 2
cos²(a) = (1 + cos(2a)) / 2
tan²(a) = (1 – cos(2a)) / (1 + cos(2a))

tan(a) = sin(2a) / (1 + cos(2a))
= (1 – cos(2a)) / sin(2a)

sin(a) = 2t / (1 + t²)
cos(a) = (1 – t²) / (1 + t²)
tan(a) = 2t / (1 – t²)