class="bi x0 y0 w1 h1"
(a
2
+ b
2
+ c
2
+ d
2
)(x
2
+ y
2
+ z
2
+ t
2
) = (ax − by − cz − dt)
2
+ (bx +
ay −dz + ct)
2
+ (cx + dy + az −dt)
2
+ (dx − cy + bz + at)
2
ax−by−cz−dt = 0 bx+ay−dz+ct = 0
cx + dy + az − dt = 0 dx − cy + bz + at = 0
a = b = c = d = 0 x = y = z = t = 0
(a
2

+ b
2
2
+ ··· + b
2
n
) − (a
1
b
1
+ a
2
b
2
+ ··· + a
n
b
n
)
2
=
= (a
1
b
2
− a
2
b
1
)

) = (a
1
+ a
2
+ ··· + a
n
)
2
a
1
= a
2
= ··· = a
n
(x − y)
2
+ (y − z)
2
+ (z − x)
2
= (y + z −
2x)
2
+ (z + x − 2y)
2
+ (x + y −2z)
2
x = y = z
(a
2

)
3
(p
2
− q
2
)
4
+ (2pq + q
2
)
4
+ (2pq + p
2
)
4
= 2(p
2
+ pq + q
2
)
4
X
2
+ XY + Y
2
= Z
3
X = q
3

n
= (iy −kx)
n
+ (iz −ky)
n
+ (ix −kz)
n
n = 0, 1, 2, 4 i i
2
= −1
x
n
+ (x + 3)
n
+ (x + 5)
n
+ (x + 6)
n
+ (x + 9)
n
+ (x +
10)
n
+ (x + 12)
n
+ (x + 15)
n
= (x + 1)
n
+ (x + 2)

+ d
2
)
(a
2
− b
2
+ c
2
− d
2
)
2
+ 2(ab − bc + dc + ad)
2
= (a
2
+ b
2
+ c
2
+ d
2
)
2
−
2(ab − ad + bc + dc)
2
(a
2

= 4(a
4
+b
4
+c
4
)+24(a
2
b
2
+b
2
c
2
+c
2
a
2
)
s = a + b + c = 2p

sym
s(s − 2b)(s − 2c) = (s − 2a)(s − 2b)(s − 2c) + 8abc

sym
a(p − a)
2
= abc − 2(p − a)(p − b)(p − c )
s = a + b + c 2δ = a
2

+ b
2
+ c
2
− ab − bc − ca)
a, b, c
(a + b + c)
3
−

sym
(a + b − c)
3
(a − b)
3
+ (b − c)
3
+ (c − a)
3
= 3(a − b)(b − c)(c − a)
[(a − b)
2
+ (b − c)
2
+ (c − a)
2
]
2
= 2[(a − b)
4

2
+b
2
+c
2
2
•
a
3
+b
3
+c
3
3
·
a
2
+b
2
+c
2
2
=
a
5
+b
5
+c
5
5

7
7
=
a
3
+b
3
+c
3
3
·
a
4
+b
4
+c
4
4
·
2n a
1
, a
2
, . . . , a
n
b
1
, b
2
, . . . , b

− a
k+1
)s
k
n
a
1
+ a
2
+ ··· + a
n
=
n
2
s
n

k=1
(s − a
k
)
2
=
n

k=1
a
2
k
Ax

= 1
a =
a
1
+ a
2
+ ··· + a
n
n
b =
b
1
+ b
2
+ ··· + b
n
n
n

k=1
a
k
b
k
= nab − (a
1
− a)
2
− (a
2

(1+
1
x − 1
)(1−
1
2x − 1
)(1+
1
3x − 1
) ···(1+
1
(2n − 1)x − 1
)(1−
1
2nx − 1
) =
=
(n + 1)x
(n + 1)x − 1
·
(n + 2)x
(n + 2)x − 1
···
(n + n)x
(n + n)x − 1
x
3
= (x ·
x
3

x
2
− 9
+ ··· +
20
x
2
− 100
= 11

1
(x − 1)(x + 10)
+
1
(x − 2)(x + 9)
+ ··· +
1
(x − 10)(x + 1)

a
b
=
c
d
ab
cd
=
(a + b)
2
(c + d)

=
1
a + b + c
x
2
y
2
z
2
a
2
b
2
+
(x
2
− a
2
)(y
2
− a
2
)(z
2
− a
2
)
a
2
(a

2
− a
2
− b
2
S
k
=
a
k
(a − b)(a − c)
+
b
k
(b − c)(b − a)
+
c
k
(c − a)(c − b)
S
−2
=
1
abc
· (
1
a
+
1
b

3
+ a
2
b + ab
2
+
b
2
c + bc
2
+ c
2
a + ca
2
S
k
=

cyclic
a
k
(a − b)(a − c)(a − d)
S
0
= S
1
= S
2
= 0; S
3

a
2
b
2
c
2
(a − d)(b − d)(c − d)
= abc + bcd + cda + dab

cyclic
a
k
(a − b)(a − c)(x − a)
k = 1, 2

cyclic
b + c + d
(a − b)(a − c)(a − d)(a − x)
=
x − a − b − c − d
(x − a)(x − b)(x − c)(x − d)

cyclic
a
k
(x − b)(x − c)
(a − b)(a − c)
= x
k
k = 0, 1, 2

a + b
·
b − c
b + c
·
c − a
c + a
= 0

cyclic
b − c
(a − b)(a − c)
= 2

sym
1
a − b

sym
b
2
+ c
2
− a
2
2bc
= 1
1
−1
1

bz + cy
x(−ax + by + cz)
=
cx + az
y(ax − by + cz)
=
ay + bx
z(ax + by −cz)
x
a(b
2
+ c
2
− a
2
)
=
y
b(c
2
+ a
2
− b
2
)
=
z
c(a
2
+ b

2
−a
2
)x = (c
2
+a
2
−b
2
)y =
(a
2
+ b
2
− c
2
)z
x
3
+ y
3
+ z
3
= (x + y)(y + z)(z + x)
1
x
+
1
y
=

k
k = 0, 1, 2, 3
x =
3
√
4 +
3
√
2 + 1
(1 +
1
x
)
3
• (a + b + c)(bc + ca + ab) = abc + (b + c)(c + a)(a + b)
• (a
2
− 1)(b
2
− 1)(c
2
− 1) + (a + bc)(b + ca)(c + ab) = (abc + 1)(a
2
+
b
2
+ c
2
+ 2abc − 1)
• (b + c − a)

+ b
5
− (a + b)
5
= −5ab(a
2
+ ab + b
2
)
• (a + b)
7
− a
7
− b
7
= 7ab(a + b)(a
2
+ ab + b
2
)
2
xy + yz + zx = 0

sym
(x + y)
2
+ 24x
2
y
2

−
2
c
| +
|b − a|
|ab|
+
b + a
ab
+
2
c
f(a, b, c) = 4max{
1
a
,
1
b
,
1
c
}
a
b + c
+
b
c + a
+
c
a + b

z
t + x + y
=
t
x + y + z
x + y = z + t
x
2
+ y
2
+ z
2
+ t
2
= (x + y)
2
+ (x − z)
2
+ (x − t)
2
ab + bc + ca = 1
(1 + a
2
)(1 + b
2
)(1 + c
2
) = [(a + b)(b + c)(c + a)]
2
a, b, c b = c a + b = c c

c
(a − b)
2
= 0
x, y, z
xy + yz + zx = 0, a =

y
2
+ yz + z
2
b =
√
z
2
+ zx + x
2
, c =

x
2
+ xy + y
2
(a + b − c)(b + c − a)(c + a − b) = 0
a, b, c, d ac + bd = (b + d + a −c)(b +
d − a + c)
(ab + cd)(ad + bc) = (ac + bd)(a
2
− ac + c
2

a
7
+ b
7
+ c
7
+ d
7
= 0
(a + b)(a + c)(a + d) = 0

cyclic
a
k
(a − x)(a − y)
(a − b)(a − c)
=
xy
abc
•

cyclic
x(y + z)
2
− 4xyz = (x + y)(y + z)(z + x)
• 1 + x + x
2
+ x
3
+ x

+ x
3
+ x
4
+ x
5
+ x
6
)
a+b(1+a)+c(1+a)(1+b)+···+l(1+a)(1+b) ···(1+k) = (1+a)(1+b) ···(1+l)−1
a = b = c = ··· = l
a + b + c = 0
(
b − c
a
+
c − a
b
+
a − b
c
)(
a
b − c
+
b
c − a
+
c
a − b

4a + 9b + 16c + 25d = 12
9a + 16b + 25c + 36d = 123
16a + 25b + 36c + 49d
a, b, c
a
b
=
b
c
=
c
a
a + b + c
a + b − c
a, b, c, d
a
4
b
+
b
4
d
=
1
b + d
a
2
+ c
2
= 1

√
3
+
2 −
√
3
√
2 −

2 −
√
3
=
√
2
3

3
√
2 − 1 =
3

1
9
−
3

2
9
+

3
1
x
+
1
y
+
1
z
= 1
3

ax
2
+ by
2
+ cz
2
=
3
√
a +
3
√
b +
3
√
c
a, b, c, d abcd = 1
a + b + c + d =


14 − 5
√
3 = 3
√
2
3

6 +

847
27
+
3

6 −

847
27
= 3

4 −
√
15 +

5 +
√
21 +

6 −

17 − 2
√
7+

5 −

17 + 2
√
7−

5 +

17 − 2
√
7
4

2 +
√
5 + 2

2 +
√
5 +
4

2 +
√
5 − 2


n 2n + 1 {2, 5, 9}
a
1
, a
2
, . . . , a
2n+1
a
2n+1
= a
1
a
1
a
2
− a
2
a
3
+ ··· + a
2n−1
a
2n
− a
2n
a
2n+1
= 0
a, b, c ∈ R
1

1
, . . . , x
n
k
S
k
= x
k
1
+ ··· + x
k
n
S
2
= S
3
= S
4
S
k
= S
1
k
x, y
(
√
x
2
+ 3 + x)(


|f + a − c − d| =
√
3 ·

|c − d| + |f − a|

a + c + e = b + d + f
n w = cos
2kπ
n + 1
+ i · sin
2kπ
n + 1
n + 1 1 1
a
k
=

cos
2kπ
n + 1

n
1 + a
1
w + a
2
w
2
+ ··· + a

2
)
• tan a + tan b =
sin(a+b)
cos a·cos 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)]
• cos(a + b) · cos(a − b) = cos
2
a − sin
2
b
• (cos a + cos b)
2
+ (sin a + sin b)
2
= 4 cos
2
a−b
2
• (cos a − cos b)
2
+ (sin a − sin b)
2

a
• cos 3a = 4 cos
3
a − 3 cos a
• tan 3a =
3 tan a−tan
3
a
1−3tg
2
a
• tan a − tan b =
sin(a−b)
cos a·cos b
• cot a + cot b =
sin(a+b)
sin a·sin b
• cot a − cot b =
sin(b−a)
sin a·sin b
tan
a
2
= 4 tan
b
2
tan
a − b
2
=

(3+cos
2
2a) cos 2a
4
• 2(sin
6
a + cos
6
a) − 3(sin
4
a + cos
4
a) + 1 = 0
•
1+sin 2a
sin a+cos a
−
1−tan
2
a
2
1+tan
2
a
2
•
√
1+cos a+
√
1−cos a

aa
1
+ bb
1
ab
1
+ a
1
b
f(x) = a sin x + b cos x x
1
, x
2
x
1
− x
2
= k · π k ∈ Z f(x
1
) = f (x
2
) = 0
f(x)
a = cos(x −α), b = sin(x−β) a
2
−2ab sin(α−
β) + b
2
= cos
2

− 1
1 + 2r cos u + r
2
=
r + cos u
r −cos v
= ±
sin u
sin v
= −
1 + r cos u
1 + r cos v
tan
u
2
· tan
v
2
= ±
r + 1
r −1
cos x = tan y, cos y = tan z, cos z = tan x
sin x = sin y = sin z =
√
5 − 1
2
sin x
a
1
=

3
sin
2
x
2
=
2a
2
− a
1
− a
3
4a
2
sin β
sin(2α + β)
=
m
n
1 +
tan α
tan β
m + n
=
1 − tan α tan β
m − n
0 < α, β <
π
2
α = β

sin(a + b + c)

cyclic
cos a
=

cyclic
tan a

cyclic
sin a − sin(a + b + c) = 4

cyclic
sin
a + b
2

cyclic
cos a − cos(a + b + c) = 4

cyclic
cos
a + b
2
sin
4
α
a
+
cos

sin(a − b) sin(a − c)
= 0

cyclic
cos a
sin(a − b) sin(a − c)
= 0

cyclic
sin a sin(b −c) cos(b + c − a) = 0

cyclic
cos a sin(b −c) sin(b + c − a) = 0

cyclic
1
sin(a − b)(a − c)
=
1
2

cyclic
cos
a−b
2
a + b + c = π

cyclic
sin 3a sin
3

a
2

cyclic
tan a =

cyclic
tan a

cyclic
tan
a
2
tan
b
2
= 1

cyclic
sin 2a = 4

cyclic
sin a

cyclic
cos
2
a = 2

cyclic

cos(α
1
+ λ) + a
2
cos(α
2
+ λ) + ··· + a
n
cos(α
n
+ λ) = 0
a
tan(α + x)
=
b
tan(α + y)
=
c
tan(α + z)

cyclic
a + b
a − b
sin
2
(x − y) = 0
(a+b) sin(x−α) = (a−b) sin(x+α ) atg
x
2
= c+b tan

+ α) = 9 tan
2
3α + 6
sin 18
0
=
√
5 − 1
4
tg15
0
= 2 −
√
3
sin 15
0
=
√
6 −
√
2
4
cos 15
0
=
√
6 −
√
2
4

√
5
8

cos 42
0
+ cos 102
0
+ cos 114
0
+ cos 174
0

2
=
3
4
tan 3
0
·tan 17
0
·tan 23
0
·tan 37
0
·tan 43
0
·tan 57
0
·tan 63

128
−1 < x < 1
6

k=0
1 − x
2
1 − 2x cos
2kπ
7
+ x
2
= 7 ·
1 + x
7
1 − x
7
1
sin
2
π
7
+
1
sin
2
2π
7
+
1

2 −
√
3 − 2
tan
2
1
0
+ tan
2
2
0
+ ··· + tan
2
89
0
= 4005
tan
6
20
0
− 33 tan
4
20
0
+ 27 tan 20
0
= 3
cos
4
π

=
1
2
cos
π
5
+
√
7
2
sin
π
5
a
1
, a
2
, . . . , a
n
+1 −1
2 sin

a
1
+
a
1
a
2
2

tan(a + b) = 3 tan a sin(2a + 2b) + sin 2a = 2 sin 2b
1
cos 290
0
+
1
√
3 sin 250
0
=
4
√
3
tan 9
0
− tan 63
0
+ tan 81
0
− tan 27
0
= 4
cos 10
0
cos 50
0
cos 70
0
=
√

50
0
+ tan
6
70
0
= 433
4 cos 36
0
+ cot 7
0
30

=
√
1 +
√
2 +
√
3 +
√
4 +
√
5 +
√
6
cos
π
12
· cos

2n + 1
=
n
2
−
1
4
 = cos
2π
n
+ i sin
2π
n
n
A
k
= a
0
+ a
1

k
+ a
2

2k
+ ··· + a
n−1

(n−1)k


n
4

tan
4
φ − ··· + A
A = (−1)
n
2
tan
n
φ n (−1)
n−1
2

n
n−1

tan
n−1
φ
sin nφ
cos
n
φ
=

n
1

φ n
0 < α ≤ π 0 < β ≤ π
cos α + cos β −cos(α + β) =
3
2
α = β =
π
3
0 < α ≤ π 0 < β ≤ π
cos α · cos β ·cos(α + β) = −
1
8
α = β =
π
3
cos θ +cos ϕ = a; sin θ + sin ϕ = b
cos(θ + ϕ) =
a
2
− b
2
a
2
+ b
2
sin(θ + ϕ) =
2ab
a
2
+ b

2
(x − a) cos θ + y sin θ = (x − a) cos θ
1
+ y sin θ
1
= a
tan
θ
2
− tan
θ
1
2
= 2l
y
2
= 2ax − (1 − l
2
)x
2
x cos θ + t sin θ = x cos ϕ + y sin ϕ = 2a
2 sin
θ
2
· sin
ϕ
2
= 1
y
2

α + sin
2
β =
sin(α + β) α + β =
π
2
cos
2
θ =
cos α
cos β
; cos
2
ϕ =
cos γ
cos β
;
tan θ
tan ϕ
=
tan α
tan γ
tan
2
α
2
· tan
2
γ
2

1
2
(β + γ)
=
tan β
tan
1
2
(α + γ)
=
tan γ
tan
1
2
(α + β)
sin(θ −β) · cos α
sin(ϕ − β) · cos β
+
cos(α + θ) · sin β
cos(ϕ − β) · sin α
= 0
tan θ ·tan α
tan ϕ · tan β
+
cos(α − β)
cos(α + β)
= 0
tan θ =
1
2

θ
m
2
+ m cos α = 2
cos θ =
sin β
sin α
, cos ϕ =
sin γ
sin α
, cos(θ −ϕ) = sin β · sin γ
tan
2
α = tan
2
β + tan
2
γ
tan α, tan β
x
2
+ πx +
√
2 = 0
sin
2
(α + β) + π sin(α + β) cos(α + β) +
√
2 cos
2

n

k=1
(x
2
− 2x cos
2kπ
2n + 1
+ 1)
x
2n+1
+ 1 = (x + 1)
n

k=1
(x
2
− 2x cos
2kπ
2n + 1
+ 1)
x
2n+1
+ 1 =
n−1

k=0
(x
2
− 2x cos

2
n
sin
π
2n + 1
· sin
2π
2n + 1
···sin
nπ
2n + 1
=
√
n
2
n
cos
π
2n + 1
· cos
2π
2n + 1
···cos
nπ
2n + 1
=
1
2
n
cos

nπ
2n + 1
=
√
2n + 1
cos α+i sin α x
n
+p
1
x
n−1
+
···+ p
n
= 0 p
1
sin α + p
2
sin 2α + ···+ p
n
sin nα = 0 p
1
, p
2
, . . . , p
n
3

cos
2π

cos
4π
9
+
3

cos
8π
9
=
3

1
2
(3
3
√
9 − 6)