สูตรตรีโกณมิติจากหนังสือคณิตศาสตร ปรนัย เล มที่ 31 โลก ...

[Pages:1] 31

sin cos tan cosec sec cot

(-, ) (-, ) (-, ) - { | cos = 0} (-, ) - { | sin = 0} (-, ) - { | cos = 0} (-, ) - { | sin = 0}

[-1, 1] [-1, 1] (-, ) (-, -1] [1, ) (-, -1] [1, ) (-, )

arcsin arccos arctan arccot arcsec arccosec

[-1, 1] [-1, 1] (-, ) (-, ) (-, -1] [1, ) (-, -1] [1, )

[-

2

,

2

]

[0, ]

(-

2

,

2

)

(0, )

[0,

2

)

(

2

,

]

[-

2

,

0)

(0,

2

]

sin2 A + cos2 A = 1

1 + tan2 A = sec2 A

1 + cot2 A = cosec2 A

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

cot(A + B) =

cot A cot B -1 cot A + cot B

sin 2A = 2 sin A cos A

sin 2A

=

2 tan A 1 + tan2 A

cos 2A = cos2 A - sin2 A = 1 - 2 sin2 A = 2 cos2 A - 1

= sin2

A 2

1 - cos A 2

sin 3A = 3 sin A - 4 sin3 A

= cos2

A 2

1 + cos A 2

cos 3A = 4 cos3 A - 3 cos A

tan 2A =

2 tan A 1 - tan2 A

cos 2A

=

1 - tan2 A 1 + tan2 A

= tan2

A 2

1 - cos A 1 + cos A

sin(n + (-1) n ) = sin sin(n + ) = (-1) n sin

cos(2n ? ) = cos

tan(n - ) = -tan

cos(n + ) = (-1) n cos tan(n + ) = tan

sin A

+

sin B

=

2 sin

A+ B 2

cos

A-B 2

sin A

-

sin B

=

2 sin

A-B 2

cos

A+B 2

cos A

+

cos B

=

2 cos

A

+ B 2

cos

A-B 2

cos A

-

cos B

=

-2 sin

A+B 2

sin

A-B 2

2 sin A + B sin B - A

2

2

2 sin A cos B = sin(A + B) + sin(A - B) 2 cos A sin B = sin(A + B) - sin(A - B) 2 cos A cos B = cos(A + B) + cos(A - B) 2 sin A sin B = cos(A - B) - cos(A + B)

 31

ABC a, b, c A, B, C

= ABC =

s(s - a)(s - b)(s - c)

s =

a+b+c 2

= 1 ab sin C = 1 ac sin B = 1 bc sin A

2

2

2

R = ABC r = ABC

= rs =

abc 4R

,

r

=

4R

sin

A 2

sin

B 2

sin

C 2

sin A a

=

sin B b

=

sin C c

=

2 abc

2R =

a sin A

=

b sin B

=

c sin C

R =

abc 4

c2 = a2 + b2 - 2ab cos C b2 = a2 + c2 - 2ac cos B a2 = b2 + c2 - 2bc cos A

cos A =

b2 + c2 - a2 2bc

cos B =

a2 + c2 - b2 2ac

cos C =

a2 + b2 - c2 2ab

sin

A 2

=

(s - b)(s - c) bc

sin

B 2

=

(s - a)(s - c) ac

sin

C 2

=

(s - a)(s - b) ab

cos A = s(s - a)

2

bc

cos B = s(s - b)

2

ac

cos C = s(s - c)

2

ab

tan

A 2

=

(s - b)(s - c) s(s - a)

tan

B 2

=

(s - a)(s - c) s(s - b)

tan

C 2

=

(s - a)(s - b) s(s - c)

ABC

( 5.)

1.

sin A

+ sin B

+ sin C

= 4 cos

A 2

cos

B 2

cos

C 2

2.

cos A

+

cos B

+ cos C

=

1 +

4 sin

A 2

sin

B 2

sin

C 2

3. tan A + tan B + tan C = tan A tan B tan C

sin, cos, tan

x1 ,

x2 , ... ,

x n

(0,

)

2

( 10.)

1.

sin x1 + sin x2 + ... + sin xn n

sin

x1

+

x2 + n

... +

xn

2.

n sin x1sin x2 ... sin xn

sin

x1

+

x2

+ ... + n

xn

3. cos x1 + cos x2 + ... + cos xn cos x1 + x2 + ... + xn

n

n

4.

n cos x1 cos x2 ... cos xn

cos x1 + x2 + ... + xn n

5.

tan x1 + tan x2 + ... + tan xn n

tan

x1

+

x2 + n

... +

xn

( x1 = x2 = ... = xn )

cos n sin

. 14.1 n

cos n =

- n + 2n-1 cosn 2n-3 cosn-2

n(n - 3) 2!

2n-5

cosn-4 -

n(n - 4)(n - 5) 3!

2n-7 cosn-6

+ - + ... n(n - 5)(n - 6)(n - 7) 4!

2n-9

cosn-8

n(n - 6)(n - 7)(n - 8)(n - 9) 5!

2n -11

cosn-10

. 13.4

sin

n

sin

2 n

sin

3 n

...

sin

(n

- 1) n

=

n 2n -1

n n > 1

. 14.17

sin

2n

sin

3 2n

sin

5 2n

...

sin

(n

- 4) 2n

sin

(n

- 2) 2n

=

1 2n -1

n

. 14.19

sin sin 3 sin 5 ... sin (n - 3) sin (n -1) =

2n 2n 2n

2n

2n

1 2n -1

n

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download