cos(-x)

Temel Formuller 2
cos(x+y)=cosx cosy ‐ sinx siny cos(x‐y)=cosx cos y + sinx sin y + ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ cos(x+y)+cos(x‐y)=2 cosx cosy 2
sin x+cos x=1 cos(x+y)=cosx cosy ‐ sinx siny (1) sin(x+y)=sinx cosy + siny cosx (2) sin(‐x)=‐ sinx (3) cos(‐x)=cos(x) (4) tan(‐x)=‐tan(x) (4.a) cosx cosy= 1 [cos(x+y) +cos(x‐y)] (7) 2
*************************** (5) den (1) i cikar cos(x‐y)=cosx cos y + sinx sin y cos(x+y)=cosx cosy ‐ sinx siny ‐ ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ cos(x‐y)‐cos(x+y)=2 sinx siny x 0 30 45 60 90 sinx 0 0.5 0.707 0.8 1 cosx 1 0.8 0.707 0.5 0 1)
x
0
0
2
sin x
x
30
1
2
0
0.5
0
30
4
2
cos x
1
45
60
2
2
sinx siny = 1 [ cos(x‐y)‐ cos(x+y)] (8) 90
3
2
2
************************************ (2) ve (6) yi topla 4
2
0.707 0.866 1
45
3
2
2
2
60
sin(x+y)=sinx cosy +siny cosx (2) sin(x‐y)=sinx cosy ‐ siny cos x (6) + ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 90
1
2
0.866 0.707 0.5
0
2
sin(x+y) + sin(x‐y) =2 sinx cosy sinx cosy = 1 [ sin(x+y)+ sin(x‐y)] (9) 2
************************ ************ (1) de x=90 koy 0
y=sin x
cos(x+90)=cosx cos90 ‐ sinx sin90 π/2
cos(x+90)=– sinx (10) 2π
π
x
************************ ************ (1) de x=‐90 koy cos(x‐90)=cosx cos(‐90) ‐ sinx sin(‐90) cos(x‐ 90)= sinx (11) ************************ ************ (2) de x=90 koy y=cos x
sin(x+90)=sinx cos90 + sin90 cosx sin(x+ 90)= cosx (12) π
π/2
2π
x
************************ ************ (2) de x=‐90 koy sin(x‐90)=sinx cos(‐90) + sin(‐90) cosx ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ (1) de y yerine –y yaz ve 3,4 bagintilarini kullan cos(x‐y)=cosx cos(‐y) ‐ sinx sin(‐y) sin(x‐ 90)= ‐cosx (13) ************************ ************ (1) de x=90, y=‐q koy cos(90‐q)=cos90 cos(‐q) – sin90 sin(‐q) cos(x‐y)=cosx cos y + sinx sin y (5) cos(90‐q)= sinq (14) (2) de y yerine –y yaz ve 3,4 bagintilarini kullan sin(x‐y)=sinx cos(‐y) ‐ sinx cos(‐y) ************************ ************ sin(x‐y)=sinx cosy ‐ siny cos x (6) ******************************** (1) ve (5) i topla (2) de x=90, y=‐q koy sin(90‐q)=sin90 cos(‐q) + cos90 sin(‐q) sin(90‐q)= cosq (15) ************************ ************ (1) de y=x koy cos(x+x)=cosx cosx ‐ sinx sinx cos(2x) = cos2x‐sin2x sin2x=1‐cos2x koy cos2x ‐sin2x= cos2x –(1‐ cos2x)= 2cos2x ‐1 2
cos(2x) = 2cos x ‐1 (16) = 1
1
1
sinq ‐ sin(3q) + sin q 2
4
4
sin3q= = ‐ 1 sin(3q) + 3 sin q (34) 4
4
veya veya sin(3q)= ‐4sin3q +3sinq (35) cos2x = 1 [ 1+cos(2x)] (17) ********************** ******* 2
************************ ************ (1) de y=x koy cos(x+x)=cosx cosx ‐ sinx sinx cos(2x) = cos2x‐sin2x cos2x=1‐sin2x koy cos2x ‐sin2x= 1‐ sin2x – sin2x = 1‐2sin2x cos(2x) =1‐ 2sin2x (18) veya sin2x = 1 [ 1‐cos(2x)] (19) 2
********************** ******* (2) de y=x koy sin(x+x)=sinx cosx + sinx cosx=2 sinx cosx sin2x =2 sinx cosx (20) ********************** ******* (16) ve(17) de da q=2x koy cos q = 2cos2 q ‐1 (21) 2
2q
cos q = 1‐2sin
2
(22) ********************** ******* (20) de q=2x koy sinq =2 sin q cos q (23) 2
2
********************** ******* 90 180 270 ‐90 ‐180 sinx 1 0 ‐1 ‐1 0 cosx 0 ‐1 0 0 ‐1 ********************** ******* sin3q=sinq sin2q (19) bagintisinda verilen sin2q yerine koy 1
2
1
2
cos3q=cosq cos2q (17) bagintisinda verilen cos2q yerine degerini koy 1
2
1
2
cos3q= cosq [ 1+cos(2q)] = [ cosq+ cosq cos(2q)] (7 bagintisinda x=q, y=2q koy 1
2
1
= [ cos(3q)+ cosq] 2
1
1
3
cos q= [ cosq+ [ cos(3q)+ cosq]] 2
2
1
3
= cos3q + cosq 4
4
cos3q= = 1 cos3q + 3 cosq (36) 4
4
cosq cos2q= [cos(q+2q) +cos(q‐2q)] cos3q=4 cos3q ‐3 cosq (37) ************************** ***** tan(x+y)= sin( x + y ) = sin x cos y + sin y cos x
cos( x + y)
cos x cos y − sin sin y
payi ve paydayi cosx cosy ifadesine bolelim sin x cos y + sin y cos x sin x cos y sin y cos x
+
cos x cos y
cos x cos y cos x cos y =
=
cos x cos y − sin sin y
cos x cos y
sin sin y
−
cos x cos y
cos x cos y cos x cos y
sin x sin y
+
tan x + tan y
cos x cos y
=
=
sin x sin y 1 − tan x tan y 1−
cos x cos y
tan( x + y ) =
tan x + tan y
1 − tan x tan y (41) sin3q= sinq [ 1‐cos(2q)] = [ sinq‐ sinq cos(2q)] *************** ***************** (9) bagintisinda x=q, y=2q koy tan(‐x)=‐tan(x) oldugu dikkate alinirsa 1
2
sinq cos(2q)= [ sin(q+2q)+ sin(q‐2q)] 1
2
= [ sin(3q)‐ sin(q) ] 1
2
1
2
sin3q= [ sinq‐ [ sin(3q)‐ sin q] ] tan( x − y ) =
tan x − tan y
1 + tan x tan y (42) *************** ***************** (41) de y=x koy tan(2 x) =
2 tan x
1 − tan 2 x (43) *************** ***************** x
x
sin x = 2 sin cos = 2
2
2
(43) de q=2x koy q
2 tan
2
tan(q) =
q
1 − tan 2
2
sin x =
sin x
sin x
1 − cos x
=
=
2
cos x
cos x
1 − sin x
2
(51) **************************** ***** sin x
tan x =
⇒
tan 2 x =
1 − sin 2 x tan2x(1‐ sin2x) = sin2x tan2x ‐ tan2x sin2x = sin2x tan2x ‐ tan2x sin2x ‐ sin2x=0 ‐ tan2x sin2x ‐ sin2x=‐ tan2x sin 2 x
1 − sin 2 x tan 2 x
tan x
⇒ sin x =
sin x =
2
1 + tan x 1 + tan 2 x
2
⇒ tan 2 x =
tan2x cos2x= 1‐ cos2x cos2x(1 +tan2x)=1 cos 2 x =
1 − cos 2 x
cos 2 x 2
cos x =
1
1 + tan x
2
1
=
x ⎞
⎛
⎜ 2 tan
⎟
2
⎜
⎟
1+
⎜ 1 − tan 2 x ⎟
⎜
⎟
2⎠
⎝
sinq =2 sin cos (52)ve (53) den sin
x
=
2
tan
x
2
1
x
cos =
2
x
x
1 + tan 2 1 + tan 2 2
2
Ucu birlestirilirse 2
basitlestirmek icin q=tan(x/2) tanimi yapalim. 1
⎛ 2q ⎞
⎟
1+ ⎜⎜
2⎟
⎝ 1− q ⎠
=
(23) den q
2
x 2
⎛
⎞
⎜
⎟
2x
1
⎟ − 1 cos x = 2cos ‐1 = 2⎜
⎜
2
x⎟
⎜ 1 + tan 2 ⎟
2⎠
⎝
x
1 − tan
2
cos x =
x
1 + tan 2 (55) 2
1
1
⇒ cos x =
2
1 + tan x 1 + tan 2 x
q
2
1 + tan 2
************ *********** (21) ve (53) den =
(53) *************** ***************** x
2
x (54) 2
(52) *************** ***************** 1 − cos 2 x
cos x
1 + tan 2
1
************************************* * (54) ve (55) icin baska bir yol. (53) ve (44) birlestirilirse ‐ sin2x (tan2x+1)= ‐ tan2x tan x =
x
2
x
2
1 + tan 2
(44) **************************** ***** tan x =
tan
tan
2
1
=
1
1 + 2q 2 + q 4
1 − 2q 2 + q 4
1+
=
2
4q
1− 2q2 + q4
=
1
1− 2q + q
4q2
+
2
4
2
4
1− 2q + q 1− 2q + q
2
4
x
1 − tan 2
1− q2
2
=
=
2
2 2
x +
q
1
2
(1 + q )
1 + tan
2 2
2
(1 − q )
1
x
2
cos x =
2 x
1 + tan
2
1 − tan 2
(56) ************************* ********* ** (52) ve (44) birlestirilirse 2 tan
sin x =
tan x
1 + tan 2 x
=
x
2
1 − tan 2
x
2
x ⎞
⎛
⎜ 2 tan ⎟
2 ⎟
1+ ⎜
⎜ 1 − tan 2 x ⎟
⎜
⎟
2⎠
⎝
2
tan x/2= t, q=tan(x/2) tanimi yapalim. =
2q
1− q2
⎛ 2q ⎞
⎟
1 + ⎜⎜
2 ⎟
⎝1− q ⎠
2
=
2q
1− q2
4q 2
1+
1 − 2q 2 + q 4
=
2q
1− q2
1 − 2q 2 + q 4
4q 2
+
2
4
1 − 2q + q 1 − 2 q 2 + q 4
=
=
‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ ‐‐‐ 2q
1− q2
Ters Trigonometrik Fonksiyonlar 1 − 2q 2 + q 4
4q 2
+
1 − 2q 2 + q 4 1 − 2q 2 + q 4
2q
1− q2
1 + 2q 2 + q 4
1 − 2q 2 + q 4
=
2q
1− q2
(1 + q 2 ) 2 (1 − q 2 ) 2
2 tan
x
2
2q 1 − q
2q
=
=
1 − q 2 1 + q 2 1 + q 2 1 + tan 2 x 2
x
2 tan
2
sin x =
2 x (57) 1 + tan
2
********************** *************** ***** tan x=t/1 denirse. =
2
‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐