Tr 3.docx

TRIGONOMETRY ROUTINE TASK III

Arranged by : Nadya Ananda Br Sembiring 4181111027

BILINGUAL MATHEMATICS EDUCATION FACULTY OF MATHEMATICS AND NATURAL SCIENCE MEDAN STATE UNIVERSITY February 2018

NOTE : OP = OQ = 1

COS (A-B) = (COS A . COS B + SIN A SIN B) PROOF : 𝑄𝑅

SIN B = SIN A =

𝑂𝑄 𝑃𝑇 𝑂𝑃

COS A = COS B = PQ2

𝑂𝑇 𝑂𝑅 𝑂𝑅 𝑂𝑄

= QR = PT = OT = OR

= OQ2+OP2 – 2OQ . OP COS (A-B) = 1 + 1 – 2.1.1 COS (A-B) = 2-2 COS (A-B)...............(1)

PQ2 =WQ2 +PM2 = (OP-OT)2 + (PT-QR) 2 =(COS B – COS A) 2 + (SIN A – SIN B) 2 =(COS2B – 2 COS A. COS B + COS 2 A ) + (SIN2 A – 2 SIN B .SIN A + SIN2B)

= (COS 2 B + SIN 2 B )+ (COS 2 A + SIN 2 A) - 2(COS A . COS B + SIN A SIN B) = 1+1 -2 (COS A . COS B + SIN A SIN B) =2-2 (COS A . COS B + SIN A SIN B)...............(2)

1 AND 2 =2-2 COS (A-B) = 2-2 (COS A . COS B + SIN A SIN B) COS (A-B) = (COS A . COS B + SIN A SIN B).......... (PROVEN)

COS (A+B) = (COS A . COS B - SIN A SIN B) PROOF : NOTE : OP = OQ = 1

𝑂𝑅

COS ( A+B) = 𝑂𝑃 = OR

COS A = SIN A =

𝑂𝑄 𝑂𝑃

𝑃𝑄 𝑂𝑃

COS B =

SIN B =

= OQ.............. 1

= PQ.................. 2

𝑂𝑆 𝑂𝑄

𝑇𝑄 𝑃𝑄

=

=

𝑂𝑆 𝐶𝑂𝑆 𝐴

𝑇𝑄 𝑆𝐼𝑁 𝐴

COS ( A+B) =

𝑶𝑹 𝑶𝑷

= OS = COS A . COS B ................. 3

= TQ = SINA.SIN B .............. 4

= OR

OR = OS – RS = COS A . COS B - SINA.SIN B

3 AND 4 COS (A+B) = (COS A . COS B - SIN A SIN B)......... PROVEN

SIN (A+B) = (SIN A . COS B + SIN B COS A) PROOF :

β α

𝐴𝑂

SIN α = SIN β =

𝑂𝐵 𝐵𝐶

COS α = COS β=

= AO = AC SIN α .............. 1

𝐴𝐶

= OB = BC SIN β .............. 2

𝑂𝐶 𝐴𝐶

𝑂𝐶 𝐵𝐶

= OC = AC COS α .............. 3 = OC = BC COS β .............. 4

LUAS SEGITIGA ABC = LUAS SEGITIGA AOC + LUAS SEGITIGA OBC 1 2 1 2 1 2

1

1

2

2

SIN (α+β) AC . BC = AO . OC +

OB . OC

1

1

2

2

SIN (α+β) AC . BC = AC SIN α BC COS β + BC SIN β . AC . COS α SIN (α+β) AC . BC =

1 2

AC . BC (SIN α . COS β + SIN β COS α) :

1 2

AC . BC

SIN (α+β) = (SIN α . COS β + SIN β COS α) ..... PROVEN

SIN (A+B) = (SIN A . COS B + SIN B COS A) PROOF :

𝑄𝑅

ROQ = SIN A =

COS A =

𝑅𝑂

𝑆𝑅

RST = SIN A = 𝑅𝑇

COS A =

𝑅𝑇

ORT = SIN B = 𝑂𝑇

COS B =

SEGITIGA TOP = SIN (A+B) = = =

𝑂𝑄 𝑂𝑅 𝑆𝑇 𝑅𝑇

𝑂𝑅 𝑂𝑇

𝑇𝑃 𝑂𝑇

NOTE : PS = RQ

𝑃𝑆+𝑆𝑇 𝑂𝑇 𝑃𝑆 𝑂𝑇 𝑃𝑆 𝑜𝑅

X X

𝑂𝑅 𝑂𝑅 𝑂𝑅 𝑜𝑇

+ +

𝑆𝑇 𝑂𝑇 𝑆𝑇 𝑅𝑇

X X

𝑅𝑇 𝑅𝑇 𝑅𝑇 𝑂𝑇

= SIN A . COS B + SIN B COS A SIN (A+B) = (SIN A . COS B + SIN B COS A)...... PROVEN

𝑻𝒂𝒏 𝒂+𝐭𝐚𝐧 𝒃

Tan (A +B ) = 𝟏−𝐭𝐚𝐧 𝒂.𝐭𝐚𝐧 𝒃

PROOF : tan (a +b ) = =

sin(𝑎+𝑏) cos(𝑎+𝑏)

(𝐒𝐈𝐍 𝐀 .𝐂𝐎𝐒 𝐁 + 𝐒𝐈𝐍 𝐁 𝐂𝐎𝐒 𝐀) (𝐂𝐎𝐒 𝐀 .𝐂𝐎𝐒 𝐁 − 𝐒𝐈𝐍 𝐀 𝐒𝐈𝐍 𝐁)

sin 𝑎.cos 𝑏 cos 𝑎 ,cos 𝑏 cos 𝑎 cos 𝑏 cos 𝑎 cos 𝑏

: cos a cos b

sin 𝑏 .cos 𝑎

+ cos 𝑎 .cos 𝑏 sin 𝑎 sin 𝑏

- cos 𝑎 cos 𝑏

𝑇𝑎𝑛 𝑎+tan 𝑏

= 1−tan 𝑎.tan 𝑏

𝑻𝒂𝒏 𝒂+𝐭𝐚𝐧 𝒃

= Tan (A +B ) = 𝟏−𝐭𝐚𝐧 𝒂.𝐭𝐚𝐧 𝒃 ........... PROVEN

𝑻𝒂𝒏 𝒂−𝐭𝐚𝐧 𝒃

Tan (A -B ) = 𝟏+𝐭𝐚𝐧 𝒂.𝐭𝐚𝐧 𝒃

PROOF : (𝐒𝐈𝐍 𝐀 .𝐂𝐎𝐒 𝐁− 𝐒𝐈𝐍 𝐁 𝐂𝐎𝐒 𝐀)

= (𝐂𝐎𝐒 𝐀 .𝐂𝐎𝐒 𝐁+ 𝐒𝐈𝐍 𝐀 𝐒𝐈𝐍 𝐁) sin 𝑎.cos 𝑏 cos 𝑎 ,cos 𝑏 cos 𝑎 cos 𝑏 cos 𝑎 cos 𝑏

: cos a cos b

sin 𝑏 .cos 𝑎

- cos 𝑎 .cos 𝑏 sin 𝑎 sin 𝑏

+ cos 𝑎 cos 𝑏

𝑇𝑎𝑛 𝑎−tan 𝑏

= 1+tan 𝑎.tan 𝑏

𝑻𝒂𝒏 𝒂−𝐭𝐚𝐧 𝒃

Tan (A -B ) = 𝟏+𝐭𝐚𝐧 𝒂.𝐭𝐚𝐧 𝒃 .......... PROVEN

Note :

DOUBLE ANGLE Sin (a+b) = sin a cos b + sin b cos a Sin (2A) = sin (A +A) = sin a cos a + sin a cos a Sin (2A) = 2 sin a cos a Cos (a+b ) = cos a cos b – sin a sin b cos(2A) = cos (A +A) =cos a cos a – sin a sin a cos(2A) =cos2a – sin2a ...... 1 Dari persamaan 1 : cos2a – sin2a = 1 cos2a = 1– sin2a sin2a = 1- cos2a

= 1- sin 2 a – sin 2 a cos(2A) =1-2sin2a.......2

= cos 2 a – sin 2 a =cos 2 a – (1-cos 2 a) cos(2A) = 2 cos 2a -1 ....... 3

B Diubah ke A

𝑻𝒂𝒏 𝒂+𝐭𝐚𝐧 𝒃

Tan (A +B ) = 𝟏−𝐭𝐚𝐧 𝒂.𝐭𝐚𝐧 𝒃

tan (2A) = tan (A +A) 𝑻𝒂𝒏 𝒂+𝐭𝐚𝐧 𝒂

=𝟏−𝐭𝐚𝐧 𝒂.𝐭𝐚𝐧 𝒂 tan 2

𝟐𝒕𝒂𝒏 𝒂

tan (2A) =𝟏−𝐭𝐚𝐧 (𝐬𝐪𝐮𝐞𝐫𝐞) 𝒂

HALF ANGLE IDENTITIES  Cos ( 2A) = 1- 2 sin 2 A 2 sin 2 A = 1- Cos ( 2A) sin 2 A =

1− Cos ( 2A)

sin A =± √ sin

Note : A diubah menjadi ½ α

𝟐 𝟏− 𝐂𝐨𝐬 ( 𝟐𝐀) 𝟐

𝟏

𝟏− 𝐂𝐨𝐬 (𝛂 )

𝟐

𝟐

α = =± √

 Cos ( 2A) = 2 cos 2 A – 1 2 cos 2 A = 1 + Cos ( 2A) cos 2 A =

1+Cos ( 2A)

cos A =± √

𝟐 𝟏+ 𝐂𝐨𝐬 ( 𝟐𝐀) 𝟐

𝟏

𝟏+ 𝐂𝐨𝐬 (𝛂 )

𝟐

𝟐

cos α = =± √

example : cos 75 cos

𝟏 𝟐

150

1

1+ Cos (150)

2

2

cos 150 =± √

1

= √ 2(1+√3 )

= √

1+√3/2 2

=

ADDITIONAL IDENTITIES