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TRIGONOMETRI - PowerPoint PPT Presentation

TRIGONOMETRI. Setelah menyaksikan tayangan ini anda dapat Menyelesaikan soal-soal yang berkaitan dengan jumlah dan selisih sudut serta sudut rangkap. Rumus jumlah dan selisih dua sudut sin(  + ) = sin.cos + cos.sin sin( - ) = sin.cos - cos.sin.

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tayangan ini anda dapat

Menyelesaikan

soal-soal yang berkaitan

dengan jumlah dan selisih sudut

serta sudut rangkap

jumlah dan selisih dua sudut

sin( + )

= sin.cos + cos.sin

sin( - )

= sin.cos - cos.sin

1. Sin 75o = ….

Bahasan:

sin( + ) = sin.cos + cos.sin

sin750 = sin(450 + 300)

= sin450cos300 + cos450sin300

= ½√2.½√3 + ½√2.½

= ¼√6 + ¼√2

= ¼√2(√2 + 1)

B

2. Diketahui sin A = cos B =

A dan B adalah sudut-sudut lancip

sin(A – B) =….

Bahasan:

sin(A – B)= sinAcosB – cosAsinB

sinA =

cosA =

?

?

cos B =

sin B =

5

3

24

25

4

7

sin A =  cos A =

cos B =  sin B =

sin(A – B) =….

= sinAcosB – cosAsinB

= x - x

=

=

jumlah dan selisih dua sudut

cos( + )

= coscos - sinsin

cos( - )

= coscos + sinsin

Bahasan:

coscos + sinsin = cos( - )

=

=

=

a. –sina.sinb b. cosa.cosb

c. sina.sinb d. 1 – tana.tanb

e. 1 + tana.tanb

= 1 – tana.tanb  jawab d

nilai cos56° + sin56°.tan28°

Bahasan:

cos56° + sin56°.tan28°

= cos56° + sin56°.

= cos56° +

=

=

=

Nilai cos56° + sin56°.tan28° = 1

= 1

ABC berlaku cosA.cosB = ½.

Maka cos(A – B) =….

Bahasan:

 siku-siku ABC; cosA.cosB = ½

makaΔABC siku-siku di C

C = 90°

A + B + C = 180°  A + B = 90°

A = 90° – B  B = 90° – A

cos(A – B)

= cosA.cosB + sinA.sinB

= ½ + sin(90 – B).sin(90-A)

= ½ + cosB.cosA

= ½ + ½

= 1

Jadi cos(A – B) = 1

jumlah dan selisih dua sudut

tan( + ) =

tan( - ) =

1. tan 105° = ….

Bahasan:

tan105° = tan(60° + 45°)

tan 105° = x

=

=

=

= -2 - √3

tan B =½. Nilai tan A= ….

Bahasan:

A + B = 135°

tan(A + B) = tan 135°

= -1

= -1

tan A + ½= -1 + ½tan A

tan A - ½tan A = -1 - ½

½tan p = -1½

3. Jika tan q =½ dan p – q = ¼π

maka tan p = ….

Bahasan:

p – q = ¼π

tan(p – q) = tan ¼π

= 1

= 1

tan p - ½ = 1 + ½tan p

tan p - ½tan p = 1 + ½

½tan p = 1½

sin2a = 2 sina.cosa

contoh: 1. sin10° = 2sin5°.cos5°

2. sin6P = 2sin3P.cos3P

3. sin t = 2sin½t.cos½t

• Diketahui cos =

• Nilai sin 2 =….

• Bahasan:

• cos =

• sin =

5

4

3

cos =

sin =

= 2. x =

5

4

3

2. Jika tan A = ½ maka sin 2A =….

Bahasan:

tan A = ½

sinA =

dancosA =

sin2A = 2 sinA.cosA

= 2x x=

1

2

maka harga sin 2x =….

Bahasan:

sinx – cosx = p

(sinx – cosx)2 = p2

sin2x – 2sinx.cosx + cos2x = p2

sin2x – 2sinx.cosx + cos2x = p2

sin2x + cos2x – 2sinx.cosx = p2

1 – sin2x = p2

1 – p2 = sin2x

Jadi, harga sin2x = 1 – p2

4. Diketahui A adalah sudut lancip

dan cos½A =

Nilai sin A = ….

Bahasan:

cos½A =

dengan phytagoras

t2 = 2x – (x + 1)

t = √x - 1

√2x

t = √x - 1

√x+ 1

√2x

t = √x - 1

√x+ 1

cos½A =  sin½A =

sinA = 2sin½A.cos½A

= 2 x x

cos 2a = cos2a – sin2a

= 2cos2a – 1

= 1 – 2sin2a

maka cos 2x =….

Bahasan:

cos2x = 1 – 2sin2x

= 1 – 2(½)2

= 1 – ½

= ½

3. Diketahui tan p = ½

maka cos 2p =….

Bahasan:

tan p = ½ 

cos2p = 1 – 2sin2p

= 1 – 2( )2

= 1 –

=

sin p =

√5

1

2

dengan cos 2A =

Nilai tan A = ….

Bahasan:

• cos 2A = 1 – 2sin2A

= 1 – 2sin2A

2sin2A = 1 – =

2sin2A

2cos2A

• cos 2A = 2cos2A – 1

= 2cos2A – 1

2cos2A = + 1 =

tan2A = =

tan2A = ½

A lancip  Jadi, tan A = ½√2

dan cos½A =

Bahasan:

cos A =2cos2½A – 1

= 2 - 1

= 2 - 1

=

cos x = 2 - 1

cos x =

cos x = 

x

√x2 – 1

1

Bahasan:

tan 2a =

Contoh: 1. tan 20° =

2. tan 10x =

maka tan 2A =….

Bahasan:

tan 2A =

=

= =

tan x =

2. Jika cos x =

maka tan 2x =….

Bahasan:

tan 2x =

=

=

13

12

5

=

=