Tolong bantu kerjakan nomor 1-6

Mapel : Matematika
Kelas : X SMA
Bab : Trigonometri

Pembahasan :
1||

Cos A = -5/13
Sin A = 12/13 (Sin + di Kuadran II)
Sin B = 4/5
Cos B = 3/5 (Cos + di Kuadran I)

Maka...
Sin A Cos B + Cos A Sin B
= (12/13)(3/5) + (-5/13)(4/5)
= 36/65 - 20/65
= 16/65

2||

Ingat Rumus !
Cos (A + B) = Cos A Cos B - Sin A Sin B

Maka...
Cos 50° Cos 10° - Sin 50° sin 10°
= Cos (50 + 10)°
= Cos 60°
= 1/2

3||

(A - B) = 30°
Sin(A - B) = Sin 30°
Sin A . Cos B - Cos A Sin B = 1/2
2/3 - Cos A Sin B = 1/2
Cos A Sin B = 2/3 - 1/2
Cos A Sin B = (4 - 3)/6
Cos A Sin B = 1/6

Maka...
Sin (A + B)
= Sin A . Cos B + Cos A Sin B
= 2/3 + 1/6
= (4 + 1)/6
= 5/6

4||

a + β = 315°
Tan (A + B) = Tan 315°
(Tan A + Tan β)/(1 - Tan A . Tan β) = -1
3/4 + Tan β = -1 + 3/4 Tan β
3/4 Tan β - Tan β = 3/4 + 1
Tan β (3/4 - 1) = 7/4
Tan β (-1/4) = 7/4
Tan β = (7/4)/(-1/4)
Tan β = -7

5||

Cos A Cos B = Sin A Sin B
Cos A Cos B - Sin A Sin B = 0
Cos (A + B) = 0

Sin A Cos B = Cos A Sin B
Sin A Cos B - Cos A Sin B = 0
Sin (A - B) = 0

Karena nilai Sin dan Cos adalah 0, maka berada apda kuadran I.
Jenis segitiga itu adalah segitiga Lancip.

6||

Tan 15° = p

Maka...
(Tan 165° - Tan 105°)/(1 + Tan 165° . Tan 105°)
= Tan (165 - 105)°
= Tan 60°
= Tan (45 + 15)°
= (Tan 45° + Tan 15°)/(1 - Tan 45° Tan 15°)
= (1 + p)/(1 - p)

[tex]\displaystyle 1\\\sin a\cos b+\cos a\sin b=\frac{12}{13}\cdot\frac35-\frac5{13}\cdot\frac45\\\sin a\cos b+\cos a\sin b=\frac{36}{65}-\frac{20}{65}\\\boxed{\boxed{\sin a\cos b+\cos a\sin b=\frac{16}{65}}}[/tex]

[tex]\displaystyle 2\\\cos50^\circ\cos10^\circ-\sin50^\circ\sin10^\circ=\cos(50+10)^\circ\\\cos50^\circ\cos10^\circ-\sin50^\circ\sin10^\circ=\cos60^\circ\\\boxed{\boxed{\cos50^\circ\cos10^\circ-\sin50^\circ\sin10^\circ=\frac12}}[/tex]

[tex]\displaystyle 3\\\sin(a-b)=\sin a\cos b-\cos a\sin b\\\sin30^\circ=\sin a\cos b-\cos a\sin b\\\frac12=\frac23-\cos a\sin b\\\cos a\sin b=\frac16\\\\\sin(a+b)=\sin a\cos b+\cos a\sin b\\\sin(a+b)=\frac23+\frac16\\\boxed{\boxed{\sin(a+b)=\frac56}}[/tex]

[tex]\displaystyle 4\\\tan (a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}\\\tan 315^\circ=\frac{\frac34+\tan b}{1-\frac34\tan b}\\-1=\frac{\frac34+\tan b}{1-\frac34\tan b}\\\frac34\tan b-1=\frac34+\tan b\\-\frac74=\frac14\tan b\\\boxed{\boxed{-7=\tan b}}[/tex]

[tex]\displaystyle 6\\\frac{\tan165^\circ-\tan105^\circ}{1+\tan165^\circ\tan105^\circ}=\tan(165^\circ-105^\circ)\\\frac{\tan165^\circ-\tan105^\circ}{1+\tan165^\circ\tan105^\circ}=\tan60^\circ\\\boxed{\boxed{\frac{\tan165^\circ-\tan105^\circ}{1+\tan165^\circ\tan105^\circ}=\sqrt3}}[/tex]