Mohon bantuan untuk menyelesaikan nomor 7 - 12

Mapel : Matematika
Kelas : X SMA
Bab : Trigonometri

Pembahasan :
7||

a + β = pi/6
a + β = 30°
Cos (a + β) = Cos 30°
Cos A Cos β - Sin A Sin β = 1/2 √3
3/4 - Sin A Sin β = 1/2 √3
Sin A Sin β = 3/4 - 1/2 √3

Maka...
Cos (a - β)
= Cos A Cos β + Sin A Sin β
= 3/4 + (3/4 - 1/2 √3)
= 3/2 - 1/2√3
= 1/2(3 - √3)

8||
Sin 105°
= Sin (90 + 15)°
= Cos 15°
= Cos (45 - 30)°
= Cos 45° Cos 30° + Sin 45° Sin 30°
= (1/2 √2)(1/2 √3) + (1/2 √2)(1/2)
= 1/4 √6 + 1/4 √2
= 1/4(√6 + √2)

Aturan Sinus
AB/Sin C = Bc/Sin A
(√2 + √6)/Sin 105° = BC/Sin 30°
1/2(√2 + √6)/[1/4(√6 + √2) = BC
BC = 2 cm

10||
Sin 15° = p
Cos 15° = √[1 - p²]

Sin 75°
= Sin (60 + 15)°
= Sin 60° Cos 15° + Cos 60° Sin 15°
= (1/2 √3) √[1 - p²] + (1/2) p
= 1/2 [√(3 - 3p²) + p]

11||

Sin C = 2/(√13)
Tan C = 2/3

Maka...
Tan (A + B) = (Tan A + Tan B)/(1 - Tan Tan B)
Tan C = (Tan A + Tan B)/(1 - 13)
2/3 = (Tan A + Tan B)/(-12)
Tan A + Tan B = -8

12|| Maksudnya Sin A Sin B kali...

Cos (A + B) = 3/5
Cos A Cos B - Sin A Sin B = 3/5

Cos (A - B) = 12/13
Cos A Cos B + Sin A Sin B = 12/13

Eliminasi...
2Sin A Sin B = (12/13 - 3/5)
2Sin A Sin B = (60 - 39)/65
2Sin A Sin B = 21/65
Sin A Sin B = 21/130

[tex]\displaystyle 7\\\cos(a+b)=\cos a\cos b-\sin a\sin b\\\cos\frac16\pi=\frac34-\sin a\sin b\\\frac12\sqrt3=\frac34-\sin a\sin b\\\sin a\sin b=\frac34-\frac12\sqrt3\\\\\cos(a-b)=\cos a\cos b+\sin a\sin b\\\cos(a-b)=\frac34+\frac34-\frac12\sqrt3\\\boxed{\boxed{\cos(a-b)=\frac12(3-\sqrt3)}}[/tex]

[tex]\displaystyle 8\\A+B+C=180^\circ\\A+45^\circ+105^\circ=180^\circ\\A=30^\circ\\\\\frac{AB}{\sin C}=\frac{BC}{\sin A}\\\frac{\sqrt2+\sqrt6}{\sin105^\circ}=\frac{BC}{\sin30^\circ}\\\frac{\sqrt2+\sqrt6}{\frac{\sqrt2+\sqrt6}{4}}=\frac{BC}{\frac12}\\\boxed{\boxed{2\,\text{cm}=BC}}[/tex]

[tex]\displaystyle 9\\\tan\theta+\tan\gamma=\frac{\sin\theta}{\cos\theta}+\frac{\sin\gamma}{\cos\gamma}\\\tan\theta+\tan\gamma=\frac{\sin\theta\cos\gamma+\cos\theta\sin\gamma}{\cos\theta\cos\gamma}\\\tan\theta+\tan\gamma=\frac{\sin(\theta+\gamma)}{\cos\theta\cos\gamma}\\p=\frac{\sin(\theta+\gamma)}{\cos\theta\cos\gamma}\\\boxed{\boxed{\frac{\cos\theta\cos\gamma}{\sin(\theta+\gamma)}=\frac{1}{p}}}[/tex]

[tex]\displaystyle 10\\\sin^215^\circ+\cos^215^\circ=1\\p^2+\cos^215^\circ=1\\\cos15^\circ=\sqrt{1-p^2}\\\\\sin75^\circ=\sin(60+15)^\circ\\\sin75^\circ=\sin60^\circ\cos15^\circ+\cos60^\circ\sin15^\circ\\\sin75^\circ=\frac12\sqrt3\cdot\sqrt{1-p^2}+\frac12\cdot p\\\boxed{\boxed{\sin75^\circ=\frac12\sqrt{3-3p^2}+\frac12p}}[/tex]

[tex]\displaystyle 11\\\sin C=\frac2{\sqrt{13}}\\\sin(180^\circ-(A+B))=\frac2{\sqrt{13}}\\\sin(A+B)=\frac2{\sqrt{13}}\\\\\tan A\tan B=13\\\frac{\sin A\sin B}{\cos A\cos B}=13\\\sin A\sin B=13\cos A\cos B\\\\\cos(A+B)=\cos A\cos B-\sin A\sin B\\\frac{3}{\sqrt{13}}=\cos A\cos B-13\cos A\cos B\\\frac{3}{\sqrt{13}}=-12\cos A\cos B\\-\frac{1}{4\sqrt{13}}=\cos A\cos B\\\\\tan A+\tan B=\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}\\\tan A+\tan B=\frac{\sin A\cos B+\cos A\sin B}{\cos A\cos B}[/tex]
[tex]\displaystyle \tan A+\tan B=\frac{\sin(A+B)}{\cos A\cos B}\\\tan A+\tan B=\frac{\frac2{\sqrt{13}}}{-\frac{1}{4\sqrt{13}}}\\\tan A+\tan B=-\frac{2}{\frac{1}{4}}\\\boxed{\boxed{\tan A+\tan B=-8}}[/tex]

[tex]\displaystyle 12\\\cos(A+B)-\cos(A-B)=-2\sin\frac12(2A)\sin\frac12(2B)\\\frac35-\frac{12}{13}=-2\sin A\sin B\\\frac{39-60}{65}=-2\sin A\sin B\\\boxed{\boxed{\frac{21}{130}=\sin A\sin B}}[/tex]