Soal dan Pembahasan Super Lengkap – Penerapan Identitas Trigonometri

Diketahui $\cos 2A = -\dfrac{7}{25}.$
Karena $180^{\circ} \leq 2A \leq 270^{\circ}$, maka dengan membagi $2$ pada ketiga ruasnya, diperoleh $90^{\circ} \leq A \leq 135^{\circ}.$
Jadi, $A$ berada di kuadran II.
Perhatikan bahwa $\cos 2A = 2 \cos^2 A-1$ sehingga kita peroleh
$\begin{aligned} \cos 2A & = -\dfrac{7}{25} \\ 2 \cos^2 A-1 & = -\dfrac{7}{25} \\ \cos^2 A & = \dfrac{9}{25} \\ \cos A & =- \dfrac35 \end{aligned}$
$\cos A$ bernilai negatif karena $A$ berada di kuadran II (ingat: SEMUA SINdikat TANgannya KOSong ).
Diketahui: $\text{sa} = 3$ dan $\text{mi} = 5$. Dengan pendekatan segitiga siku-siku dan rumus Pythagoras, diperoleh
$\text{de} = \sqrt{5^2-3^2} = 4.$
Dari sini, diperoleh
$\begin{aligned} \sin A & = \dfrac{\text{de}}{\text{mi}} = \dfrac45 \\ \tan A & = -\dfrac{\text{de}}{\text{sa}} = -\dfrac43 \\ \csc A & = \dfrac{\text{mi}}{\text{de}} = \dfrac54 \end{aligned}$
Perhatikan bahwa hanya $\sin$ dan kebalikannya, $\csc$, yang bernilai positif di kuadran II.
Berdasarkan uraian di atas, opsi jawaban yang tepat adalah E.

Diketahui
$\sin A = \dfrac{\text{de}}{\text{mi}} = \dfrac35$ dan
$\cos B = \dfrac{\text{sa}}{\text{mi}} = – \dfrac35.$
Kuadran saat sinus sudut bernilai positif dan kosinus sudut bernilai negatif adalah kuadran II. Jadi, $A$ dan $B$ terletak di kuadran II.
Dengan menggunakan pendekatan segitiga siku-siku dan Teorema Pythagoras (Tripel Pythagoras: $3, 4, 5),$ kita peroleh
$\cos A = -\dfrac45$ dan $\sin B = \dfrac45.$
Selanjutnya, dengan menggunakan identitas jumlah sudut:
$\boxed{\begin{aligned} \sin (x + y) & = \sin x \cos y + \cos x \sin y \\ \sin 2x & = 2 \sin x \cos x \\ \cos 2x & = \cos^2 x-\sin^2 x \end{aligned}}$
diperoleh
$$\begin{aligned} \sin (2A+B) & = \sin 2A \cos B + \cos 2A \sin B \\ & = (2 \sin A \cos A) \cos B + (\cos^2 A-\sin^2 A) \sin B \\ & = 2 \cdot \dfrac35 \cdot \left(-\dfrac45\right) \cdot \left(-\dfrac35\right) + \left(\left(-\dfrac45\right)^2-\left(\dfrac35\right)\right)^2 \cdot \dfrac45 \\ & = \dfrac{72}{125} + \left(\dfrac{16}{25}-\dfrac{9}{25}\right) \cdot \dfrac45 \\ & = \dfrac{72}{125} + \dfrac{28}{125} \\ & = \dfrac{100}{125} = \dfrac45 \end{aligned}$$Jadi, nilai dari $\boxed{\sin (2A+B) = \dfrac45}$
(Jawaban B)

Gunakan identitas jumlah & selisih sudut.
$$\boxed{\cos A-\cos B = -2 \sin \dfrac12(A+B) \sin \dfrac12(A-B)}$$Dengan demikian, dapat kita tuliskan
$$\begin{aligned} \cos 265^{\circ}-\cos 95^{\circ} & = -2 \sin \dfrac12(265^{\circ}+95^{\circ}) \sin \dfrac12(265^{\circ}-95^{\circ}) \\ & = -2 \sin \dfrac12(360^{\circ}) \sin \dfrac12(170^{\circ}) \\ & = -2 \sin 180^{\circ} \sin 85^{\circ} \\ & = -2 \cdot 0 \cdot \sin 85^{\circ} \\ & = 0 \end{aligned}$$Jadi, nilai dari $\boxed{\cos 265^{\circ}-\cos 95^{\circ} = 0}$
(Jawaban C)

Gunakan identitas jumlah & selisih sudut.
$$\boxed{\sin A-\sin B = 2 \cos \dfrac12(A+B) \sin \dfrac12(A-B)}$$Dengan demikian, dapat kita tuliskan
$$\begin{aligned} \sin 75^{\circ}-\sin 165^{\circ} & = 2 \cos \dfrac12(75^{\circ}+165^{\circ}) \sin \dfrac12(75^{\circ}-165^{\circ}) \\ & = 2 \cos \dfrac12(240^{\circ}) \sin \dfrac12(-90^{\circ}) \\ & = 2 \cos 120^{\circ} (-\sin 45^{\circ}) \\ & = 2 \cdot \left(-\dfrac12\right) \cdot \left(-\dfrac12\sqrt2\right) \\ & = \dfrac12\sqrt2 \end{aligned}$$Jadi, nilai dari $\boxed{\sin 75^{\circ}-\sin 165^{\circ} = \dfrac12\sqrt2}$
(Jawaban D)

Diketahui $\cos x = \dfrac35$ dengan $x$ berada di kuadran pertama.
Diketahui: $\text{sa} = 3$ dan $\text{mi} = 5$.
Dengan pendekatan segitiga siku-siku dan rumus Pythagoras, diperoleh
$\text{de} = \sqrt{5^2-3^2} = 4.$
Dari sini, diperoleh
$\sin x = \dfrac{\text{de}}{\text{mi}} = \dfrac45.$
Selanjutnya, gunakan identitas trigonometri berikut.
$$\boxed{\begin{aligned} \sin A + \sin B & = 2 \sin \dfrac12(A+B) \cos \dfrac12(A-B) \\ \sin 2x & = 2 \sin x \cos x \end{aligned}}$$Kita akan memperoleh
$$\begin{aligned} \sin 3x + \sin x & = 2 \sin \dfrac12(3x+x) \cos \dfrac12(3x-x) \\ & = 2 \sin 2x \cos x \\ & = 2(2 \sin x \cos x) \cos x \\ & = 4 \sin x \cos^2 x \\ & = 4 \cdot \dfrac45 \cdot \left(\dfrac35\right)^2 \\ & = \dfrac{144}{125} \end{aligned}$$Jadi, nilai dari $\boxed{\sin 3x + \sin x = \dfrac{144}{125}}$
(Jawaban E)

Gunakan identitas trigonometri berikut.
$$\boxed{\begin{aligned} \sin^2 x + \cos^2 x & = 1 \\ \cos (A-B) &= \cos A \cos B + \sin A \sin B \end{aligned}}$$Diketahui $\sin A + \sin B = 1.$
Kuadratkan kedua ruas,
$\begin{aligned} (\sin A + \sin B)^2 & = 1^2 \\ \color{blue}{\sin^2 A + 2 \sin A \sin B + \sin^2 B} & \color{blue}{= 1} \end{aligned}$
Diketahui $\cos A+ \cos B = \dfrac{\sqrt5}{\sqrt3}.$
Kuadratkan kedua ruas,
$$\begin{aligned} (\cos A + \cos B)^2 & = \left(\dfrac{\sqrt5}{\sqrt3}\right)^2 \\ \color{red}{\cos^2 A + 2 \cos A \cos B + \cos^2 B} & \color{red}{=\dfrac53}\end{aligned}$$Jumlahkan kedua persamaan yang diberi warna biru dan merah di atas.
$$\begin{aligned} (\sin^2 A + \cos^2 A) + 2 \sin A \sin B + 2 \cos A \cos B) + (\sin^2 B + \cos^2 B) & = 1+\dfrac53 \\ \cancel{1} + 2(\cos A \cos B + \sin A \sin B) + 1 & = \cancel{1}+\dfrac53 \\ 2(\cos A \cos B + \sin A \sin B) & = \dfrac23 \\ \cos A \cos B + \sin A \sin B & = \dfrac13 \\ \cos (A-B) & = \dfrac13 \end{aligned}$$Jadi, nilai dari $\boxed{\cos(A-B)=\dfrac13}$
(Jawaban E)

Diketahui bahwa $x-y = \dfrac{\pi}{6}$ dan $\color{red}{\tan x = 3 \tan y}$ dengan $x, y$ lancip.
Dengan menggunakan identitas selisih sudut pada tangen, kita peroleh
$\begin{aligned} \tan (x-y) & = \dfrac{\color{red}{\tan x}-\tan y}{1+\color{red}{\tan x} \tan y} \\ \tan \dfrac{\pi}{6} &= \dfrac{\color{red}{3 \tan y}-\tan y}{1+\color{red}{3 \tan y} \tan y} \\ \dfrac{1}{\sqrt3} & = \dfrac{2 \tan y}{1 + 3 \tan^2 y} \\ 1+3 \tan^2 y & = 2\sqrt3 \tan y \\ (\sqrt3 \tan y-1)^2 & = 0 \\ \sqrt3 \tan y-1 & = 0 \\ \tan y & = \dfrac{1}{\sqrt3} \\ \Rightarrow y & = \dfrac{\pi}{6} \end{aligned}$
Karena $x-y=\dfrac{\pi}{6}$ dan $y=\dfrac{\pi}{6}$, berarti $x=\dfrac{\pi}{3}$.
Jadi, nilai $\boxed{x+y=\dfrac{\pi}{3}+\dfrac{\pi}{6} = \dfrac{\pi}{2}}$
(Jawaban A)

Diketahui:
$\begin{aligned} \sin \alpha & = \dfrac35 \\ \cos \beta & = \dfrac{12}{13} \\ \alpha, \beta~&\text{lancip}. \end{aligned}$
Nilai sinus untuk sudut $\beta$ dan kosinus untuk sudut $\alpha$ akan bernilai positif karena $\alpha, \beta$ keduanya di kuadran pertama.
Perhatikan bahwa
$\begin{aligned} \cos \alpha & = + \sqrt{1-\sin^2 \alpha} \\ & = \sqrt{1-\left(\dfrac35\right)^2} \\ & = \sqrt{1-\dfrac{9}{25}} \\ & = \sqrt{\dfrac{16}{25}} = \dfrac45 \end{aligned}$
dan
$\begin{aligned} \sin \beta & = + \sqrt{1-\cos^2 \beta} \\ & = \sqrt{1-\left(\dfrac{12}{13}\right)^2} \\ & = \sqrt{1-\dfrac{144}{169}} \\ & = \sqrt{\dfrac{25}{169}} = \dfrac{5}{13} \end{aligned}$
Dengan menggunakan identitas jumlah sudut sinus, kita peroleh
$\begin{aligned} \sin (\alpha + \beta) & = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ & = \dfrac35 \cdot \dfrac{12}{13} + \dfrac45 \cdot \dfrac{5}{13} \\ & = \dfrac{36}{65}+\dfrac{20}{65} = \dfrac{56}{65} \end{aligned}$
Jadi, nilai $\boxed{\sin (\alpha + \beta) = \dfrac{56}{65}}$
(Jawaban A)

Diketahui $\cos x = \dfrac{12}{13}$.
Dengan menggunakan identitas Pythagoras, diperoleh
$\begin{aligned} \sin x & = \sqrt{1-\cos^2 x} \\ & = \sqrt{1-\left(\dfrac{12}{13}\right)^2} \\ & = \sqrt{\dfrac{25}{169}} = \dfrac{5}{13} \end{aligned}$
Perhatikan bahwa $\tan \dfrac12x$ dapat ditentukan dengan menggunakan identitas setengah sudut.
$\begin{aligned} \tan \dfrac12x & = \dfrac{1-\cos x}{\sin x} \\ & = \dfrac{1-\dfrac{12}{13}}{\dfrac{5}{13}} \\ & = \dfrac{\dfrac{1}{\cancel{13}}}{\dfrac{5}{\cancel{13}}} = \dfrac15 \end{aligned}$
Jadi, nilai dari $\boxed{\tan \dfrac12x = \dfrac15}$
(Jawaban B)

Gunakan identitas sudut ganda berikut.
$\boxed{\begin{aligned} \cos 2A & = 1-2 \sin^2 A \\ \sin 2A & = 2 \sin A \cos A \end{aligned}}$
Kita akan memperoleh
$$\begin{aligned} \dfrac{1+\cos 2A}{\sin 2A} & = \dfrac{1 + (1-2 \sin^2 A)}{2 \sin A \cos A} \\ & = \dfrac{2-2 \sin^2 A}{2 \sin A \cos A} \\ & = \dfrac{2(1 -\sin^2 A)}{2 \sin A \cos A} \\ & = \dfrac{2 \cos^2 A}{2 \sin A \cos A} \\ & = \dfrac{\cos A}{\sin A} = \cot A \end{aligned}$$Jadi, bentuk lain dari $\dfrac{1+\cos 2A}{\sin 2A}$ adalah $\boxed{\cot A}$
(Jawaban C)

Diketahui:
$\begin{aligned} \tan \alpha & = 1 \\ \tan \beta & = \dfrac{1}{3} \\ \alpha, \beta~&\text{lancip}. \end{aligned}$
Nilai sinus dan kosinus untuk sudut $\beta$ dan sudut $\alpha$ akan bernilai positif karena $\alpha, \beta$ keduanya di kuadran pertama.
Karena $\tan \alpha = \dfrac{\text{de}}{\text{sa}} = \dfrac{1}{1}$, maka
$\begin{aligned} \sin \alpha & = \dfrac{ \text{de}}{\text{mi}} = + \dfrac{1}{\sqrt{1+1}} = \dfrac{1}{\sqrt2} \\ \cos \alpha & = \dfrac{ \text{sa}}{\text{mi}} = + \dfrac{1}{\sqrt{1+1}} = \dfrac{1}{\sqrt2} \end{aligned}$
Karena $\tan \beta = \dfrac{\text{de}}{\text{sa}} = \dfrac{1}{3}$, maka
$\begin{aligned} \sin \beta & = \dfrac{ \text{de}}{\text{mi}} = + \dfrac{1}{\sqrt{1+3^2}} = \dfrac{1}{\sqrt{10}} \\ \cos \beta & = \dfrac{ \text{sa}}{\text{mi}} = +\dfrac{3}{\sqrt{1+3^2}} = \dfrac{3}{\sqrt{10}} \end{aligned}$
Dengan menggunakan identitas selisih sudut sinus, kita peroleh
$$\begin{aligned} \sin (\alpha- \beta) & = \sin \alpha \cos \beta-\cos \alpha \sin \beta \\ & = \dfrac{1}{\sqrt{2}} \cdot \dfrac{3}{\sqrt{10}}-\dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{\sqrt{10}} \\ & = \dfrac{3}{\sqrt{20}}-\dfrac{1}{\sqrt{20}} = \dfrac{2}{\sqrt{20}} = \dfrac15\sqrt5 \end{aligned}$$Jadi, nilai $\boxed{\sin (\alpha-\beta) = \dfrac15\sqrt5}$
(Jawaban B)

Perhatikan bahwa $\sin (\alpha-\beta) = \dfrac16$ dapat ditulis kembali dalam bentuk lain menggunakan identitas selisih sudut sinus, yakni
$\color{red}{\sin \alpha \cos \beta-\cos \alpha \sin \beta = \dfrac16}.$
Dari persamaan $\tan \alpha-\tan \beta = \dfrac15$, kita peroleh
$\begin{aligned} \dfrac{\sin \alpha}{\cos \alpha}-\dfrac{\sin \beta}{\cos \beta} & = \dfrac15 \\ \dfrac{\color{red}{\sin \alpha \cos \beta-\cos \alpha \sin \beta}}{\cos \alpha \cos \beta} & = \dfrac15 \\ \dfrac{\color{red}{\frac16}}{\cos \alpha \cos \beta} & = \dfrac15 \\ \cos \alpha \cos \beta & = \dfrac56 \end{aligned}$
Jadi, nilai dari $\boxed{\cos \alpha \cos \beta=\dfrac56}$
(Jawaban D)

Diketahui $\sin x + \cos x = -\dfrac15$.
Kuadratkan kedua ruas untuk mendapatkan
$$\begin{aligned} (\sin x + \cos x)^2 & = \left(-\dfrac15\right)^2 \\ \color{red}{\sin^2 x} + \color{blue}{2 \sin x \cos x} + \color{red}{\cos^2 x} & = \dfrac{1}{25} \\ 1 + \sin 2x & = \dfrac{1}{25} \\ \sin 2x & = \dfrac{1}{25}-1 \\ &= -\dfrac{24}{25} \end{aligned}$$Catatan:
$\boxed{\begin{aligned} \color{red}{\sin^2 x + \cos^2 x} & = \color{red}{1} \\ \color{blue}{2 \sin x \cos x} & = \color{blue}{\sin 2x} \end{aligned}}$
Jadi, nilai dari $\boxed{\sin 2x = -\dfrac{24}{25}}$
(Jawaban A)

Dari persamaan
$\color{blue}{\sin \theta + \cos \theta = \dfrac12}$,
kuadratkan kedua ruasnya untuk mendapatkan
$\begin{aligned} (\sin \theta + \cos \theta)^2 & = (\dfrac12)^2 \\ \color{red}{\sin^2 \theta} + 2 \sin \theta \cos \theta + \color{red}{\cos^2 \theta} & = \dfrac14 \\ \color{red}{1} + 2 \sin \theta \cos \theta & = \dfrac14 \\ 2 \sin \theta \cos \theta & = -\dfrac34 \\ \sin \theta \cos \theta & = -\dfrac38 \end{aligned}$
Gunakan rumus pemfaktoran:
$a^3+b^3 = (a+b)^3-3ab(a+b)$
Untuk $a = \sin \theta$ dan $b = \cos \theta$, kita peroleh
$$\begin{aligned} \sin^3 \theta + \cos^3 \theta & = (\color{blue}{\sin \theta + \cos \theta})^3-3 \sin \theta \cos \theta(\color{blue}{\sin \theta + \cos \theta}) \\ & = \left(\dfrac12\right)^3-3\left(-\dfrac38\right)\left(\dfrac12\right) \\ & = \dfrac18 + \dfrac{9}{16} = \dfrac{11}{16} \end{aligned}$$Jadi, nilai dari $\boxed{\sin^3 \theta + \cos^3 \theta = \dfrac{11}{16}}$
(Jawaban E)

Ubah terlebih dahulu persamaan matriks di atas dalam bentuk persamaan aljabar biasa dengan melakukan perkalian matriks.
Kita akan memperoleh dua persamaan, yaitu
$$\begin{cases} \tan x \cos^2 x + \sin x \cos x & = \dfrac{a}{2} && (\cdots 1) \\ \cos^2 x + \tan x \sin x \cos x & = \dfrac{b}{2} && (\cdots 2) \end{cases}$$Tinjau persamaan $(2)$.
Karena $\tan x = \dfrac{\sin x}{\cos x}$ (identitas perbandingan) dan $\cos^2 x = 1-\sin^2 x$ (Identitas Pythagoras), maka kita peroleh
$$\begin{aligned} (1-\sin^2 x)+\dfrac{\sin x}{\cancel{\cos x}} \cdot \sin x \cdot \cancel{\cos x} & = \dfrac{b}{2} \\ 1-\sin^2 x + \sin^2 x & = \dfrac{b}{2} \\ 1 & = \dfrac{b}{2} \\ b & = 2 \end{aligned}$$Karena diketahui bahwa $b=2a$ dan kita dapatkan $b=2$, maka $a=1$.
Substitusi $a=1$ pada persamaan $(1)$.
$$\begin{aligned} \tan x \cos^2 x + \sin x \cos x & = \dfrac12 \\ \dfrac{\sin x}{\cancel{\cos x}} \cdot \cancelto{\cos x}{\cos^2 x} + \sin x \cos x & = \dfrac12 \\ \sin x \cos x + \sin x \cos x & = \dfrac12 \\ \color{red}{2 \sin x \cos x} & = \dfrac12 \\ \color{red}{\sin 2x} & = \dfrac12 \end{aligned}$$Diketahui $0 \leq x \leq \pi.$
Sinus bernilai $\dfrac12$ ketika sudutnya $30^{\circ}$ atau $150^{\circ}$. Untuk itu, kita tuliskan
$\sin 2x = \sin 30^{\circ}$ yang berarti $x = 15^{\circ}$ dan
$\sin 2x = \sin 150^{\circ}$ yang berarti $x = 75^{\circ}.$
Jadi, nilai $x$ yang memenuhi berdasarkan pilihan jawaban yang tersedia adalah $\boxed{15^{\circ}}$
(Jawaban E)

Diketahui $\sin \alpha = \dfrac{\sqrt3}{3} = \dfrac{\text{de}}{\text{mi}}.$
Dengan menggunakan Teorema Pythagoras, kita peroleh
$\text{sa} = \sqrt{(3)^2-(\sqrt3)^2} = \sqrt6$
Untuk itu, didapat
$\begin{aligned} \cos \alpha & = \dfrac{\text{sa}}{\text{mi}} = \dfrac{\sqrt6}{3} \\ \cot \alpha & = \dfrac{\text{sa}}{\text{de}} = \dfrac{\sqrt6}{\sqrt3} = \sqrt2 \end{aligned}$
Perhatikan bahwa $\tan \left(\dfrac12 \pi-\alpha\right) = \cot \alpha$.
Dengan demikian,
$$\begin{aligned} \tan \left(\dfrac12 \pi-\alpha\right)+3 \cos \alpha & = \cot \alpha + 3 \cos \alpha \\ & = \sqrt2 + \cancel{3} \cdot \dfrac{\sqrt6}{\cancel{3}} \\ & = \sqrt6 + \sqrt2 \end{aligned}$$Jadi, nilai dari $\boxed{\tan \left(\dfrac12 \pi-\alpha\right)+3 \cos \alpha = \sqrt6 + \sqrt2}$
(Jawaban C)

Dengan menggunakan identitas sudut ganda:
$\boxed{\sin 2A = 2 \sin A \cos A}$
diperoleh bahwa
$\begin{aligned} & 2 \cos \left(\dfrac14 \pi + x\right) \sin \left(\dfrac14 \pi + x\right) \\ & = \sin 2\left(\dfrac14 \pi + x\right) \\ & = \sin \left(\dfrac12 \pi + 2x\right) \\ & = \cos 2x \end{aligned}$
Jadi, nilai dari $$\boxed{2 \cos \left(\dfrac14 \pi + x\right) \sin \left(\dfrac14 \pi + x\right) = \cos 2x}$$(Jawaban E)

Gunakan identitas jumlah fungsi sinus:
$$\boxed{\sin A + \sin B = 2 \sin \dfrac{A+B}{2} \cos \dfrac{A+B}{2}}$$Kita akan memperoleh
$$\begin{aligned} & \color{red}{\sin 160^{\circ} + \sin 140^{\circ}}-\cos 10^{\circ} \\ & = 2 \sin \dfrac{160^{\circ}+140^{\circ}}{2} \cos \dfrac{160^{\circ}-140^{\circ}}{2}-\cos 10^{\circ} \\ & = 2 \sin 150^{\circ} \cos 10^{\circ}-\cos 10^{\circ} \\ & = 2 \cdot \dfrac12 \cdot \cos 10^{\circ}-\cos 10^{\circ} = 0 \end{aligned}$$Jadi, nilai dari $\boxed{\sin 160^{\circ} + \sin 140^{\circ}-\cos 10^{\circ} = 0}$
(Jawaban C)

Gunakan identitas Pythagoras dan identitas relasi sudut di kuadran pertama berikut.
$\boxed{\begin{aligned} \cos^2 x & = 1-\sin^2 x \\ \sin x & = \cos (90^{\circ}-x) \end{aligned}}$
Kita akan memperoleh
$$\begin{aligned} & \cos^2 30^{\circ} + \cos^2 40^{\circ}+\cos^2 50^{\circ} +\cos^2 60^{\circ} \\ & = (1-\cancel{\sin^2 30^{\circ}}) + (1-\bcancel{\sin^2 40^{\circ}}) + \bcancel{\sin^2 40^{\circ}} + \cancel{\sin^2 30^{\circ}} \\ & = 1+1=2 \end{aligned}$$Jadi, nilai dari $$\boxed{\cos^2 30^{\circ} + \cos^2 40^{\circ}+\cos^2 50^{\circ} +\cos^2 60^{\circ}=2}$$(Jawaban A)

Diberikan $\triangle ABC$ (siku-siku). Diketahui $\sin A \sin B = 0,5$.
Diketahui juga bahwa siku-sikunya di titik $A$, berarti dapat ditulis
$\sin 90^{\circ} \sin B = 0,5.$
Karena $\sin 90^{\circ} = 1$, maka haruslah $\sin B = 0,5.$
Dengan menggunakan relasi sudut sinus dan kosinus, diperoleh
$\begin{aligned} \cos (A+B) & = \cos (90^{\circ}+B) \\ & = -\sin B = -0,5 \end{aligned}$
Jadi, nilai dari $\boxed{\cos (A+B)=-0,5}$
(Jawaban D)

Perhatikan bahwa $\cos A \cos B = \dfrac13$. Apabila $A = 90^{\circ}$ atau $B = 90^{\circ}$, maka persamaan tersebut tidak berlaku sebab $\cos 90^{\circ} = 0$. Artinya, sudut siku-sikunya di $C$ ditulis $\angle C = 90^{\circ}$.
Ini juga berarti $B = 90^{\circ}-A$.
Dengan demikian, diperoleh
$\begin{aligned} \cos A \cos B & = \dfrac13 \\ \cos A \cos (90^{\circ}-A) & = \dfrac13 \\ \cos A \sin A & = \dfrac13 \\ \text{Kalikan}~2~\text{pada kedua ruas} \\ 2 \cos A \sin A & = \dfrac23 \\ \sin 2A & = \dfrac23 \end{aligned}$
Catatan: Ingat bahwa
$\boxed{\begin{aligned} \cos (90^{\circ}-A) & = \sin A \\ 2 \sin A \cos A & = \sin 2A \end{aligned}}$
Dengan menggunakan pendekatan segitiga siku-siku seperti gambar,
diperoleh $\boxed{\cos 2A = \dfrac{\text{sa}}{\text{mi}} = \dfrac13\sqrt5}$
(Jawaban E)

Perhatikan bahwa dengan mengalikan kedua ruas dengan $2$, kita peroleh
$\begin{aligned} \dfrac12x+y =\dfrac{\pi}{4} \Rightarrow x + 2y & = \dfrac{\pi}{2} \\ x & = \dfrac{\pi}{2}-2y \end{aligned}$
Selanjutnya, gunakan identitas sudut ganda untuk tangen.
$\boxed{\begin{aligned} \tan (90^{\circ}-\theta) & = \cot \theta \\ \tan 2\theta & = \dfrac{2 \tan \theta}{1-\tan^2 \theta} \end{aligned}}$
Untuk kotangen, balik saja posisi pembilang dan penyebutnya.
Dengan demikian,
$\begin{aligned} \tan x & = \tan \left(\dfrac{\pi}{2}-2y\right) \\ & = \cot 2y \\ & = \dfrac{1-\tan^2 y}{2 \tan y} \end{aligned}$
Jadi, diperoleh $\boxed{\tan x = \dfrac{1-\tan^2 y}{2 \tan y}}$
(Jawaban A)

Diketahui $\sin (3x+2y) = \dfrac13$ dan $\cos (3x-4y) = \dfrac34$.
Misalkan $\alpha = 3x + 2y$ dan $\beta = 3x-4y$ sehingga
$$\begin{aligned} 6y & = (3x+2y)-(3x-4y) = \alpha-\beta \\ 9x &= 2(3x+2y)+(3x-4y) = 2\alpha+\beta \end{aligned}$$

Perhatikan bahwa $\sin \alpha = \dfrac13$ sehingga dengan menggunakan pendekatan segitiga siku-siku, diperoleh $\cos \alpha = \dfrac{2\sqrt2}{3}$. Begitu juga untuk $\cos \beta = \dfrac34$, diperoleh $\sin \beta = \dfrac{\sqrt7}{4}$.
Selanjutnya, kita peroleh
$\begin{aligned} \sin 6y & = \sin (\alpha-\beta) \\ & = \sin \alpha \cos \beta+\cos \alpha \sin \beta \\ & = \dfrac13 \cdot \dfrac34-\dfrac{\sqrt7}{4} \cdot \dfrac{2\sqrt2}{3} \\ & = \dfrac{3-2\sqrt{14}}{12} \end{aligned}$
dan
$$\begin{aligned} \cos 9x & = \cos (2\alpha+\beta) \\ & = \cos 2\alpha \cos \beta-\sin 2\alpha \sin \beta \\ & = (\cos^2 \alpha-\sin^2 \alpha) \cos \beta-2 \sin \alpha \cos \alpha \sin \beta \\ & = \left(\left(\dfrac{2\sqrt2}{3}\right)^2-\left(\dfrac13\right)^2\right) \cdot \dfrac34-\cancel{2} \cdot \dfrac13 \cdot \dfrac{2\sqrt2}{3} \cdot \dfrac{\sqrt7}{\cancelto{2}{4}} \\ & = \dfrac{7}{12}-\dfrac{2\sqrt{14}}{18} = \dfrac{21-4\sqrt{14}}{36} \end{aligned}$$Dengan demikian, didapat
$\begin{aligned} \dfrac{\sin 6y}{\cos 9x} & = \dfrac{3-2\sqrt{14}}{\cancel{12}} \times \dfrac{\cancelto{3}{36}}{21-4\sqrt{14}} \\ & = \dfrac{9-6\sqrt{14}}{21-4\sqrt{14}} \end{aligned}$
Jadi, nilai dari $\dfrac{\sin 6y}{\cos 9x}$ adalah $\boxed{\dfrac{9-6\sqrt{14}}{21-4\sqrt{14}}}$
(Jawaban A)

Gunakan identitas trigonometri berikut.
$$\boxed{\begin{aligned} 2 \cos x \cos y & = \cos (x+y) + \cos (x-y) \\ \cos x + \cos y & = 2 \cos \dfrac12(x+y) \cos \dfrac12(x-y) \end{aligned}}$$Untuk itu, dapat kita peroleh
$$\begin{aligned} & 20 \cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ} \\ & = 10(2 \cos 20^{\circ} \cos 40^{\circ}) \cos 80^{\circ} \\ & = 10(\cos 60^{\circ} + \cos (-20^{\circ})) \cos 80^{\circ} \\ & = 10 \cos 60^{\circ} \cos 80^{\circ} + 10 \cos 20^{\circ} \cos 80^{\circ} \\ & = 10 \left(\dfrac12\right) \cos 80^{\circ} + 5(2 \cos 20^{\circ} \cos 80^{\circ}) \\ & = 5 \cos 80^{\circ} + 5(\cos 100^{\circ} + \cos (-60^{\circ}) \\ & = 5(\cos 80^{\circ} + \cos 100^{\circ}) + \dfrac52 \\ & = 5(2 \cos \dfrac12(80^{\circ}+100^{\circ}) \cos \dfrac12(80^{\circ}-100^{\circ}) + \dfrac52 \\ & = 5(2 \cos 90^{\circ} \cos 10^{\circ}) + \dfrac52 \\ & = 5(2(0) \cos 10^{\circ}) + \dfrac52 \\ & = \dfrac52 \end{aligned}$$Jadi, nilai dari $\boxed{20 \cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ} = \dfrac52}$
(Jawaban C)

Diketahui $\sin (40+x)^{\circ} = \dfrac{\text{de}}{\text{mi}} = \dfrac{a}{1}$ dengan $(40^{\circ}+x)$ berada di kuadran pertama.
Berdasarkan rumus Pythagoras, diperoleh
$\text{sa} = \sqrt{1^2-a^2} = \sqrt{1-a^2}$
sehingga
$\begin{aligned} \cos (40^{\circ} + x) & = \dfrac{\text{sa}}{\text{mi}} \\ & = \dfrac{\sqrt{1-a^2}}{1} \\ & = \sqrt{1-a^2} \end{aligned}$
Dengan demikian,
$$\begin{aligned} \cos (70^{\circ}+x) & = \color{red}{\cos ((40^{\circ}+x)+30^{\circ})} \\ & = \cos (40^{\circ}+x) \cos 30^{\circ}-\sin (40^{\circ}+x) \sin 30^{\circ} \\ & = \sqrt{1-a^2} \cdot \dfrac12\sqrt3-a \cdot \dfrac12 \\ & = \dfrac12\sqrt{3(1-a^2)}-\dfrac12a \\ & = \dfrac{\sqrt{3(1-a^2)}-a}{2} \end{aligned}$$Catatan: gunakan rumus jumlah dan selisih sudut.
Jadi, nilai dari $\boxed{\cos (70^{\circ}+x) = \dfrac{\sqrt{3(1-a^2)}-a}{2}}$
(Jawaban B)

Diketahui $\cos (x+15^{\circ}) = a$ dengan $0^{\circ} \leq x \leq 30^{\circ}.$
Perhatikan bahwa
$\begin{aligned} \color{red}{\cos (2x+30^{\circ})} & = \cos 2(x+15^{\circ}) \\ & = 2 \cos^2 (x+15^{\circ})-1 \\ & = \color{red}{2a^2-1} \end{aligned}$
Dengan demikian,
$$\begin{aligned} \color{blue}{\sin (2x+30^{\circ})} & = \sqrt{1-\cos^2 (2x+30^{\circ})} \\ & = \sqrt{1-(2a^2-1)^2} \\ & = \sqrt{1-(4a^4-4a^2+1)} \\ & = \sqrt{4a^2(1-a^2)} \\ & = \color{blue}{2a\sqrt{1-a^2}} \end{aligned}$$Selanjutnya, dengan menggunakan identitas jumlah sudut kosinus diperoleh
$$\begin{aligned} \cos (2x+60^{\circ}) & = \cos ((2x+30^{\circ})+30^{\circ}) \\ & = \color{red}{\cos (2x+30^{\circ})} \cdot \cos 30^{\circ}-\color{blue}{\sin (2x+30^{\circ})} \cdot \sin 30^{\circ} \\ & = \color{red}{(2a^2-1)} \cdot \dfrac12\sqrt3-\color{blue}{2a\sqrt{1-a^2}} \cdot \dfrac12 \\ & = \dfrac12\sqrt3(2a^2-1)-a\sqrt{1-a^2} \end{aligned}$$Jadi, nilai dari
$$\boxed{\cos (2x+60^{\circ}) = \dfrac12\sqrt3(2a^2-1)-a\sqrt{1-a^2}}$$(Jawaban B)

Perhatikan sketsa gambar berikut.
Dari gambar, diketahui bahwa
$\begin{aligned} |AB| & = 3 \\ |AC| & = 5 \\ |BF| & = 2 \\ |CD| & = 1 \end{aligned}$
Catatan: Notasi seperti $|AB|$ artinya panjang $AB$.
Dengan menggunakan Teorema Pythagoras, diperoleh
$\begin{aligned} |AF| & = \sqrt{|AB|^2+|BF|^2} \\ & = \sqrt{3^2+2^2} = \sqrt{13} \\ |AD| & = \sqrt{|AC|^2+|CD|^2} \\ & = \sqrt{5^2+1^2} = \sqrt{26} \end{aligned}$
sehingga berlaku perbandingan trigonometri
$\begin{aligned} \sin \alpha & = \dfrac{|CD|}{|AD|} = \dfrac{1}{\sqrt{26}} \\ \cos \alpha & = \dfrac{|AC|}{|AD|} = \dfrac{5}{\sqrt{26}} \\ \sin \beta & = \dfrac{|AB|}{|AF|} = \dfrac{3}{\sqrt{13}} \\ \cos \beta & = \dfrac{|BF|}{|AF|} = \dfrac{2}{\sqrt{13}} \end{aligned}$
Dengan menggunakan identitas selisih sudut sinus, diperoleh
$$\begin{aligned} \sin (\beta-\cos \alpha) & = \sin \beta \cos \alpha-\sin \alpha \cos \beta \\ & = \dfrac{3}{\sqrt{13}} \cdot \dfrac{5}{\sqrt{26}}- \dfrac{1}{\sqrt{26}} \cdot \dfrac{2}{\sqrt{13}} \\ & = \dfrac{15}{\sqrt{13^2 \cdot 2}}-\dfrac{2}{\sqrt{13^2 \cdot 2}} \\ & = \dfrac{13}{13\sqrt2} = \dfrac12\sqrt2 \end{aligned}$$Karena $\alpha, \beta$ lancip, maka $(\alpha-\beta)$ haruslah lancip sehingga kemungkinan nilai untuk $(\beta-\alpha)$ adalah $45^{\circ}$.
(Jawaban B)

Substitusi $\cos^2 \theta = 1-\sin^2 \theta$ pada persamaan yang diberikan untuk memperoleh
$\begin{aligned} & \sin^2 \theta + (1-\sin^2 \theta)^2 \\ & = \sin^2 \theta + (1-2 \sin^2 \theta + \sin^4 \theta) \\ & = 1- \sin^2 \theta + \sin^4 \theta \\ & = 1-(\sin^2 \theta)(1-\sin^2 \theta) \\ & = 1-\sin^2 \theta \cos^2 \theta \\ & = 1-(\sin \theta \cos \theta)^2 \\ & = 1-\left(\dfrac12 \sin 2\theta\right)^2 = 1-\dfrac14 \sin^2 2\theta \end{aligned}$
Catatan: $2 \sin x \cos x = \sin 2x$.
Karena nilai $\sin \theta$ berada pada interval $-1 \leq \sin \theta \leq 1$, maka untuk itu, berlaku
$\begin{array}{rcccl} -1 & \leq & \sin 2\theta & \leq & 1 \\ 0 & \leq & \sin^2 2\theta & \leq & 1 \\ 0 & \leq & \dfrac14 \sin^2 2\theta & \leq & 1 \\ -\dfrac14 & \leq & -\dfrac14 \sin^2 2\theta & \leq & 0 \\ 1-\dfrac14 & \leq & 1-\dfrac14 \sin^2 2\theta & \leq & 1+0 \\ \dfrac34 & \leq & \dfrac14 \sin^2 2\theta & \leq & 1 \end{array}$
Jadi, untuk semua nilai real $\theta$, berlaku $\boxed{\dfrac34 \leq \alpha \leq 1}$
(Jawaban D)

Jawaban a)
Berdasarkan identitas sudut ganda:
$\boxed{\cos 2x = 1-2 \sin^2 x}$
kita peroleh bahwa
$\begin{aligned} 1-2 \sin^2 22,5^{\circ} & = \cos 2(22,5^{\circ}) \\ & = \cos 45^{\circ} \\ & = \dfrac12\sqrt2 \end{aligned}$
Jawaban b)
Berdasarkan identitas perkalian:
$$\boxed{2 \sin A \cos B = \sin (A+B)+\sin (A-B)}$$kita peroleh bahwa
$$\begin{aligned} & 2 \cos 97,5^{\circ} \sin 52,5^{\circ} \\ & = \sin (52,5^{\circ}+97,5^{\circ}) + \sin (52,5^{\circ}-97,5^{\circ}) \\ & = \sin (150^{\circ}) + \sin (-45^{\circ}) \\ & = \sin 150^{\circ}- \sin 45^{\circ} \\ & = \dfrac12-\dfrac12\sqrt2 \\ & = \dfrac12(1-\sqrt2) \end{aligned}$$Jawaban c)
Berdasarkan identitas perkalian:
$$\boxed{2 \sin A \cos B = \sin (A+B)+\sin (A-B)}$$kita peroleh bahwa
$\begin{aligned} & 6 \sin 165^{\circ} \cos 15^{\circ} \\ & = 3(2 \sin 165^{\circ} \cos 15^{\circ}) \\ & = 3(\sin (165^{\circ} + 15^{\circ}) + \sin (165^{\circ} -15^{\circ}) \\ & = 3(\sin 180^{\circ} + \sin 150^{\circ}) \\ & = 3\left(0 + \dfrac12\right) = \dfrac32 \end{aligned}$
Jawaban d)
Berdasarkan identitas selisih fungsi sinus dan kosinus:
$$\boxed{\begin{aligned} \sin A-\sin B & = 2 \cos \dfrac{A+B}{2} \sin \dfrac{A+B}{2} \\ \cos A-\cos B & = -2 \sin \dfrac{A+B}{2} \sin \dfrac{A+B}{2} \end{aligned}}$$kita peroleh bahwa
$\begin{aligned} & \dfrac{\cos 75^{\circ}-\cos 15^{\circ}}{\sin 75^{\circ}-\sin 15^{\circ}} \\ & = \dfrac{-2 \sin \dfrac{75^{\circ}+15^{\circ}}{2} \cos \dfrac{75^{\circ}-15^{\circ}}{2}}{2 \cos \dfrac{75^{\circ}+15^{\circ}}{2} \sin \dfrac{75^{\circ}-15^{\circ}}{2}} \\ & = \dfrac{-\cancel{2} \sin 45^{\circ} \bcancel{\sin 30^{\circ}}}{\cancel{2} \cos 45^{\circ} \bcancel{\sin 30^{\circ}}} \\ & = \dfrac{-\cancel{\dfrac12\sqrt2}}{\cancel{\dfrac12\sqrt2}} \\ & = -1 \end{aligned}$

Berdasarkan identitas jumlah sudut sinus:
$\boxed{\sin (x+y) = \sin x \cos y + \cos x \sin y}$
kita peroleh
$$\begin{aligned} \sin \left(A+\dfrac{\pi}{3}\right) & = \sin A \\ \sin A \cos \dfrac{\pi}{3} + \cos A \sin \dfrac{\pi}{3} & = \sin A \\ \sin A \cdot \dfrac12 + \cos A \cdot \dfrac12\sqrt3 & = \sin A \\ \dfrac12\sin A-\sin A & = -\dfrac12\sqrt3 \cos A \\ \bcancel{-\dfrac12} \sin A & = \bcancel{-\dfrac12}\sqrt3 \cos A \\ \sin A & = \sqrt3 \cos A \\ \dfrac{\sin A}{\cos A} = \tan A & = \sqrt3 \end{aligned}$$Jadi, terbukti bahwa $\tan A = \sqrt3$.

Berdasarkan identitas jumlah dan selisih sudut kosinus:
$\boxed{\cos (x \pm y) = \cos x \cos y \mp \sin x \sin y}$
kita peroleh
$$\begin{aligned} 3 \cos \left(A+\dfrac{\pi}{4}\right) & = \cos \left(A-\dfrac{\pi}{4}\right) \\ 3 \left(\cos A \cos \dfrac{\pi}{4}-\sin A \sin \dfrac{\pi}{4}\right) & = \cos A \cos \dfrac{\pi}{4}+\sin A \sin \dfrac{\pi}{4} \\ 3 \left(\cos A \cdot \dfrac12\sqrt2 -\sin A \cdot \dfrac12\sqrt2\right) & = \cos A \cdot \dfrac12\sqrt2 +\sin A \cdot \dfrac12\sqrt2 \\ \dfrac32\sqrt2 \cos A-\dfrac32\sqrt2 \sin A & = \dfrac12\sqrt2 \cos A+ \dfrac12\sqrt2 \sin A \\ \dfrac32\sqrt2 \cos A-\dfrac12\sqrt2 \cos A & = \dfrac12\sqrt2 \sin A+\dfrac32\sqrt2 \sin A \\ \bcancel{\sqrt2} \cos A & = 2 \cdot \bcancel{\sqrt2} \sin A \\ \cos A & = 2 \sin A \\ \dfrac{\sin A}{\cos A} = \tan A & = \dfrac12 \end{aligned}$$Jadi, terbukti bahwa $\tan A = \dfrac12$.

Jawaban a)
Diketahui persamaan
$$\cos x = \cos 100^{\circ} \cos 35^{\circ}-\sin 100^{\circ} \sin 35^{\circ}.$$Gunakan identitas jumlah sudut kosinus.
$\boxed{\cos (A+B) = \cos A \cos B-\sin A \sin B}$
Untuk $A = 100^{\circ}$ dan $B = 35^{\circ}$, kita peroleh bahwa
$\begin{aligned} & \cos 100^{\circ} \cos 35^{\circ}-\sin 100^{\circ} \sin 35^{\circ} \\ & = \cos (100^{\circ} + 35^{\circ}) \\ & = \cos 135^{\circ} = -\dfrac12\sqrt2 \end{aligned}$
Jadi, nilai $x = 135^{\circ}$. Selain itu, nilai $x$ bisa juga $225^{\circ}$ karena $\cos 225^{\circ} = -\dfrac12\sqrt2$.
Jawaban b)
Diketahui persamaan
$\sin x = \sin 115^{\circ} \cos 5^{\circ} + \cos 115^{\circ} \sin 5^{\circ}$
Gunakan identitas jumlah sudut sinus.
$$\boxed{\sin (A+B) = \sin A \cos B-\cos A \sin B}$$Untuk $A = 115^{\circ}$ dan $B = 5^{\circ}$, kita peroleh bahwa
$\begin{aligned} & \sin 115^{\circ} \cos 5^{\circ} + \cos 115^{\circ} \sin 5^{\circ} \\ & = \sin (115^{\circ} + 5^{\circ}) \\ & = \sin 120^{\circ} = \dfrac12\sqrt3 \end{aligned}$
Jadi, nilai $x = 120^{\circ}$. Selain itu, nilai $x$ bisa juga $60^{\circ}$ karena $\sin 60^{\circ} = \dfrac12\sqrt3$.
Jawaban c)
Diketahui persamaan
$\tan 3x = \dfrac{\tan \dfrac{7}{12}\pi-\tan \dfrac{\pi}{12}}{1+\tan \dfrac{7}{12}\pi \tan \dfrac{\pi}{12}}$
Gunakan identitas selisih sudut tangen.
$\boxed{\tan (A-B) = \dfrac{\tan A-\tan B}{1+\tan A \tan B}}$
Untuk $A = \dfrac{7}{12}\pi$ dan $B = \dfrac{\pi}{12}$, kita peroleh bahwa
$$\begin{aligned} \dfrac{\tan \dfrac{7}{12}\pi-\tan \dfrac{\pi}{12}}{1+\tan \dfrac{7}{12}\pi \tan \dfrac{\pi}{12}} & = \tan \left(\dfrac{7}{12}\pi-\dfrac{\pi}{12}\right) \\ & = \tan \dfrac12\pi = \tan 3x \end{aligned}$$Berdasarkan persamaan dasar trigonometri, kita dapatkan bentuk berikut.
$\begin{aligned} 3x & = \dfrac12\pi + k \cdot \pi \\ x & = \dfrac16\pi + k \cdot \dfrac{\pi}{3} \end{aligned}$
Untuk $k = 0, 1, 2, 3, 4, 5$, berturut-turut diperoleh $x = \dfrac16\pi$, $x = \dfrac12\pi$, $x = \dfrac56\pi$, $x = \dfrac76\pi$, $\dfrac32\pi$, dan $x = \dfrac{11}{6}\pi$.

Karena $\sin A = \dfrac35$, maka dengan menggunakan identitas Pythagoras, diperoleh
$\begin{aligned} \cos a &= \sqrt{1-\sin^2 a} \\ & = \sqrt{1-\left(\dfrac35\right)^2} \\ & = \sqrt{1-\dfrac{9}{25}} = \sqrt{\dfrac{16}{25}} = \dfrac45 \end{aligned}$
Dengan prinsip yang sama, dari $\cos b = \dfrac{5}{13}$, kita peroleh
$\begin{aligned} \sin b & = \sqrt{1-\cos^2 b} \\ & = \sqrt{1-\left(\dfrac{5}{13}\right)^2} \\ & = \sqrt{1-\dfrac{25}{169}} = \sqrt{\dfrac{144}{169}} = \dfrac{12}{13} \end{aligned}$
Perhatikan bahwa $a+b+c=180^{\circ}$ sehingga $c=180^{\circ}-(a+b)$. Untuk itu, didapat
$\begin{aligned} \sin c & = \sin (180^{\circ}-(a+b)) \\ & = \sin (a+b) \\ & = \sin a \cos b + \cos a \sin b \\ & = \dfrac35 \cdot \dfrac{5}{13} + \dfrac45 \cdot \dfrac{12}{13} \\ & = \dfrac{15}{65}+\dfrac{48}{65} = \dfrac{63}{65} \end{aligned}$
Jadi, nilai dari $\boxed{\sin c = \dfrac{63}{65}}$

Kita akan menentukan nilai $\tan x$ dan $\tan y$ terlebih dahulu dengan memperhatikan gambar di atas.
Berdasarkan rumus Pythagoras, panjang $KM$ dirumuskan oleh
$\begin{aligned} KM & = \sqrt{KL^2+LM^2} \\ & = \sqrt{24^2+7^2} \\ & = \sqrt{625} = 25~\text{cm} \end{aligned}$
Juga berdasarkan rumus Pythagoras, panjang $KN$ dirumuskan oleh
$\begin{aligned} KN & = \sqrt{KM^2-MN^2} \\ & = \sqrt{25^2-15^2} \\ & = \sqrt{400} = 20~\text{cm} \end{aligned}$
Dengan demikian,
$\tan x = \dfrac{LM}{KL} = \dfrac{7}{24}$
$\tan y = \dfrac{MN}{KN} = \dfrac{15}{20} = \dfrac34.$
Gunakan identitas jumlah sudut tangen.
$\begin{aligned} \tan (x+y) & = \dfrac{\tan x + \tan y}{1-\tan x \tan y} \\ & = \dfrac{\dfrac{7}{24} + \dfrac34}{1-\dfrac{7}{24} \cdot \dfrac34} \\ & = \dfrac{\dfrac{25}{24}}{1-\dfrac{21}{96}} \\ & = \dfrac{\cancel{25}}{\cancel{24}} \cdot \dfrac{\cancelto{4}{96}}{\cancelto{3}{75}} \\ & = \dfrac43 \end{aligned}$
Jadi, nilai dari $\boxed{\tan (x+y) = \dfrac{4}{3}}$

Untuk membuktikan pernyataan di atas, kita akan menggunakan identitas jumlah sudut tangen.
$$\begin{aligned} \tan (x+y) & = \dfrac{\tan x + \tan y }{1-\tan x \tan y} \\ \tan \dfrac54\pi & = \dfrac{\tan x + \tan y }{1-\tan x \tan y} \\ 1 & = \dfrac{\tan x + \tan y }{1-\tan x \tan y} \\ 1-\tan x \tan y & = \tan x + \tan y \\ \tan x + \tan y + \tan x \tan y-1 & = 0 \\ (\tan x + 1)(\tan y +1)-2 & = 0 \\ (\tan x + 1)(\tan y +1) & = 2 \end{aligned}$$Jadi, terbukti bahwa $\boxed{(\tan x + 1)(\tan y +1)=2}$

Gunakan identitas jumlah sudut sinus.
$$\begin{aligned} \sin (x+60^{\circ}) & = \sin x \\ \sin x \cos 60^{\circ} + \cos x \sin 60^{\circ} & = \sin x \\ \dfrac12 \sin x + \dfrac12\sqrt3 \cos x & = \sin x \\ \cancel{\dfrac12}\sqrt3 \cos x & = \cancel{\dfrac12} \sin x \\ \sqrt3 \cos x & = \sin x \\ \dfrac{\sin x}{\cos x} = \tan x & = \sqrt3 \end{aligned}$$Jadi, terbukti bahwa $\tan x = \sqrt3$.

Diketahui
$\begin{aligned} p & = \sin x + \sin y \\ q & = \cos x + \cos y \end{aligned}$
Akan dibuktikan bahwa $\dfrac12(p^2+q^2) = 1+\cos (x-y)$ dengan menggunakan sejumlah identitas trigonometri berikut.
$$\boxed{\begin{aligned} \sin^2 x + \cos^2 x & = 1 \\ \cos (x-y) & = \cos x \cos y + \sin x \sin y \end{aligned}}$$Kita akan memperoleh
$$\begin{aligned} \dfrac12(p^2+q^2) & = \dfrac12((\sin x+\sin y)^2+(\cos x + \cos y)^2) \\ & = \dfrac12(\color{red}{\sin^2 x} + 2 \sin x \sin y + \color{blue}{\sin^2 y} + \color{red}{\cos^2 x} + 2 \cos x \cos y + \color{blue}{\cos^2 y}) \\ & = \dfrac12(2(\cos x \cos y + \sin x \sin y)+\color{red}{1}+\color{blue}{1}) \\ & = (\cos x \cos y + \sin x \sin y)+1 \\ & = 1+\cos (x-y) \end{aligned}$$Jadi, terbukti bahwa $\dfrac12(p^2+q^2) = 1+\cos (x-y)$.

Jawaban a)
Dengan menggunakan identitas trigonometri berikut:
$$\boxed{\begin{aligned} \sin (A \pm B) & = \sin A \cos B \pm \cos A \sin B \\ \sin^2 A + \cos^2 A & = 1 \end{aligned}}$$diperoleh
$$\begin{aligned} \sin (x+y) \sin (x-y) & =(\sin x \cos y + \cos x \sin y)(\sin x \cos y-\cos x \sin y) \\ & = \sin^2 x \cos^2 y- \bcancel{\sin x \sin y \cos x \cos y} + \bcancel{\sin x \sin y \cos x \cos y} -\cos^2 x \sin^2 y \\ & = \sin^2 x \cos^2 y-\cos^2 x \sin^2 y \\ & = \sin^2 x(1-\sin^2 y)-(1-\sin^2 x) \sin^2 y \\ & = \sin^2 x-\cancel{\sin^2 x \sin^2 y}-\sin^2 y+\cancel{\sin^2 x \sin^2 y} \\ & = \sin^2 x-\sin^2 y \end{aligned}$$Jadi, terbukti bahwa $\sin (x+y) \sin (x-y) = \sin^2 x-\sin^2 y$.
Jawaban b)
Dengan menggunakan identitas trigonometri berikut:
$$\boxed{\begin{aligned} \sin (A \pm B) & = \sin A \cos B \pm \cos A \sin B \\ \sin^2 A + \cos^2 A & = 1 \end{aligned}}$$diperoleh
$$\begin{aligned} \cos (x+y) \cos (x-y) & =(\cos x \cos y- \sin x \sin y)(\cos x \cos y+\sin x \sin y) \\ & = \cos^2 x \cos^2 y+\bcancel{\sin x \sin y \cos x \cos y}- \bcancel{\sin x \sin y \cos x \cos y} -\sin^2 x \sin^2 y \\ & = \cos^2 x \cos^2 y-\sin^2 x \sin^2 y \\ & = \cos^2 x(1-\sin^2 y)-(1-\cos^2 x) \sin^2 y \\ & = \cos^2 x-\cancel{\cos^2 x \sin^2 y}-\sin^2 y+\cancel{\cos^2 x \sin^2 y} \\ & = \cos^2 x-\sin^2 y \end{aligned}$$Jadi, terbukti bahwa $\cos (x+y) \cos (x-y) = \cos^2 x-\sin^2 y$.

Gunakan hubungan relasi sudut dan identitas trigonometri berikut.
$\begin{aligned} \cos x & = \sin (90^{\circ}-x) \\ \sin^2 x + \cos^2 x & = 1 \end{aligned}$
Kita dapat tuliskan
$$\begin{aligned} & \cos^2 1^{\circ} + \cos^2 2^{\circ} + \cdots + \cos^2 14^{\circ} + \cos^2 76^{\circ} + \cos^2 77^{\circ} + \cdots + \cos^2 90^{\circ} \\ & = \cos^2 1^{\circ} + \cos^2 2^{\circ} + \cdots + \cos^2 14^{\circ} + \sin^2 (90^{\circ}-76^{\circ}) + \sin^2 (90^{\circ}-77^{\circ}) + \cdots + \sin^2 (90^{\circ}-89^{\circ}) + \cos^2 90^{\circ} \\ & = (\cos^2 1^{\circ} +\sin^2 1^{\circ}) + (\cos^2 2^{\circ} + \sin^2 2^{\circ}) + \cdots + (\cos^2 14^{\circ} + \sin^2 14^{\circ}) + \color{red}{\cos^2 90^{\circ}} \\ & = \underbrace{1+1+\cdots+1}_{\text{ada 14}} + \color{red}{0} \\ & = 14 \end{aligned}$$Jadi, hasil perhitungan ekspresi trigonometri tersebut adalah $\boxed{14}$

Gunakan sifat logaritma berikut.
$\boxed{\log_a b + \log_a c = \log_a bc}$
Gunakan juga hubungan komplemen sudut pada tangen.
$\boxed{\tan x \tan (90^{\circ}-x) = \tan x \cot x = 1}$
Kita peroleh
$$\begin{aligned} & \log_{10} (\tan 1^{\circ}) + \log_{10} (\tan 2^{\circ}) + \log_{10} (\tan 3^{\circ}) + \cdots + \log_{10} (\tan 88^{\circ}) + \log_{10} (\tan 89^{\circ}) \\ & = \log_{10} (\tan 1^{\circ} \cdot \tan 2^{\circ} \cdot \tan 3^{\circ} \cdots \tan 88^{\circ} \cdot \tan 89^{\circ}) \\ & = \log_{10} ((\tan 1^{\circ} \tan 89^{\circ}) \cdot (\tan 2^{\circ} \tan 88^{\circ}) \cdots (\tan 44^{\circ} \tan 46^{\circ}) \cdot \tan 45^{\circ}) \\ & = \log_{10} (1 \cdot 1 \cdots 1 \cdot 1) \\ & = \log_{10} 1 = 0 \end{aligned}$$Jadi, hasil dari perhitungannya adalah $\boxed{0}$

Pada persamaan yang diberikan di atas, kalikan kedua ruas dengan $\sin \left(\dfrac{x}{2^{2010}}\right)$, lalu gunakan identitas sudut ganda $\sin x \cos x = \dfrac12 \sin 2x$ secara berulang-ulang sampai $2010$ kali.
$$\begin{aligned} \dfrac{1}{\sin \left(\dfrac{x}{2^{2010}}\right)} & = 2^{2010} \sqrt2 \cos \left(\dfrac{x}{2}\right) \cos \left(\dfrac{x}{2^2}\right) \cdots \cos \left(\dfrac{x}{2^{2010}}\right) \\ 1 & = 2^{2010} \sqrt2 \cos \left(\dfrac{x}{2}\right) \cos \left(\dfrac{x}{2^2}\right) \cdots \color{red}{\cos \left(\dfrac{x}{2^{2010}}\right) \sin \left(\dfrac{x}{2^{2010}}\right)} \\ 1 & = \dfrac12 \cdot 2^{2010} \sqrt2 \cos \left(\dfrac{x}{2}\right) \cos \left(\dfrac{x}{2^2}\right) \cdots \color{red}{\cos \left(\dfrac{x}{2^{2009}}\right) \sin \left(\dfrac{x}{2^{2009}}\right)} \\ 1 & = \left(\dfrac12\right)^2 \cdot 2^{2010} \sqrt2 \cos \left(\dfrac{x}{2}\right) \cos \left(\dfrac{x}{2^2}\right) \cdots \cos \color{red}{\left(\dfrac{x}{2^{2008}}\right) \sin \left(\dfrac{x}{2^{2008}}\right)} \\ \vdots & ~~~~~~~~~~~\vdots~~~~~~~~~~~~\vdots~~~~~~~~~~~~\vdots \\ 1 & = \left(\dfrac12\right)^{2009} \cdot 2^{2010} \sqrt2 \cos \left(\dfrac{x}{2}\right) \sin \left(\dfrac{x}{2}\right) \\ 1 & = \cancel{\left(\dfrac12\right)^{2010}} \cdot \cancel{2^{2010}} \sqrt2 \sin x \\ 1 & = \sqrt2 \sin x \\ \dfrac{1}{\sqrt2} & = \sin x \end{aligned}$$Karena $0 \leq x \leq \pi$, maka kita peroleh $x = \dfrac{\pi}{4}$ atau $x = \dfrac{3\pi}{4}$.