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Matematika Dasar Trigonometri (👊 Soal Dari Berbagai Sumber 👊)

Desember 08, 2014 Tambah Komentar Edit

eksplorasiDing... Dong... waktunya istirahat... Ding... Dong... it's time to have break... soal latihan uji kompetensi buku matematikaJika $\dfrac{cos\ \theta}{1-sin\ \theta}=a$ maka $tan\ \dfrac{1}{2}\ \theta=...$Jika $\dfrac{cos\ \theta}{1-sin\ \theta}=a$ untuk $\theta\neq \dfrac{\pi}{2}$ maka $tan\ \dfrac{1}{2}\ \theta=...$
$\begin{align}
(A).\ & \dfrac{a}{a+1} \\
(B).\ & \dfrac{1}{a+1} \\
(C).\ & \dfrac{a+1}{a-1} \\
(D).\ & \dfrac{a-1}{a+1} \\
(E).\ & 2a^{2}
\end{align}$1. Soal UM UGM 2006Jika $\dfrac{cos\ \theta}{1-sin\ \theta}=a$ untuk $\theta\neq \frac{\pi}{2}+2k\pi$ maka $tan\ \dfrac{1}{2}\ \theta=...$
$\begin{align}
(A).\ & \dfrac{a}{a+1} \\
(B).\ & \dfrac{1}{a+1} \\
(C).\ & \dfrac{a+1}{a-1} \\
(D).\ & \dfrac{a-1}{a+1} \\
(E).\ & \dfrac{a}{a-1}
\end{align}$Alternatif Pembahasan: show

Untuk mengerjakan soal diatas, ada beberapa data yang kita perlukan, yaitu:

$cos\ 2\theta=cos^{2}\theta-sin^{2}\theta$
$cos\ 4\theta=cos^{2}2\theta-sin^{2}2\theta$
$cos\ \theta=cos^{2}\frac{1}{2}\theta-sin^{2}\frac{1}{2}\theta$
$sin^{2}\theta+cos^{2}\theta=1$
$sin^{2}\frac{1}{2}\theta+cos^{2}\frac{1}{2}\theta=1$
$sin\ 2\theta=2\ sin\ \theta\ cos\ \theta$
$sin\ 4\theta=2\ sin\ 2\theta\ cos\ 2\theta$
$sin\ \theta=2\ sin\ \frac{1}{2}\theta\ cos\ \frac{1}{2}\theta$

Data-data yang kita peroleh diatas kita substituskan ke soal,
$a=\dfrac{cos\ \theta}{1-sin\ \theta}$
$a=\dfrac{cos^{2}\frac{1}{2}\theta-sin^{2}\frac{1}{2}\theta}{\left ( sin^{2}\frac{1}{2}\theta+cos^{2}\frac{1}{2}\theta \right )\left ( 2\ sin\ \frac{1}{2}\theta\ cos\ \frac{1}{2}\theta \right )}$
$a=\dfrac{\left (cos\frac{1}{2}\theta-sin\frac{1}{2}\theta \right )\left ( cos\frac{1}{2}\theta+sin\frac{1}{2}\theta \right )}{\left (cos\frac{1}{2}\theta-sin\frac{1}{2}\theta \right )^{2}}$
$a=\dfrac{\left (cos\frac{1}{2}\theta+sin\frac{1}{2}\theta \right )}{\left (cos\frac{1}{2}\theta-sin\frac{1}{2}\theta \right )}$
$a\left (cos\frac{1}{2}\theta-sin\frac{1}{2}\theta \right )=\left (cos\frac{1}{2}\theta+sin\frac{1}{2}\theta \right )$
$a\ cos\frac{1}{2}\theta-a\ sin\frac{1}{2}\theta=cos\frac{1}{2}\theta+sin\frac{1}{2}\theta$
$a\ cos\frac{1}{2}\theta-cos\frac{1}{2}\theta=a\ sin\frac{1}{2}\theta+sin\frac{1}{2}\theta$
$cos\frac{1}{2}\theta\left (a\ -1 \right )=sin\frac{1}{2}\theta \left (a\ +1 \right )$
$\dfrac{\left (a\ -1 \right )}{\left (a\ +1 \right )}=\frac{sin\frac{1}{2}\theta}{cos\frac{1}{2}\theta}$
$\dfrac{\left (a\ -1 \right )}{\left (a\ +1 \right )}=tan\frac{1}{2}\theta$

2. Soal UM UGM 2009JIka $sin\ A=\sqrt{2pq}$ dan $tan\ A=\dfrac{\sqrt{2pq}}{p-q}$, maka $p^{2}+q^{2}=\cdots$
$\begin{align}
(A).\ & -1 \\
(B).\ & 0 \\
(C).\ & \dfrac{1}{4} \\
(D).\ & \dfrac{1}{2} \\
(E).\ & 1
\end{align}$Alternatif Pembahasan: show

Dari soal ada beberapa data yang bisa kita ambil dan kembangkan yaitu:
$ sin\ A =\sqrt{2pq}$
ruas kiri dan kanan persamaan diatas sama-sama dikuadratkan menjadi $ sin^{2}A=2pq$.

Berikutnya diketahui $ tan\ A=\frac{\sqrt{2pq}}{p-q}$
$ \frac{sin\ A}{cos\ A}=\frac{\sqrt{2pq}}{p-q}$
$ \frac{sin\ A}{cos\ A}=\frac{sin\ A}{p-q}$
diperoleh persamaan $ cos\ A=p-q$

Ruas kiri dan kanan persamaan $ cos\ A=p-q$ sama-sama dikuadratkan, menjadi:
$ \left (p-q \right )^{2}=cos^{2}A$
$ p^{2}+q^{2}-2pq=cos^{2}A$
$ p^{2}+q^{2}=cos^{2}+2pqA$
$ p^{2}+q^{2}=cos^{2}+sin^{2}A$
$ p^{2}+q^{2}=1$

$\therefore$ Pilihan yang sesuai adalah $(E).\ 1$

3. SIMAK UI 2015 Kode 302 (👊 Soal Lengkap 👊)Diketahui $sin(40^{\circ}+ \alpha)=b$, dengan $0 \lt \alpha \lt 45^{\circ}$. Nilai dari $cos(10^{\circ}+\alpha)=\cdots$
$\begin{align}
(A).\ & \dfrac{1}{2} \left(\sqrt{3(1-b^{2})}+b \right) \\
(B).\ & \dfrac{1}{2} \left(\sqrt{(1-b^{2})}-b \right) \\
(C).\ & \dfrac{1}{2} \left(b-\sqrt{(1-b^{2})} \right) \\
(D).\ & \dfrac{1}{2} \left(b-\sqrt{3(1-b^{2})} \right) \\
(E).\ & \dfrac{1}{2} \left(\sqrt{3(1-b^{2})}-b \right)
\end{align}$Alternatif Pembahasan: show

Kita coba kerjakan dengan memisalkan $m=40^{\circ}+ \alpha$ sehingga $m-40^{\circ}=\alpha$.
$sin(40^{\circ}+ \alpha)=b$
$sin\ m=b$
dengan menggunakan identitas dasar trigonometri atau defenisi trigonometri pada segitiga siku-siku, kita peroleh:
$ sin^{2}\ m+cos^{2}\ m=1$
$ b^{2}+cos^{2}\ m=1$
$ cos\ m=\pm \sqrt{1-b^{2}}$
Karena $m$ pada kwadran satu maka $ cos\ m=\sqrt{1-b^{2}}$

$\begin{align}
& cos(10^{\circ}+\alpha) \\
& = cos(10^{\circ}+m-40^{\circ}) \\
& = cos(m-30^{\circ}) \\
& = cos\ m\ cos\ 30^{\circ} + sin\ m\ sin\ 30^{\circ} \\
& = \sqrt{1-b^{2}} \left(\dfrac{1}{2}\sqrt{3} \right) + b\ \left(\dfrac{1}{2} \right) \\
& = \dfrac{1}{2} \left( \sqrt{1-b^{2}} \left(\sqrt{3} \right) + b \right) \\
& = \dfrac{1}{2} \left( \sqrt{3 \left(1-b^{2} \right)} + b \right)
\end{align}$

$ \therefore $ Pilihan yang sesuai adalah $(E)\ \dfrac{1}{2} \left(\sqrt{3(1-b^{2})}-b \right)$

4. SIMAK UI 2015 Kode 354 (👊 Soal Lengkap 👊)Jika $cos(2A)=-\dfrac{\sqrt{5}}{3}$, maka $9\left(sin^{6}A+cos^{6}A \right)=\cdots$
$\begin{align}
(A).\ & 2 \\
(B).\ & 3 \\
(C).\ & 4 \\
(D).\ & 5 \\
(E).\ & 6
\end{align}$Alternatif Pembahasan: show

Kita ketahui bahwa identitas aljabar atau identitas dasar trigonometri sebagai berikut;

$ sin^{2}(2A)+cos^{2}(2A)=1$
$ sin(2A)=2 sin\ A\ cos\ A$
$(a+b)^{2}=a^{2}+b^{2}+2ab$
$(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)$
$a^{3}+b^{3}=(a+b)^{3}-3ab(a+b)$
$a^{6}+b^{6}=(a^{2}+b^{2})^{3}-3a^{2}b^{2} \left( a^{2}+b^{2} \right)$

$\begin{align}
& \left(sin^{6}A+cos^{6}A \right) \\
& = (sin^{2}A+cos^{2}A)^{3}-3sin^{2}A\ cos^{2}A \left( sin^{2}A+cos^{2}A \right) \\
& = (1)^{3}-3sin^{2}A\ cos^{2}A \left( 1 \right) \\
& = 1-3sin^{2}A\ cos^{2}A \\
& = 1-3(sin\ A\ cos\ A)^{2}
\end{align}$
Dari data yang kita punya yaitu $cos(2A)=-\dfrac{\sqrt{5}}{3}$ kita peroleh
$sin(2A)=\pm \sqrt{1-cos^{2}(2A)}$
$sin(2A)=\pm \sqrt{1-\left(-\dfrac{\sqrt{5}}{3} \right)^{2}}$
$sin(2A)=\pm \sqrt{1-\left(\dfrac{5}{9} \right) }$
$sin(2A)=\pm \sqrt{\dfrac{4}{9}}$
$2 sin\ A\ cos\ A=\pm \dfrac{2}{3}$
$sin\ A\ cos\ A=\pm \dfrac{1}{3}$
$\begin{align}
& 9 \left(sin^{6} A+cos^{6} A \right) \\
& = 9 \left(1 -3 \left( sin\ A\ cos\ A\ \right)^{2} \right) \\
& = 9 \left(1 -3 \left( \pm \dfrac{1}{3} \right)^{2} \right) \\
& = 9 \left(1 -3 \left( \dfrac{1}{9} \right) \right) \\
& = 9 \left(1 - \dfrac{1}{3} \right) \\
& = 9 \left( \dfrac{2}{3} \right) \\
& = 6
\end{align}$

$ \therefore $ Pilihan yang sesuai adalah $(E)\ 6$

5. SIMAK UI 2015 Kode 354 (👊 Soal Lengkap 👊)Bentuk identitas trigonometri berikut yang benar adalah...
$\begin{align}
(1).\ & \dfrac{cos\ x+sin\ x}{cos\ x-sin\ x}=tan\ 2x-sec\ 2x \\
(2).\ & tan\ \left( \frac{\pi}{4} + x \right) = \dfrac{1-sin(-2x)}{cos(2x)} \\
(3).\ & \dfrac{1+2sin^{2}x}{2cot(\frac{\pi}{4}+x)cos^{2}(\frac{\pi}{4}-x)}=-1 \\
(4).\ & \dfrac{cot^{2}2x-1}{2cot\ 2x}-cos\ 8x\ cot\ 4x=sin\ 8x
\end{align}$Alternatif Pembahasan: show

Dengan menggunakan beberapa identitas dasar trigonometri dan manipulais alajabar, maka akan kita peroleh;
$\dfrac{cos\ x+sin\ x}{cos\ x-sin\ x}$
$=\dfrac{cos\ x+sin\ x}{cos\ x-sin\ x} \times \dfrac{cos\ x+sin\ x}{cos\ x+sin\ x}$
$=\dfrac{cos^{2}x+sin^{2}x+2sin\ x\ cos\ x}{cos^{2}x-sin^{2}x}$
$=\dfrac{1+2sin\ x\ cos\ x}{cos(2x)}$
$=\dfrac{1}{cos(2x)}+\dfrac{2sin\ x\ cos\ x}{cos(2x)}$
$=sec(2x)+\dfrac{sin(2x)}{cos(2x)}$
$=sec(2x)+tan(2x) \neq tan\ 2x-sec\ 2x $
Pernyataan $(1)$ Salah.

$\dfrac{1-sin(-2x)}{cos(2x)}$
$=\dfrac{1+sin(2x)}{cos(2x)}$
$=\dfrac{1}{cos(2x)}+\dfrac{sin(2x)}{cos(2x)}$
$=sec(2x)+tan(2x)$
$=\dfrac{cos\ x+sin\ x}{cos\ x-sin\ x}$
$=\dfrac{1+\dfrac{sin\ x}{cos\ x}}{1-\dfrac{sin\ x}{cos\ x}}$
$=\dfrac{1+tan\ x}{1-tan\ x}$
$=\dfrac{tan\ \frac{\pi}{4}+tan\ x}{tan\ \frac{\pi}{4}-tan\ x}$
$=tan\ \left( \frac{\pi}{4} + x \right)$
Pernyataan $(2)$ Benar.

$\dfrac{1+2sin^{2}x}{2cot(\frac{\pi}{4}+x)cos^{2}(\frac{\pi}{4}-x)}$
$=\dfrac{1+2sin^{2}x}{2 \cdot \dfrac{cos\ x -sin\ x}{cos\ x +sin\ x} \left(\frac{1}{2}\sqrt{2} \right)^{2} (cos\ x +sin\ x)}$
$=\dfrac{1+2sin^{2}x}{(cos\ x -sin\ x)(cos\ x +sin\ x)}$
$=\dfrac{1+2sin^{2}x}{cos^{2} x -sin^{2} x}$
$=\dfrac{1+2sin^{2}x}{1-2sin^{2} x} \neq -1$
Pernyataan $(3)$ Salah.

$\dfrac{cot^{2}2x-1}{2cot\ 2x}-cos\ 8x\ cot\ 4x $
$=\dfrac{cot^{2}2x-1}{2cot\ 2x}-cos\ 8x\ cot\ 4x $
$=\dfrac{cot\ 2x}{2} - \dfrac{1}{2cot\ 2x}-cos\ 8x\ cot\ 4x $
$=\dfrac{cos\ 2x}{2sin\ 2x} - \dfrac{sin\ 2x}{2cos\ 2x}-cos\ 8x\ \dfrac{cos\ 4x}{sin\ 4x} $
$=\dfrac{2cos^{2} 2x-2sin^{2} 2x}{4sin\ 2x\ cos\ 2x} -\dfrac{cos\ 8x\ cos\ 4x}{sin\ 4x} $
$=\dfrac{2cos\ 4x}{2 sin\ 4x} -\dfrac{cos\ 8x\ cos\ 4x}{sin\ 4x} $
$=\dfrac{cos\ 4x-cos\ 8x\ cos\ 4x}{sin\ 4x} $
$=\dfrac{cos\ 4x\left (1-cos\ 8x\ \right )}{sin\ 4x} $
$=\dfrac{cos\ 4x\left (sin^{2}4x+cos^{2}4x-cos^{2}4x+sin^{2}4x \right )}{sin\ 4x} $
$=\dfrac{cos\ 4x\ 2sin^{2}4x}{sin\ 4x} $
$=2sin 4x\ cos\ 4x\ $
$=sin\ 8x$
Pernyataan $(4)$ Benar.

$ \therefore $ Pilihan yang sesuai adalah $(C)\ (2)\ \text{dan}\ (4)$

Jika engkau tidak sanggup menahan lelahnya belajar, Maka engkau harus menanggung pahitnya kebodohan" ___pythagoras

lembar jawaban penilaian harian matematika,

lembar jawaban penilaian akhir semester matematika,

presentasi hasil diskusi matematika atau

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