Levi Rizki Saputra Notes

Sudut 15 Derajat

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# Menggunakan Pengurangan Sudut

sin⁡(15°)=sin⁡(45°−30°)=sin⁡45°cos⁡30°−cos⁡45°sin⁡30°=22⋅32−22⋅12=24(3−1)cos⁡(15°)=cos⁡(45°−30°)=cos⁡(45°)cos⁡(30°)+sin⁡(45°)sin⁡(30°)=22⋅32+22⋅12=24(3+1)tan⁡(15°)=tan⁡(45°−30°)=tan⁡(45°)−tan⁡(30°)1+tan⁡(45°)⋅tan⁡(30°)=1−331+1⋅33=(3−33)(3+33)=3−33⋅33+3=3−33+3×3−33−3=32−2⋅3⋅3+(3)232−(3)2=9−63+39−3=12−636=6(2−3)6=2−3\begin{aligned} \sin(15\degree) &= \sin(45\degree - 30\degree) \\ &= \sin 45\degree \cos 30\degree - \cos 45\degree \sin 30\degree \\ &= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} \\ &= \frac{\sqrt{2}}{4}(\sqrt{3} - 1) \\ \cos( 15\degree ) & =\cos( 45\degree -30\degree )\\ & =\cos( 45\degree )\cos( 30\degree ) +\sin( 45\degree )\sin( 30\degree )\\ & =\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} +\frac{\sqrt{2}}{2} \cdot \frac{1}{2}\\ & =\frac{\sqrt{2}}{4}\left(\sqrt{3} +1\right) \\ \tan( 15\degree ) & =\tan( 45\degree -30\degree )\\ & =\frac{\tan( 45\degree ) -\tan( 30\degree )}{1+\tan( 45\degree ) \cdot \tan( 30\degree )}\\ & =\frac{1-\dfrac{\sqrt{3}}{3}}{1+1\cdot \dfrac{\sqrt{3}}{3}}\\ & =\frac{\left(\dfrac{3-\sqrt{3}}{3}\right)}{\left(\dfrac{3+\sqrt{3}}{3}\right)}\\ & =\frac{3-\sqrt{3}}{3} \cdot \frac{3}{3+\sqrt{3}}\\ & =\frac{3-\sqrt{3}}{3+\sqrt{3}} \times \frac{3-\sqrt{3}}{3-\sqrt{3}}\\ & =\frac{3^{2} -2\cdot 3\cdot \sqrt{3} +\left(\sqrt{3}\right)^{2}}{3^{2} -\left(\sqrt{3}\right)^{2}}\\ & =\frac{9-6\sqrt{3} +3}{9-3}\\ & =\frac{12-6\sqrt{3}}{6}\\ & =\frac{6\left( 2-\sqrt{3}\right)}{6}\\ & =2-\sqrt{3} \end{aligned}

# Menggunakan Perkalian Sudut

sin⁡(15°)=sin⁡(30°×12)=1−cos⁡(30°)2=1−322=(2−32)2=2−34\begin{aligned} \sin (15\degree ) & =\sin (30\degree \times \frac{1}{2} )\\ & =\sqrt{\frac{1-\cos( 30\degree )}{2}}\\ & =\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}\\ & =\sqrt{\frac{\left(\frac{2-\sqrt{3}}{2}\right)}{2}}\\ & =\sqrt{\frac{2-\sqrt{3}}{4}} \end{aligned}

Menyempurnakan kuadrat dari 2−34\frac{2-\sqrt{3}}{4} menjadi (a−b)2=a2−2ab+b2(a-b)^2=a^2 - 2ab + b^2

Mencoba 2−32-\sqrt{3}. Misal −2ab=3-2ab = \sqrt{3}. Tidak bisa

Mengalikan percahan

2−34×22=4−228\frac{2-\sqrt{3}}{4} \times \frac{2}{2} =\frac{4-2\sqrt{2}}{8}

Mencoba 4−224-2\sqrt{2}.

Variabel Nilai
−2ab-2ab 232\sqrt{3}
aa 11
bb 3\sqrt{3}
a2−2ab+b2a^2 - 2ab + b^2 4−234-2\sqrt{3}

Bisa.

2−34=4−238=a2−2ab+b28=(a−b)28=(1−3)28=±(1−3)18=±(1−3)4⋅2=±(1−3)22=±(1−3)22×22=±(1−3)224=±(1−3)24 \begin{align*} \sqrt{\frac{2-\sqrt{3}}{4}}&=\sqrt{\frac{4-2\sqrt{3}}{8}}\\ &=\sqrt{\frac{a^{2} -2ab+b^{2}}{8}}\\ &=\sqrt{\frac{( a-b)^{2}}{8}}\\ &=\sqrt{\frac{\left( 1-\sqrt{3}\right)^{2}}{8}}\\ &=\pm \left( 1-\sqrt{3}\right)\sqrt{\frac{1}{8}}\\ &=\pm \frac{\left( 1-\sqrt{3}\right)}{\sqrt{4\cdot 2}}\\ &=\pm \frac{\left( 1-\sqrt{3}\right)}{2\sqrt{2}}\\ &=\pm \frac{\left( 1-\sqrt{3}\right)}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\\ &=\pm \frac{\left( 1-\sqrt{3}\right)\sqrt{2}}{2\sqrt{4}}\\ &=\pm \frac{\left( 1-\sqrt{3}\right)\sqrt{2}}{4} \end{align*}

Ada dua kasus.

2−34={(1−3)24=−×++=−−(1−3)24=(3−1)24=+×++=+\sqrt{\frac{2-\sqrt{3}}{4}} =\begin{cases} \dfrac{\left( 1-\sqrt{3}\right)\sqrt{2}}{4} =\dfrac{-\times +}{+} & =-\\ \dfrac{-\left( 1-\sqrt{3}\right)\sqrt{2}}{4} =\dfrac{\left(\sqrt{3} -1\right)\sqrt{2}}{4} =\dfrac{+\times +}{+} & =+ \end{cases}

Karena 15°15\degree berada di Kuadran 1, maka sin⁡(15°)\sin(15\degree) harus positif, maka.

sin⁡(15°)=(3−1)24\sin(15\degree) = \frac{\left(\sqrt{3} -1\right)\sqrt{2}}{4}

cos⁡(15°)=cos⁡(30°×12)=1+cos⁡(30°)2=1+322=(2+32)2=2+34×22=4+238=3+1+238=(3)2+12+2⋅1⋅34.2=(3+1)24.2=±(3+1)22×22=±(3+1)224=±24(3+1)={24(3+1)=+×++=+−24(3+1)=−×+×++=−\begin{aligned} \cos( 15\degree ) & =\cos\left( 30\degree \times \frac{1}{2}\right)\\ & =\sqrt{\frac{1+\cos( 30\degree )}{2}}\\ & =\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}\\ & =\sqrt{\frac{\left(\frac{2+\sqrt{3}}{2}\right)}{2}}\\ & =\sqrt{\frac{2+\sqrt{3}}{4}} \times \frac{\sqrt{2}}{\sqrt{2}}\\ & =\sqrt{\frac{4+2\sqrt{3}}{8}}\\ & =\sqrt{\frac{3+1+2\sqrt{3}}{8}}\\ & =\sqrt{\frac{\left(\sqrt{3}\right)^{2} +1^{2} +2\cdot 1\cdot \sqrt{3}}{4.2}}\\ & =\sqrt{\frac{\left(\sqrt{3} +1\right)^{2}}{4.2}}\\ & =\pm \frac{\left(\sqrt{3} +1\right)}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\\ & =\pm \frac{\left(\sqrt{3} +1\right)\sqrt{2}}{2\sqrt{4}}\\ & =\pm \frac{\sqrt{2}}{4}\left(\sqrt{3} +1\right)\\ & =\begin{cases} \dfrac{\sqrt{2}}{4}\left(\sqrt{3} +1\right) =\dfrac{+\times +}{+} & =+\\ \dfrac{-\sqrt{2}}{4}\left(\sqrt{3} +1\right) =\dfrac{-\times +\times +}{+} & =- \end{cases} \end{aligned}

Nilai cos⁡(15°)\cos(15\degree) harus bernilai positif jadi:

cos⁡(15°)=24(3+1)\cos( 15\degree ) =\frac{\sqrt{2}}{4}\left(\sqrt{3} +1\right)

tan⁡(15°)=tan⁡(30°×12)=1−cos⁡(30°)1+cos⁡(30°)=1−321+32=(2−32)(2+32)=(2−32)(22+3)=2−32+3×2−32−3=(2−3)222−(3)2=(2−3)24−3=(2−3)21=±(2−3)={2−3=+−(2−3)=−\begin{aligned} \tan( 15\degree ) & =\tan\left( 30\degree \times \frac{1}{2}\right)\\ & =\sqrt{\frac{1-\cos( 30\degree )}{1+\cos( 30\degree )}}\\ & =\sqrt{\frac{1-\dfrac{\sqrt{3}}{2}}{1+\dfrac{\sqrt{3}}{2}}}\\ & =\sqrt{\frac{\left(\dfrac{2-\sqrt{3}}{2}\right)}{\left(\dfrac{2+\sqrt{3}}{2}\right)}}\\ & =\sqrt{\left(\frac{2-\sqrt{3}}{2}\right)\left(\frac{2}{2+\sqrt{3}}\right)}\\ & =\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}} \times \frac{\sqrt{2-\sqrt{3}}}{\sqrt{2-\sqrt{3}}}\\ & =\sqrt{\frac{\left( 2-\sqrt{3}\right)^{2}}{2^{2} -\left(\sqrt{3}\right)^{2}}}\\ & =\sqrt{\frac{\left( 2-\sqrt{3}\right)^{2}}{4-3}}\\ & =\sqrt{\frac{\left( 2-\sqrt{3}\right)^{2}}{1}}\\ & =\pm \left( 2-\sqrt{3}\right)\\ & =\begin{cases} 2-\sqrt{3} & =+\\ -\left( 2-\sqrt{3}\right) & =- \end{cases} \end{aligned}

Nilai tan⁡(15°)\tan(15\degree) harus bernilai positif jadi:

tan⁡(15°)=2−3\tan(15\degree) = 2-\sqrt{3}

# Nilai tan Menurut sin dan cos

tan⁡(15°)=sin⁡(15°)cos⁡(15°)=24(3−1)24(3+1)=3−13+1×3−13−1=(3)2−2⋅1⋅3+12(3)2−12=3+1−233−1=4−232=2(2−3)2=2−3\begin{aligned} \tan( 15\degree ) & =\frac{\sin( 15\degree )}{\cos( 15\degree )}\\ & =\frac{\frac{\sqrt{2}}{4} (\sqrt{3} -1)}{\frac{\sqrt{2}}{4}\left(\sqrt{3} +1\right)}\\ & =\frac{\sqrt{3} -1}{\sqrt{3} +1} \times \frac{\sqrt{3} -1}{\sqrt{3} -1}\\ & =\frac{\left(\sqrt{3}\right)^{2} -2\cdot 1\cdot \sqrt{3} +1^{2}}{\left(\sqrt{3}\right)^{2} -1^{2}}\\ & =\frac{3+1-2\sqrt{3}}{3-1}\\ & =\frac{4-2\sqrt{3}}{2}\\ & =\frac{2\left( 2-\sqrt{3}\right)}{2}\\ & =2-\sqrt{3} \end{aligned}

# Kesimpulan

Nilai fungsi trigonometri sudut 15°15\degree

sin⁡(15°)=24(3−1)cos⁡(15°)=24(3+1)tan⁡(15°)=2−3\begin{aligned} \sin(15\degree) &= \frac{\sqrt{2}}{4}(\sqrt{3} - 1) \\ \cos( 15\degree ) &=\frac{\sqrt{2}}{4}\left(\sqrt{3} +1\right) \\ \tan( 15\degree ) &=2-\sqrt{3} \end{aligned}