Jika α dan β adalah sudut-sudut lancip dengan tan α=247 dan cotan α=125, hitunglah: c. cos (21α+β) .

Pertanyaan

Jika $\alpha$ dan $\beta$ adalah sudut-sudut lancip dengan $\tan\;\alpha=\frac7{24}$ dan $\mathrm{cotan}\;\alpha=\frac5{12}$ , hitunglah:

c. $\cos\;\left(\frac12\alpha+\beta\right)$ .

D. Rajib

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Jawaban terverifikasi

Jawaban

nilai dari $\cos\;\left(\frac12\alpha+\beta\right)$ adalah $\begin{array}{rcl}&&\frac{23}{130}\sqrt2\end{array}$

Pembahasan

Ingat kembali rumus:

$sin space A over 2 equals plus-or-minus square root of fraction numerator 1 minus cos space A over denominator 2 end fraction end root cos space A over 2 equals plus-or-minus square root of fraction numerator 1 plus cos space A over denominator 2 end fraction end root$

$\begin{array}{rcl}\sin\;\mathrm A&=&\frac{\mathrm{depan}}{\mathrm{miring}}\\\cos\;\mathrm A&=&\frac{\mathrm{samping}}{\mathrm{miring}}\\\tan\;\mathrm A&=&\frac{\mathrm{depan}}{\mathrm{samping}}\\\mathrm{cotan}\;\mathrm A&=&\frac{\mathrm{samping}}{\mathrm{depan}}\end{array}$

$\cos\;\left(A+B\right)=\cos\;A\;\cos\;B-\sin\;A\;\sin\;B$

Pada soal di atas, terdapat kekeliruan pada penulisan soal, yaitu $co\tan\;\alpha=\frac5{12}$ , kita asumsikan, $co\tan\;\beta=\frac5{12}$ .

Pertama untuk sudut $\alpha$ :

$\begin{array}{rcl}&&\begin{array}{c}\tan\;\mathrm\alpha=\frac7{24}\\\frac{\mathrm{depan}}{\mathrm{samping}}=\frac7{24}\end{array}\\&&\begin{array}{c}\Rightarrow\mathrm{depan}=7\\\Rightarrow\mathrm{samping}=24\end{array}\end{array}$

Sehingga dengan menggunakan teorema Pytahgoras:

$\begin{array}{rcl}\mathrm{miring}&=&\sqrt{\mathrm{depan}^2+\mathrm{samping}^2}\\&=&\sqrt{7^2+24^2}\\&=&\sqrt{625}\\&=&25\end{array}$

Sehingga diperoleh:

$table attributes columnalign right center left columnspacing 0px end attributes row cell sin space straight alpha end cell equals cell fraction numerator depan space over denominator miring end fraction end cell row blank equals cell 7 over 25 end cell row cell cos space straight alpha end cell equals cell samping over miring end cell row blank equals cell 24 over 25 end cell end table$

Untuk sudut $\beta$ :

$\begin{array}{l}\begin{array}{c}\begin{array}{rcl}\mathrm{cotan}\;\mathrm\beta&=&\frac5{12}\\\frac{\mathrm{samping}}{\mathrm{depan}}&=&\frac5{12}\end{array}\end{array}\\\begin{array}{l}\Rightarrow\mathrm{samping}=5\\\Rightarrow\mathrm{depan}=12\end{array}\end{array}$

Sehingga dengan menggunakan teorema Pytahgoras:

$\begin{array}{rcl}\mathrm{miring}&=&\sqrt{\mathrm{samping}^2+\mathrm{depan}^2}\\&=&\sqrt{5^2+12^2}\\&=&\sqrt{169}\\&=&13\end{array}$

Sehingga diperoleh:

$table attributes columnalign right center left columnspacing 0px end attributes row cell sin space beta end cell equals cell fraction numerator depan space over denominator miring end fraction end cell row blank equals cell 12 over 13 end cell row cell cos space beta end cell equals cell samping over miring end cell row blank equals cell 5 over 13 end cell end table$

karena diketahui bahwa sudut $\alpha$ dan $\beta$ adalah sudut-sudut lancip maka sudut $\frac\alpha2$ juga sudut lancip dan berada pada kuadran I, sehingga:

$table attributes columnalign right center left columnspacing 0px end attributes row cell cos space open parentheses 1 half alpha plus beta close parentheses end cell equals cell cos space 1 half alpha space cos space beta space minus sin space 1 half alpha space sin space beta end cell row blank equals cell square root of fraction numerator 1 plus cos space alpha over denominator 2 end fraction end root cos space beta space minus square root of fraction numerator 1 minus cos space alpha over denominator 2 end fraction end root sin space beta end cell row blank equals cell square root of fraction numerator 1 plus begin display style 24 over 25 end style over denominator 2 end fraction end root times 5 over 13 minus square root of fraction numerator 1 minus begin display style 24 over 25 end style over denominator 2 end fraction end root times 12 over 13 end cell row blank equals cell square root of fraction numerator begin display style fraction numerator 25 plus 24 over denominator 25 end fraction end style over denominator 2 end fraction end root times 5 over 13 minus square root of fraction numerator begin display style fraction numerator 25 minus 24 over denominator 25 end fraction end style over denominator 2 end fraction end root times 12 over 13 end cell row blank equals cell square root of fraction numerator begin display style 49 end style over denominator 2 times 25 end fraction end root times 5 over 13 minus square root of fraction numerator begin display style 1 end style over denominator 2 times 25 end fraction end root times 12 over 13 end cell row blank equals cell 7 over 5 square root of 1 half end root times 5 over 13 minus 1 fifth times square root of 1 half end root times 12 over 13 end cell row blank equals cell 35 over 65 square root of 1 half end root minus 12 over 65 square root of 1 half end root end cell row blank equals cell fraction numerator 35 minus 12 over denominator 65 end fraction square root of 1 half end root end cell row blank equals cell 23 over 65 square root of 1 half end root end cell row blank equals cell 23 over 65 open parentheses 1 half square root of 2 close parentheses end cell row blank equals cell 23 over 130 square root of 2 end cell end table$

Jadi, nilai dari $\cos\;\left(\frac12\alpha+\beta\right)$ adalah $\begin{array}{rcl}&&\frac{23}{130}\sqrt2\end{array}$