Pembahasan Matematika IPA SIMAK UI 2015 #9

Soal

Jika diketahui \(\sin{\theta} – \cos{\theta}= \frac{\sqrt{5} – \sqrt{3}}{2}\)
dan \(\cos^3{\theta} – \sin^3{\theta}= \frac{1}{a}(b\sqrt{5}-c\sqrt{3})\), dengan \(a,b,c\) bilangan asli, maka \(\ldots\)

(1) \( b – c > 0\)

(2) \( a – b = 7\)

(3) \( a – 2b + c = 0\)

(4) \( a + b + c = 12\)

Jawab

Karena \(\sin{\theta} – \cos{\theta}= \frac{\sqrt{5} – \sqrt{3}}{2}\), maka \(\cos{\theta} – \sin{\theta}= \frac{\sqrt{3} – \sqrt{5}}{2}\),
sehingga berlaku $$ \begin{align} \left( \cos{\theta} – \sin{\theta} \right)^2 &= \left( \dfrac{\sqrt{3} – \sqrt{5}}{2} \right)^2\\ \cos^2{\theta} + \sin^2{\theta} – 2 \sin{\theta}\cos{\theta} &= \dfrac{ 3 + 5 – 2\sqrt{15} }{4}\\ 1 – 2 \sin{\theta}\cos{\theta} &= \dfrac{8 – 2\sqrt{15}}{4}\\ 1 – \dfrac{8 – 2\sqrt{15}}{4} &= 2 \sin{\theta}\cos{\theta}\\ \dfrac{4 – (8 – 2\sqrt{15})}{4} &= 2 \sin{\theta}\cos{\theta}\\ \dfrac{2\sqrt{15} – 4}{4} &= 2 \sin{\theta}\cos{\theta}\\ \dfrac{\sqrt{15} – 2}{4} &= \sin{\theta}\cos{\theta}\\ \end{align} $$

Kemudian ingat bahwa \(a^3 – b^3 = (a – b)(a^2 + b^2 + ab)\), sehingga kita peroleh, $$ \begin{align} \cos^3{\theta} – \sin^3{\theta} &= \left(\cos{\theta} – \sin{\theta} \right) \left( \cos^2{\theta} + \sin^2{\theta} + \sin{\theta}\cos{\theta} \right)\\ &= -\left( \dfrac{\sqrt{5} – \sqrt{3}}{2} \right) \left( 1 + \dfrac{\sqrt{15} – 2}{4} \right)\\ &= \left( \dfrac{\sqrt{3} – \sqrt{5}}{2} \right)\left( \dfrac{4 + \sqrt{15} – 2}{4} \right)\\ &= \left( \dfrac{\sqrt{3} – \sqrt{5}}{2} \right)\left( \dfrac{2 + \sqrt{15}}{4} \right)\\ &= \dfrac{ 2\sqrt{3} – 2\sqrt{5} + 3\sqrt{5} – 5\sqrt{3}}{8}\\ \dfrac{1}{a}(b\sqrt{5}-c\sqrt{3}) &= \dfrac{\sqrt{5} – 3\sqrt{3}}{8}\\ \end{align} $$

Dengan demikian \(a = 8, b = 1, c = 3\).
Oleh karena itu \(b – c = 1 – 3 =-2\), nomor 1 salah. Jadi nomor 3 juga salah. Di samping itu \(a – b = 8 – 1 = 7\), nomor 2 benar dan \(a+b+c = 8 + 1 + 3 = 12\), nomor 4 benar. Jadi jawabannya adalah C

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