RUKO 2 Trigonometri Analitika no 1-30

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Sobat hudamath, di postingan kali ini akan kita bahas RUKO 2 Buku Sukino Matematika Peminatan Kelas XI Kurikulum 2013 di soal-soal no tertentu saja. Untuk pembahasan no yang lainnya bisa sobat download di bagian bawah postingan ini. 

Langsung saja ya

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16.  \({\cos ^4}3\theta  - {\sin ^4}3\theta  =  \ldots \)
        A.  \(\cos 2\theta \)
        B.  \(\cos 3\theta \)
        C.  \(\sin 3\theta \)
        D.  \(\sin 6\theta \)
        E.  \(\cos 6\theta \)

        Jawab            : E
        Pembahasan  :
        \({\cos ^4}3\theta  - {\sin ^4}3\theta \)
        \( = {\left( {{{\cos }^2}3\theta } \right)^2} - {\left( {{{\sin }^2}3\theta } \right)^2}\)
       \( = \left( {{{\cos }^2}3\theta  + {{\sin }^2}3\theta } \right)\left( {{{\cos }^2}3\theta  - {{\sin }^2}3\theta } \right)\)
        \( = 1.\left( {{{\cos }^2}3\theta  - {{\sin }^2}3\theta } \right)\)
        \( = \cos 6\theta \)

27. Nilai dari \(\left( {\cos \frac{\pi }{7}\cos \frac{{2\pi }}{7}\cos \frac{{4\pi }}{7}} \right)\) adalah...
         A. \( - \frac{1}{8}\)
         B.  \( - \frac{1}{4}\)
         C.  0
         D.  \(\frac{1}{2}\)
         E.  \(\frac{1}{3}\)

        Jawab            : A
        Pembahasan  :
        Ingat, \(\sin 2A = 2\sin A\cos A \to \cos A = \frac{{\sin 2A}}{{2\sin A}}\)
        \(\cos \frac{\pi }{7}\cos \frac{{2\pi }}{7}\cos \frac{{4\pi }}{7} = \frac{{\sin 2\left( {\frac{\pi }{7}} \right)}}{{2\sin \left( {\frac{\pi }{7}} \right)}}.\frac{{\sin 2\left( {\frac{{2\pi }}{7}} \right)}}{{2\sin \left( {\frac{{2\pi }}{7}} \right)}}.\frac{{\sin 2\left( {\frac{{4\pi }}{7}} \right)}}{{2\sin \left( {\frac{{4\pi }}{7}} \right)}}\)
         \( = \frac{{\sin \left( {\frac{{2\pi }}{7}} \right)}}{{8\sin \left( {\frac{\pi }{7}} \right)}}.\frac{{\sin \left( {\frac{{4\pi }}{7}} \right)}}{{\sin \left( {\frac{{2\pi }}{7}} \right)}}.\frac{{\sin \left( {\frac{{8\pi }}{7}} \right)}}{{\sin \left( {\frac{{4\pi }}{7}} \right)}}\)
         \( = \frac{{\sin \left( {\pi  + \frac{\pi }{7}} \right)}}{{8\sin \left( {\frac{\pi }{7}} \right)}}\)
         \( = \frac{{\sin \pi \cos \frac{\pi }{7} + \cos \pi \sin \frac{\pi }{7}}}{{8\sin \left( {\frac{\pi }{7}} \right)}}\)
         \( = \frac{{0.\cos \frac{\pi }{7} + \left( { - 1} \right)\sin \frac{\pi }{7}}}{{8\sin \left( {\frac{\pi }{7}} \right)}}\)
         \( = \frac{{ - \sin \frac{\pi }{7}}}{{8\sin \left( {\frac{\pi }{7}} \right)}}\)
         \( =  - \frac{1}{8}\)

28.   Nilai dari \(\left( {\sin \frac{\pi }{{14}}\sin \frac{{3\pi }}{{14}}\sin \frac{{9\pi }}{{14}}} \right)\) adalah...
         A.  \(\frac{1}{{16}}\)
         B.  \(\frac{1}{{8}}\)
         C.  \(\frac{1}{{4}}\)
         D.  \(\frac{1}{{2}}\)
         E.  1

        Jawab            : B
        Pembahasan  :
         Ingat, \(\sin x = \cos \left( {\frac{\pi }{2} - x} \right)\) sehingga:
         \(\sin \frac{{3\pi }}{{14}} = \cos \left( {\frac{\pi }{2} - \frac{{3\pi }}{{14}}} \right) = \cos \left( {\frac{{4\pi }}{{14}}} \right)\)
         \(\sin \frac{{9\pi }}{{14}} = \cos \left( {\frac{\pi }{2} - \frac{{9\pi }}{{14}}} \right) = \cos \left( { - \frac{{2\pi }}{{14}}} \right) = \cos \left( {\frac{{2\pi }}{{14}}} \right)\)
         Ingat pula bahwa \(2\sin A\cos A = \sin 2A\)

         Jadi:
         \(\sin \frac{\pi }{{14}}\sin \frac{{3\pi }}{{14}}\sin \frac{{9\pi }}{{14}} = \sin \frac{\pi }{{14}}\cos \frac{{4\pi }}{{14}}\cos \frac{{2\pi }}{{14}}\)
         \( = \sin \frac{\pi }{{14}}\cos \frac{{4\pi }}{{14}}\cos \frac{{2\pi }}{{14}} \times \frac{{2\cos \frac{\pi }{{14}}}}{{2\cos \frac{\pi }{{14}}}}\)
          \( = \frac{{\left( {2\sin \frac{\pi }{{14}}\cos \frac{\pi }{{14}}} \right)\cos \frac{{4\pi }}{{14}}\cos \frac{{2\pi }}{{14}}}}{{2\cos \frac{\pi }{{14}}}}\)
          \( = \frac{{\left( {\sin \frac{{2\pi }}{{14}}} \right)\cos \frac{{4\pi }}{{14}}\cos \frac{{2\pi }}{{14}}}}{{2\cos \frac{\pi }{{14}}}} \times \frac{2}{2}\)
          \( = \frac{{\left( {2\sin \frac{{2\pi }}{{14}}\cos \frac{{2\pi }}{{14}}} \right)\cos \frac{{4\pi }}{{14}}}}{{4\cos \frac{\pi }{{14}}}}\)
          \( = \frac{{\left( {\sin \frac{{4\pi }}{{14}}} \right)\cos \frac{{4\pi }}{{14}}}}{{4\cos \frac{\pi }{{14}}}} \times \frac{2}{2}\)
         \( = \frac{{\left( {2\sin \frac{{4\pi }}{{14}}\cos \frac{{4\pi }}{{14}}} \right)}}{{8\cos \frac{\pi }{{14}}}}\)
           \( = \frac{{\sin \frac{{8\pi }}{{14}}}}{{8\cos \frac{\pi }{{14}}}}\)
           \( = \frac{{\cos \left( {\frac{\pi }{2} - \frac{{8\pi }}{{14}}} \right)}}{{8\cos \frac{\pi }{{14}}}}\)
           \( = \frac{{\cos \left( { - \frac{\pi }{{14}}} \right)}}{{8\cos \frac{\pi }{{14}}}}\)
           \( = \frac{{\cos \frac{\pi }{{14}}}}{{8\cos \frac{\pi }{{14}}}}\)
           \( = \frac{1}{8}\)



29.   Nilai \(\left( {\tan 9^\circ  - \tan 27^\circ  - \tan 63^\circ  + \tan 81^\circ } \right)\) sama dengan...
         A.  1
         B.  2
         C.  3
         D.  4
         E.  5

        Jawab            : D
        Pembahasan  :
         Ingat, \(\cot x = \tan \left( {90^\circ  - x} \right)\) sehingga:
         \(\tan 9^\circ  - \tan 27^\circ  - \tan 63^\circ  + \tan 81^\circ \)
         \( = \tan 9^\circ  - \tan 27^\circ  - \tan \left( {90^\circ  - 27^\circ } \right) + \tan \left( {90^\circ  - 9^\circ } \right)\)
         \( = \tan 9^\circ  - \tan 27^\circ  - \cot 27^\circ  + \cot 9^\circ \)
         \( = \tan 9^\circ  + \cot 9^\circ  - \left( {\tan 27^\circ  + \cot 27^\circ } \right)\)
         \( = \frac{{\sin 9^\circ }}{{\cos 9^\circ }} + \frac{{\cos 9^\circ }}{{\sin 9^\circ }} - \left( {\frac{{\sin 27^\circ }}{{\cos 27^\circ }} + \frac{{\cos 27^\circ }}{{\sin 27^\circ }}} \right)\)
         \( = \frac{{{{\sin }^2}9^\circ  + {{\cos }^2}9^\circ }}{{\sin 9^\circ \cos 9^\circ }} - \left( {\frac{{{{\sin }^2}27^\circ  + {{\cos }^2}27^\circ }}{{\sin 27^\circ \cos 27^\circ }}} \right)\)
         \( = \frac{1}{{\sin 9^\circ \cos 9^\circ }} - \frac{1}{{\sin 27^\circ \cos 27^\circ }}\)
         \( = \frac{2}{{2\sin 9^\circ \cos 9^\circ }} - \frac{2}{{2\sin 27^\circ \cos 27^\circ }}\)
         \( = \frac{2}{{\sin 18^\circ }} - \frac{2}{{\sin 54^\circ }}\)
         \( = \frac{{2\sin 54^\circ  - 2\sin 18^\circ }}{{\sin 18^\circ \sin 54^\circ }}\)
         \( = \frac{{2\left( {\sin 54^\circ  - \sin 18^\circ } \right)}}{{\sin 18^\circ \sin 54^\circ }}\)
             ingat, \(\sin A - \sin B = 2\cos \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right)\) sehingga:
         \( = \frac{{2\left( {2\cos \left( {36^\circ } \right)\sin \left( {18^\circ } \right)} \right)}}{{\sin 18^\circ \sin 54^\circ }}\)
         \( = \frac{{4\cos 36^\circ }}{{\sin 54^\circ }}\)
         = 4

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