RUKO 2 Trigonometri Analitika No 31-60
Assalamu'alaikum warahmatullahi wabarakaatuh
Mohon maaf sobat hudamath,pembahasan RUKO 2 Buku Sukino Matematika Peminatan Kelas XI Kurikulum 2013 no 31-60 nya agak telat. Mudah-mudahan belum telat-telat banget ya... :D
Di postingan ini hanya dibahas beberapa soal saja, selebihnya untuk pembahasan no yang lainnya bisa sobat download di bagian bawah postingan ini.
Di postingan ini hanya dibahas beberapa soal saja, selebihnya untuk pembahasan no yang lainnya bisa sobat download di bagian bawah postingan ini.
Langsung saja ya
.
.
.
31. Nilai dari \(\left( {\cos \left( {\frac{\pi }{5}} \right)\cos \left( {\frac{{2\pi }}{5}} \right)\cos \left( {\frac{{4\pi }}{5}} \right)\cos \left( {\frac{{8\pi }}{5}} \right)} \right)\) adalah...
A. \( - \frac{1}{{16}}\)
B. \( - \frac{1}{8}\)
C. 0
D. \(\frac{1}{{16}}\)
E. \(\frac{1}{8}\)
Jawab : A
Pembahasan :
Ingat, \(\sin 2A = 2\sin A\cos A \to \cos A = \frac{{\sin 2A}}{{2\sin A}}\)
\(\cos \left( {\frac{\pi }{5}} \right)\cos \left( {\frac{{2\pi }}{5}} \right)\cos \left( {\frac{{4\pi }}{5}} \right)\cos \left( {\frac{{8\pi }}{5}} \right)\)
\( = \frac{{\sin 2\left( {\frac{\pi }{5}} \right)}}{{2\sin \left( {\frac{\pi }{5}} \right)}}.\frac{{\sin 2\left( {\frac{{2\pi }}{5}} \right)}}{{2\sin \left( {\frac{{2\pi }}{5}} \right)}}.\frac{{\sin 2\left( {\frac{{4\pi }}{5}} \right)}}{{2\sin \left( {\frac{{4\pi }}{5}} \right)}}.\frac{{\sin 2\left( {\frac{{8\pi }}{5}} \right)}}{{2\sin \left( {\frac{{8\pi }}{5}} \right)}}\)
\( = \frac{{\sin \left( {\frac{{2\pi }}{5}} \right)}}{{16\sin \left( {\frac{\pi }{5}} \right)}}.\frac{{\sin \left( {\frac{{4\pi }}{5}} \right)}}{{\sin \left( {\frac{{2\pi }}{5}} \right)}}.\frac{{\sin \left( {\frac{{8\pi }}{5}} \right)}}{{\sin \left( {\frac{{4\pi }}{5}} \right)}}.\frac{{\sin \left( {\frac{{16\pi }}{5}} \right)}}{{\sin \left( {\frac{{8\pi }}{5}} \right)}}\)
\( = \frac{{\sin \left( {\frac{{16\pi }}{5}} \right)}}{{16\sin \left( {\frac{\pi }{5}} \right)}}\)
\( = \frac{{\sin \left( {3\pi + \frac{\pi }{5}} \right)}}{{16\sin \left( {\frac{\pi }{5}} \right)}}\)
\( = \frac{{\sin \left( {\pi + \frac{\pi }{5}} \right)}}{{16\sin \left( {\frac{\pi }{5}} \right)}}\)
\( = \frac{{\sin \pi \cos \left( {\frac{\pi }{5}} \right) + \cos \pi \sin \left( {\frac{\pi }{5}} \right)}}{{16\sin \left( {\frac{\pi }{5}} \right)}}\)
\( = \frac{{0.\cos \left( {\frac{\pi }{5}} \right) + \left( { - 1} \right)\sin \left( {\frac{\pi }{5}} \right)}}{{16\sin \left( {\frac{\pi }{5}} \right)}}\)
\( = \frac{{ - \sin \left( {\frac{\pi }{5}} \right)}}{{16\sin \left( {\frac{\pi }{7}} \right)}}\)
\( = - \frac{1}{{16}}\)
37. \(\sin 6x = \ldots \)
A. \(4\sin x - 3{\sin ^3}x\)
B. \(4{\sin ^3}2x - 3\sin 2x\)
C. \(4{\sin ^3}x - 3\sin x\)
D. \(3\sin x - 4{\sin ^3}x\)
E. \(3\sin 2x - 4{\sin ^3}2x\)
Jawab : E
Pembahasan :
\(\sin 6x = \sin \left( {4x + 2x} \right)\)
\( = \sin 4x\cos 2x + \cos 4x\sin 2x\)
\( = 2\sin 2x\cos 2x\cos 2x + \left( {1 - 2{{\sin }^2}2x} \right)\sin 2x\)
\( = 2\sin 2x{\cos ^2}2x + \left( {1 - 2{{\sin }^2}2x} \right)\sin 2x\)
\( = 2\sin 2x\left( {1 - {{\sin }^2}2x} \right) + \left( {1 - 2{{\sin }^2}2x} \right)\sin 2x\)
\( = 2\sin 2x - 2{\sin ^3}2x + \sin 2x - 2{\sin ^3}2x\)
\( = 3\sin 2x - 4{\sin ^3}2x\)
41. \(\cos 6A - 2\cos 4A - \cos 2A + 2 = \ldots \)
A. \(64{\cos ^2}A{\sin ^4}A\)
B. \(64{\sin ^2}A{\cos ^4}A\)
C. \(32{\sin ^2}A{\cos ^4}A\)
D. \(32{\cos ^2}A{\sin ^4}A\)
E. \(16{\sin ^2}A{\cos ^4}A\)
Jawab : D
Pembahasan :
Ingat, \(\cos A - \cos B = - 2\sin \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right)\) sehingga:
\(\cos 6A - 2\cos 4A - \cos 2A + 2\)
\( = \left( {\cos 6A - \cos 2A} \right) - 2\cos 4A + 2\)
\( = \left( { - 2\sin 4A\sin 2A} \right) - 2\left( {\cos 4A - 1} \right)\)
\( = \left( { - 2.2\sin 2A\cos 2A\sin 2A} \right) - 2\left( {\cos 4A - 1} \right)\)
\( = \left( { - 4{{\sin }^2}2A\cos 2A} \right) + 2\left( {1 - \cos 4A} \right)\)
\( = - 4{\sin ^2}2A\cos 2A + 2\left( {2{{\sin }^2}2A} \right)\)
\( = - 4{\sin ^2}2A\cos 2A + 4{\sin ^2}2A\)
\( = 4{\sin ^2}2A\left( {1 - \cos 2A} \right)\)
\( = 4{\left( {2\sin A\cos A} \right)^2}\left( {2{{\sin }^2}A} \right)\)
\( = 4\left( {4{{\sin }^2}A{{\cos }^2}A} \right)\left( {2{{\sin }^2}A} \right)\)
\( = 32{\cos ^2}A{\sin ^4}A\)
47. \({\sin ^2}\left( {\frac{\pi }{8} + \frac{A}{2}} \right) - {\sin ^2}\left( {\frac{\pi }{8} - \frac{A}{2}} \right) = \ldots \)
A. \(2\sin \frac{\pi }{8}\)
B. \(\sin A\sin \frac{\pi }{8}\)
C. \(\frac{1}{2}\sin A\)
D. \(\frac{1}{{\sqrt 2 }}\sin A\)
E. \(\frac{1}{{\sqrt 2 }}\cos A\)
Jawab : D
Pembahasan :
Misal \(\frac{\pi }{8} = x\) dan \(\frac{A}{2} = y\)
\({\sin ^2}\left( {\frac{\pi }{8} + \frac{A}{2}} \right) - {\sin ^2}\left( {\frac{\pi }{8} - \frac{A}{2}} \right)\)
\( = {\sin ^2}\left( {x + y} \right) - {\sin ^2}\left( {x - y} \right)\)
\( = \left( {\sin \left( {x + y} \right) + \sin \left( {x - y} \right)} \right)\left( {\sin \left( {x + y} \right) - \sin \left( {x - y} \right)} \right)\)
\( = \left( {\sin x\cos y + \cos x\sin y + \sin x\cos y - \cos x\sin y} \right)\left( {\sin x\cos y + \cos x\sin y - \left( {\sin x\cos y - \cos x\sin y} \right)} \right)\)
\( = 2\sin x\cos y.2\cos x\sin y\)
\( = 2\sin x\cos x.2\cos y\sin y\)
\( = \sin 2x\sin 2y\)
\( = \sin 2\left( {\frac{\pi }{8}} \right)\sin A\)
\( = \sin \frac{\pi }{4}\sin A\)
\( = \frac{1}{{\sqrt 2 }}\sin A\)
56. Diketahui bahwa \(\sqrt[3]{{{{\sin }^2}x}} + \sqrt[3]{{{{\cos }^2}x}} = \sqrt[3]{2}\) maka \({\cos ^2}2x = \ldots \)
A. \(\frac{2}{{27}}\)
B. \(\frac{8}{{27}}\)
C. \(\frac{9}{{27}}\)
D. \(\frac{25}{{27}}\)
E. 1
Jawab : D
Pembahasan :
\(\sqrt[3]{{{{\sin }^2}x}} + \sqrt[3]{{{{\cos }^2}x}} = \sqrt[3]{2}\)
\({\left( {{{\sin }^2}x} \right)^{\frac{1}{3}}} + {\left( {{{\cos }^2}x} \right)^{\frac{1}{3}}} = {\left( 2 \right)^{\frac{1}{3}}}\)
\({\left( {{{\left( {{{\sin }^2}x} \right)}^{\frac{1}{3}}} + {{\left( {{{\cos }^2}x} \right)}^{\frac{1}{3}}}} \right)^3} = {\left( {{{\left( 2 \right)}^{\frac{1}{3}}}} \right)^3}\)
\({\sin ^2}x + 3{\left( {{{\left( {{{\sin }^2}x} \right)}^{\frac{1}{3}}}} \right)^2}{\left( {{{\cos }^2}x} \right)^{\frac{1}{3}}} + 3.{\left( {{{\sin }^2}x} \right)^{\frac{1}{3}}}{\left( {{{\left( {{{\cos }^2}x} \right)}^{\frac{1}{3}}}} \right)^2} + {\cos ^2}x = 2\)
\(1 + 3{\left( {\left( {{{\sin }^4}x} \right)\left( {{{\cos }^2}x} \right)} \right)^{\frac{1}{3}}} + 3{\left( {\left( {{{\cos }^4}x} \right)\left( {{{\sin }^2}x} \right)} \right)^{\frac{1}{3}}} = 2\)
\(3\left( {{{\left( {{{\sin }^4}x{{\cos }^2}x} \right)}^{\frac{1}{3}}} + {{\left( {{{\cos }^4}x{{\sin }^2}x} \right)}^{\frac{1}{3}}}} \right) = 1\)
\({\left( {{{\sin }^4}x{{\cos }^2}x} \right)^{\frac{1}{3}}} + {\left( {{{\cos }^4}x{{\sin }^2}x} \right)^{\frac{1}{3}}} = \frac{1}{3}\)
\({\left( {{{\sin }^2}x{{\left( {\sin x\cos x} \right)}^2}} \right)^{\frac{1}{3}}} + {\left( {{{\cos }^2}x{{\left( {\sin x\cos x} \right)}^2}} \right)^{\frac{1}{3}}} = \frac{1}{3}\)
\({\left( {{{\sin }^2}x{{\left( {\frac{1}{2}\sin 2x} \right)}^2}} \right)^{\frac{1}{3}}} + {\left( {{{\cos }^2}x{{\left( {\frac{1}{2}\sin 2x} \right)}^2}} \right)^{\frac{1}{3}}} = \frac{1}{3}\)
\({\left( {{{\sin }^2}x.\frac{1}{4}.{{\sin }^2}2x} \right)^{\frac{1}{3}}} + {\left( {{{\cos }^2}x.\frac{1}{4}.{{\sin }^2}2x} \right)^{\frac{1}{3}}} = \frac{1}{3}\)
\({\left( {{{\sin }^2}x} \right)^{\frac{1}{3}}}{\left( {\frac{1}{4}.{{\sin }^2}2x} \right)^{\frac{1}{3}}} + {\left( {{{\cos }^2}x} \right)^{\frac{1}{3}}}{\left( {\frac{1}{4}.{{\sin }^2}2x} \right)^{\frac{1}{3}}} = \frac{1}{3}\)
\({\left( {\frac{1}{4}.{{\sin }^2}2x} \right)^{\frac{1}{3}}}\left( {{{\left( {{{\sin }^2}x} \right)}^{\frac{1}{3}}} + {{\left( {{{\cos }^2}x} \right)}^{\frac{1}{3}}}} \right) = \frac{1}{3}\)
\({\left( {\frac{1}{4}.{{\sin }^2}2x} \right)^{\frac{1}{3}}}{\left( 2 \right)^{\frac{1}{3}}} = \frac{1}{3}\)
\({\left( {{{\left( {\frac{1}{4}.{{\sin }^2}2x} \right)}^{\frac{1}{3}}}{{\left( 2 \right)}^{\frac{1}{3}}}} \right)^3} = {\left( {\frac{1}{3}} \right)^3}\)
\(\frac{1}{2}.{\sin ^2}2x = \frac{1}{{27}}\)
\({\sin ^2}2x = \frac{2}{{27}}\)
\(1 - {\cos ^2}2x = \frac{2}{{27}}\)
\({\cos ^2}2x = \frac{{27}}{{27}} - \frac{2}{{27}}\)
\({\cos ^2}2x = \frac{{25}}{{27}}\)
.
.
.
Wassalamu'alaikum warahmatullahi wabarakaatuh
.
.
.
31. Nilai dari \(\left( {\cos \left( {\frac{\pi }{5}} \right)\cos \left( {\frac{{2\pi }}{5}} \right)\cos \left( {\frac{{4\pi }}{5}} \right)\cos \left( {\frac{{8\pi }}{5}} \right)} \right)\) adalah...
A. \( - \frac{1}{{16}}\)
B. \( - \frac{1}{8}\)
C. 0
D. \(\frac{1}{{16}}\)
E. \(\frac{1}{8}\)
Jawab : A
Pembahasan :
Ingat, \(\sin 2A = 2\sin A\cos A \to \cos A = \frac{{\sin 2A}}{{2\sin A}}\)
\(\cos \left( {\frac{\pi }{5}} \right)\cos \left( {\frac{{2\pi }}{5}} \right)\cos \left( {\frac{{4\pi }}{5}} \right)\cos \left( {\frac{{8\pi }}{5}} \right)\)
\( = \frac{{\sin 2\left( {\frac{\pi }{5}} \right)}}{{2\sin \left( {\frac{\pi }{5}} \right)}}.\frac{{\sin 2\left( {\frac{{2\pi }}{5}} \right)}}{{2\sin \left( {\frac{{2\pi }}{5}} \right)}}.\frac{{\sin 2\left( {\frac{{4\pi }}{5}} \right)}}{{2\sin \left( {\frac{{4\pi }}{5}} \right)}}.\frac{{\sin 2\left( {\frac{{8\pi }}{5}} \right)}}{{2\sin \left( {\frac{{8\pi }}{5}} \right)}}\)
\( = \frac{{\sin \left( {\frac{{2\pi }}{5}} \right)}}{{16\sin \left( {\frac{\pi }{5}} \right)}}.\frac{{\sin \left( {\frac{{4\pi }}{5}} \right)}}{{\sin \left( {\frac{{2\pi }}{5}} \right)}}.\frac{{\sin \left( {\frac{{8\pi }}{5}} \right)}}{{\sin \left( {\frac{{4\pi }}{5}} \right)}}.\frac{{\sin \left( {\frac{{16\pi }}{5}} \right)}}{{\sin \left( {\frac{{8\pi }}{5}} \right)}}\)
\( = \frac{{\sin \left( {\frac{{16\pi }}{5}} \right)}}{{16\sin \left( {\frac{\pi }{5}} \right)}}\)
\( = \frac{{\sin \left( {3\pi + \frac{\pi }{5}} \right)}}{{16\sin \left( {\frac{\pi }{5}} \right)}}\)
\( = \frac{{\sin \left( {\pi + \frac{\pi }{5}} \right)}}{{16\sin \left( {\frac{\pi }{5}} \right)}}\)
\( = \frac{{\sin \pi \cos \left( {\frac{\pi }{5}} \right) + \cos \pi \sin \left( {\frac{\pi }{5}} \right)}}{{16\sin \left( {\frac{\pi }{5}} \right)}}\)
\( = \frac{{0.\cos \left( {\frac{\pi }{5}} \right) + \left( { - 1} \right)\sin \left( {\frac{\pi }{5}} \right)}}{{16\sin \left( {\frac{\pi }{5}} \right)}}\)
\( = \frac{{ - \sin \left( {\frac{\pi }{5}} \right)}}{{16\sin \left( {\frac{\pi }{7}} \right)}}\)
\( = - \frac{1}{{16}}\)
37. \(\sin 6x = \ldots \)
A. \(4\sin x - 3{\sin ^3}x\)
B. \(4{\sin ^3}2x - 3\sin 2x\)
C. \(4{\sin ^3}x - 3\sin x\)
D. \(3\sin x - 4{\sin ^3}x\)
E. \(3\sin 2x - 4{\sin ^3}2x\)
Jawab : E
Pembahasan :
\(\sin 6x = \sin \left( {4x + 2x} \right)\)
\( = \sin 4x\cos 2x + \cos 4x\sin 2x\)
\( = 2\sin 2x\cos 2x\cos 2x + \left( {1 - 2{{\sin }^2}2x} \right)\sin 2x\)
\( = 2\sin 2x{\cos ^2}2x + \left( {1 - 2{{\sin }^2}2x} \right)\sin 2x\)
\( = 2\sin 2x\left( {1 - {{\sin }^2}2x} \right) + \left( {1 - 2{{\sin }^2}2x} \right)\sin 2x\)
\( = 2\sin 2x - 2{\sin ^3}2x + \sin 2x - 2{\sin ^3}2x\)
\( = 3\sin 2x - 4{\sin ^3}2x\)
41. \(\cos 6A - 2\cos 4A - \cos 2A + 2 = \ldots \)
A. \(64{\cos ^2}A{\sin ^4}A\)
B. \(64{\sin ^2}A{\cos ^4}A\)
C. \(32{\sin ^2}A{\cos ^4}A\)
D. \(32{\cos ^2}A{\sin ^4}A\)
E. \(16{\sin ^2}A{\cos ^4}A\)
Jawab : D
Pembahasan :
Ingat, \(\cos A - \cos B = - 2\sin \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right)\) sehingga:
\(\cos 6A - 2\cos 4A - \cos 2A + 2\)
\( = \left( {\cos 6A - \cos 2A} \right) - 2\cos 4A + 2\)
\( = \left( { - 2\sin 4A\sin 2A} \right) - 2\left( {\cos 4A - 1} \right)\)
\( = \left( { - 2.2\sin 2A\cos 2A\sin 2A} \right) - 2\left( {\cos 4A - 1} \right)\)
\( = \left( { - 4{{\sin }^2}2A\cos 2A} \right) + 2\left( {1 - \cos 4A} \right)\)
\( = - 4{\sin ^2}2A\cos 2A + 2\left( {2{{\sin }^2}2A} \right)\)
\( = - 4{\sin ^2}2A\cos 2A + 4{\sin ^2}2A\)
\( = 4{\sin ^2}2A\left( {1 - \cos 2A} \right)\)
\( = 4{\left( {2\sin A\cos A} \right)^2}\left( {2{{\sin }^2}A} \right)\)
\( = 4\left( {4{{\sin }^2}A{{\cos }^2}A} \right)\left( {2{{\sin }^2}A} \right)\)
\( = 32{\cos ^2}A{\sin ^4}A\)
47. \({\sin ^2}\left( {\frac{\pi }{8} + \frac{A}{2}} \right) - {\sin ^2}\left( {\frac{\pi }{8} - \frac{A}{2}} \right) = \ldots \)
A. \(2\sin \frac{\pi }{8}\)
B. \(\sin A\sin \frac{\pi }{8}\)
C. \(\frac{1}{2}\sin A\)
D. \(\frac{1}{{\sqrt 2 }}\sin A\)
E. \(\frac{1}{{\sqrt 2 }}\cos A\)
Jawab : D
Pembahasan :
Misal \(\frac{\pi }{8} = x\) dan \(\frac{A}{2} = y\)
\({\sin ^2}\left( {\frac{\pi }{8} + \frac{A}{2}} \right) - {\sin ^2}\left( {\frac{\pi }{8} - \frac{A}{2}} \right)\)
\( = {\sin ^2}\left( {x + y} \right) - {\sin ^2}\left( {x - y} \right)\)
\( = \left( {\sin \left( {x + y} \right) + \sin \left( {x - y} \right)} \right)\left( {\sin \left( {x + y} \right) - \sin \left( {x - y} \right)} \right)\)
\( = \left( {\sin x\cos y + \cos x\sin y + \sin x\cos y - \cos x\sin y} \right)\left( {\sin x\cos y + \cos x\sin y - \left( {\sin x\cos y - \cos x\sin y} \right)} \right)\)
\( = 2\sin x\cos y.2\cos x\sin y\)
\( = 2\sin x\cos x.2\cos y\sin y\)
\( = \sin 2x\sin 2y\)
\( = \sin 2\left( {\frac{\pi }{8}} \right)\sin A\)
\( = \sin \frac{\pi }{4}\sin A\)
\( = \frac{1}{{\sqrt 2 }}\sin A\)
56. Diketahui bahwa \(\sqrt[3]{{{{\sin }^2}x}} + \sqrt[3]{{{{\cos }^2}x}} = \sqrt[3]{2}\) maka \({\cos ^2}2x = \ldots \)
A. \(\frac{2}{{27}}\)
B. \(\frac{8}{{27}}\)
C. \(\frac{9}{{27}}\)
D. \(\frac{25}{{27}}\)
E. 1
Jawab : D
Pembahasan :
\(\sqrt[3]{{{{\sin }^2}x}} + \sqrt[3]{{{{\cos }^2}x}} = \sqrt[3]{2}\)
\({\left( {{{\sin }^2}x} \right)^{\frac{1}{3}}} + {\left( {{{\cos }^2}x} \right)^{\frac{1}{3}}} = {\left( 2 \right)^{\frac{1}{3}}}\)
\({\left( {{{\left( {{{\sin }^2}x} \right)}^{\frac{1}{3}}} + {{\left( {{{\cos }^2}x} \right)}^{\frac{1}{3}}}} \right)^3} = {\left( {{{\left( 2 \right)}^{\frac{1}{3}}}} \right)^3}\)
\({\sin ^2}x + 3{\left( {{{\left( {{{\sin }^2}x} \right)}^{\frac{1}{3}}}} \right)^2}{\left( {{{\cos }^2}x} \right)^{\frac{1}{3}}} + 3.{\left( {{{\sin }^2}x} \right)^{\frac{1}{3}}}{\left( {{{\left( {{{\cos }^2}x} \right)}^{\frac{1}{3}}}} \right)^2} + {\cos ^2}x = 2\)
\(1 + 3{\left( {\left( {{{\sin }^4}x} \right)\left( {{{\cos }^2}x} \right)} \right)^{\frac{1}{3}}} + 3{\left( {\left( {{{\cos }^4}x} \right)\left( {{{\sin }^2}x} \right)} \right)^{\frac{1}{3}}} = 2\)
\(3\left( {{{\left( {{{\sin }^4}x{{\cos }^2}x} \right)}^{\frac{1}{3}}} + {{\left( {{{\cos }^4}x{{\sin }^2}x} \right)}^{\frac{1}{3}}}} \right) = 1\)
\({\left( {{{\sin }^4}x{{\cos }^2}x} \right)^{\frac{1}{3}}} + {\left( {{{\cos }^4}x{{\sin }^2}x} \right)^{\frac{1}{3}}} = \frac{1}{3}\)
\({\left( {{{\sin }^2}x{{\left( {\sin x\cos x} \right)}^2}} \right)^{\frac{1}{3}}} + {\left( {{{\cos }^2}x{{\left( {\sin x\cos x} \right)}^2}} \right)^{\frac{1}{3}}} = \frac{1}{3}\)
\({\left( {{{\sin }^2}x{{\left( {\frac{1}{2}\sin 2x} \right)}^2}} \right)^{\frac{1}{3}}} + {\left( {{{\cos }^2}x{{\left( {\frac{1}{2}\sin 2x} \right)}^2}} \right)^{\frac{1}{3}}} = \frac{1}{3}\)
\({\left( {{{\sin }^2}x.\frac{1}{4}.{{\sin }^2}2x} \right)^{\frac{1}{3}}} + {\left( {{{\cos }^2}x.\frac{1}{4}.{{\sin }^2}2x} \right)^{\frac{1}{3}}} = \frac{1}{3}\)
\({\left( {{{\sin }^2}x} \right)^{\frac{1}{3}}}{\left( {\frac{1}{4}.{{\sin }^2}2x} \right)^{\frac{1}{3}}} + {\left( {{{\cos }^2}x} \right)^{\frac{1}{3}}}{\left( {\frac{1}{4}.{{\sin }^2}2x} \right)^{\frac{1}{3}}} = \frac{1}{3}\)
\({\left( {\frac{1}{4}.{{\sin }^2}2x} \right)^{\frac{1}{3}}}\left( {{{\left( {{{\sin }^2}x} \right)}^{\frac{1}{3}}} + {{\left( {{{\cos }^2}x} \right)}^{\frac{1}{3}}}} \right) = \frac{1}{3}\)
\({\left( {\frac{1}{4}.{{\sin }^2}2x} \right)^{\frac{1}{3}}}{\left( 2 \right)^{\frac{1}{3}}} = \frac{1}{3}\)
\({\left( {{{\left( {\frac{1}{4}.{{\sin }^2}2x} \right)}^{\frac{1}{3}}}{{\left( 2 \right)}^{\frac{1}{3}}}} \right)^3} = {\left( {\frac{1}{3}} \right)^3}\)
\(\frac{1}{2}.{\sin ^2}2x = \frac{1}{{27}}\)
\({\sin ^2}2x = \frac{2}{{27}}\)
\(1 - {\cos ^2}2x = \frac{2}{{27}}\)
\({\cos ^2}2x = \frac{{27}}{{27}} - \frac{2}{{27}}\)
\({\cos ^2}2x = \frac{{25}}{{27}}\)
.
.
.
Soal dan pembahasan LKS yang lain cek disini.
Kritik dan saran silahkan berikan di komentar, termasuk jika ada salah hitung dan atau salah ketik.
Terimakasih
Semoga bermanfaat 😊
Wassalamu'alaikum warahmatullahi wabarakaatuh
Yang no 49 kk nggak ada?
ReplyDeleteMakasih bang
ReplyDelete