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Showing posts with label Pembuktian Matematika. Show all posts
Showing posts with label Pembuktian Matematika. Show all posts

Bukti Identitas |cosh (z)|^2=sinh^2 (x) + cos^2 (y)

Bukti identitas $|cosh (z)|^2=sinh^2 (x) + cos^2 (y)$, ini dipertanyakan oleh salah satu teman saya yang kebetulan sedang mengambil mata kuliah Analisis Kompleks pada program studi pendidikan matematika, Universitas Lakidende, Unaaha. Agar dapat bermanfaat bagi pembaca blog ini, saya menulis buktinya di sini. Sebelumnya terima kasih telah berkunjung!

Tulisan ini diperuntuhkan bagi mahasiswa yang sedang mencari cara memuktikan identitas tersebut. Entah itu tugas dari dosen atau kebutuhan mahasiswa sendiri. Sehingga, bagi Anda yang sedang atau telah mengambil mata kuliah Analisis Kompleks, bukalah kembali buka Anda yang membahas tentang fungsi trigonometri, fungsi hiperbolik, dan modulus pada pelajaran analisis kompleks karena kali ini hanya akan dibuktiktikan identitas di atas saja, tidak membahas materi-materi yang disebutkan sebelumnya. Pada bukti di bawah ini, saya hanya memberikan ide cara membuktikannya, selebihnya Anda tinggal mempelajarinya mengapa langkah-langkah yang ada bisa terjadi. Itu adalah tugas Anda. 

Untuk membuktikan kesamaan di atas, dapat dilakukan dengan cara merubah salah satu ruas (ruas kiri atau ruas kanan) sehingga sama dengan ruas lainnya menggunakan kesamaan-kesamaan yang telah diketahui atau dibuktikan sebelumnya. 

Perhatikan kesamaannya, dari ruas kiri yaitu $|cosh (z)|^2$ akan ditujukkan $|cosh (z)|^2=sinh^2 (x) + cos^2 (y)$ sebagai berikut.

$ \begin{align} |cosh (z)|^2 & = (cosh (z))(cosh ( \overline{z})) \\ & = (cosh (x+iy))(cosh (x-iy)) \\ & = (cosh (x) cosh (iy) + sinh (x) sinh (iy)) (cosh (x) cosh (iy) - sinh (x) sinh (iy)) \\ & = (cosh (x) cos (y) + sinh (x) i sin (y))(cosh (x) cos (y) - sinh (x) i sin (y)) \\ & = (cosh (x) cos (y))^2 - (sinh (x) i sin (y))^2 \\ & = (cosh (x) cos (y))^2 + (sinh (x) sin (y))^2 \\ & = cos^2 (y) (cosh^2 (x) - sinh^2 (x)) + sinh^2 (x) cos^2 (y1 + sinh^2 (x) (sin^2 (y) + cos^2 (y)) - sinh^2 (x) cos^2 (y) \\ & = cos^2 (y) . (1) + sinh^2 (x) . (1) \\ & = cos^2 (y) + sinh^2 (x) \end{align} $.

Kita peroleh ruas kanannya yaitu $sinh^2 (x) + cos^2 (y)$. Karena ruas kiri samadengan ruas kanan maka kita telah membuktikan bahwa $|cosh (z)|^2=sinh^2 (x) + cos^2 (y)$. Demikian bukti singkat ini, semoga dapat bermanfaat bagi pembaca. 

Bukti Turunan Cos x = -Sin x

Membuktikan turunan cos x = -sin x pada dasarnya sama dengan Bukti Turunan Sin x = Cos x.

Secara matematis Turunan Cos x = -Sin x dituliskan dengan :

Turunan suatu fungsi didefinisikan sebagai:
$f'(x)=\lim_{h \rightarrow 0}\frac{f(x+h) - f(x) }{h}$

Menurut definis Turunan dari suatu fungsi f(x) di atas maka:

$\begin{align}f'(x) &=\lim_{h \rightarrow 0}\frac{f(x+h) - f(x) }{h} \\ & =\lim_{h \rightarrow 0} \frac{Cos \quad (x +h) - cos \quad x }{h} \\ & =\lim_{h \rightarrow 0} \frac{Cosx \quad Cosh - Sinx \quad sinh - Cosx }{h} \\ & =\lim_{h \rightarrow 0} (Cosx . \frac{cosh-1}{h} - Sin x. \frac{sin h}{h} ) \\ & =\lim_{h \rightarrow 0} (-Cosx . \frac{1-cos h}{h} - Sin x. \frac{sin h}{h} ) \\ & =-Cosx [\lim_{h \rightarrow 0} \frac{1-cos h}{h}] - Sin x [\lim_{h \rightarrow 0} \frac{sin h}{h} \\ & =Cos x(0) - Sin x(1) \\ & =-Sin x \end{align}$

$\lim_{h \rightarrow 0} \frac{1-cos h}{h}=0 \\
\lim_{h \rightarrow 0} \frac{sin h}{h}$

Terbukti bahwa Dx [Cos x] = -Sin x

Jika rumus tidak terbaca gunakan browser yang support Script Math.Jax seperti Operamini, Thanks !

Bukti Turunan Sin x = Cos X

Bukti Turunan $Sin \ x = \ Cos x$. Pada kesempatan ini kita akan membuktikan bahwa turunan $\sin x $ adalah $cos \ x $. Secara matematis Turunan $\sin x = \cos x $dituliskan dengan :

$f'(x) =\lim_{h \rightarrow 0}\frac{f(x+h) - f(x) }{h}$

Menurut definis Turunan dari suatu fungsi f(x) di atas maka:

$\begin{align} f'(x) &= \lim_{h \rightarrow 0}\frac{f(x+h) - f(x) }{h} \\ & =\lim_{h \rightarrow 0} \frac{sin(x +h) - sin x }{h} \\ & =\lim_{h \rightarrow 0} \frac{sinx cosh + cosx sinh - sinx }{h} \\ & =\lim_{h \rightarrow 0} (-sinx. \frac{1-cosh}{h} + cosx. \frac{sin h}{h} ) \\ & =(-sinx) [\lim_{h \rightarrow 0} \frac{1-cos h}{h}] + (cos x) [\lim_{h \rightarrow 0} \frac{sin h}{h} \\ & =(-sin x)(0) + (cos x)(1) \\ & =cos x \end{align}$
Terbukti bahwa $Dx [\sin x] = \cos x$
Jika rumus tidak terbaca gunakan browser yang support Script Math Jax seperti Google Chroome, Operamini, dll. Baca Juga: Bukti Turunan Cos x = -Sin x

Pembuktian Untuk Bilangan yang Dipangkat Nol(0) Sama dengan Satu(1) a^0=1

Untuk membuktikan kebenaran bahwa bilangan apapun yang bukan 0 (nol) dipangkat dengan 0 (nol) hasilnya adalah 1 (satu). Kita harus merujuk pada definisi pangkat bahwa

Misalkan :

Bagaimana jika

Jawabannya adalah


Pembuktian Kesamaan Trigonometri sin a cosb=1/2. [sin(a+b) + sin (a-b)]

Pembuktian Kesamaan Trigonometri sin a cosb=1/2. [sin(a+b) + sin (a-b)] dengan memanfaatkan kesaman-kesamaan yang ada dalam trigonometri yang tentunya menggunakan persamaan lain yang sebelumnya telah dibuktikan kebenarannya.

Pada postingan saya sebelumnya membahas masalah Pembuktian Kesamaan Trigonometri sinx+siny=2 sin 1/2 (x+y)cos1/2 (x-y)  yang menggunakan teknik pembuktian tidak formal. Pembuktian formal dalam matematika merupakan pembuktian yang menggunakan kaidah-kaidah inferensi berupa hukum-hukum, dalil-dalil atau definisi.

Kali ini kita akan membuktikan Kesamaan Trigonometri sin a cosb=1/2. [sin(a+b) + sin (a-b)] dengan menggunakan kesamaan yang telah ada yaitu :
sin (a+b)=sin a cos b +cos a sin b
sin (a-b)=sin a cos b - cos a sin b  +
sin(a+b) + sin(a-b)=2 sin a cos b

<=> 2 sin a cos b = sin (a+b) + sin (a-b)
<=>sin a cos b = 1/2. [sin(a+b) + sin(a-b)]
TERBUKTI bahwa sin a cos b = 1/2. [sin(a+b) + sin(a-b)]

Demikian untuk Kesamaan Trigonometri sin a cosb=1/2. [sin(a+b) + sin (a-b) ] jika ada yang ingin ditanyakan silahkan berkomentar.

Salam MKB :D

Pembuktian Kesamaan Trigonometri sinx+siny=2 sin 1/2 (x+y)cos1/2 (x-y)

Pembuktian Turunan Trigonometri sinx+siny=2 sin 1/2 (x+y)cos1/2 (x-y) dengan cara yang sederhana ini tidak menggunakan definisi atau teorema apapun. Ini hanya sebuah metode aljabar dalam menunjukkan suatu persamaan. Misalkan 2+2=4 untuk menunjukkan kebenaran tersebut dengan metode yang akan kita coba untuk membuktikan kesamaan trigonometri tersebut, yaitu dengan cara menunjukkan ruas kiri sama dengan ruas kanan atau sebaliknya. Pembuktian ini lebih tepat kepada intuisi saja bukan sebuah bukti yang fakultatif atau formal.

Ilustrasi :
sin90+sin90=2sin1/2 (90+90)cos1/2 (90-90)
(i) Untuk sin 90+sin90=1+1=2              ket: sin 90=1
(ii) Untuk 2sin 1/2 (90+90) cos 1/2 (90-90)
=2 sin 1/2(180) cos 1/2(0)
=2 sin 90 cos 0
=2. 1 . 1
=2                        Ket: cos 0=1
Ilustrasi di atas bukannlah bukti karena contoh bukannlah bukti. Karena itu hanyalah metode induktif, metode ini tidak bisa digunakan untuk pembuktian, hanya kepada hal-hal tertentu saja misalnya untuk membuktikan barisan bilangan asli pembuktian ini dinamakan induksi matematika.

Pembuktian Kesamaan Trigonometri sinx+siny=2 sin 1/2 (x+y)cos1/2 (x-y) dengan bukti informalnya yaitu:

Misal: x=a+b
Maka: x+y=2a  => a=1/2(x+y)
x-y=2b   => b=1/2(x-y)
Jadi: sin x+ sin y=2 sin 1/2(x+y) cos1/2(x-y)
=2 sin a cos b
=2 . 1/2. [sin (a+b) +  sin (a-b) ]
= sin (a+b) sin (a-b)
=sin x + sin y TERBUKTI
Catatan: Untuk pembuktian sin a cos b=1/2[sin (a+b) + sin (a-b)] akan di tunjukkan pada postingan berikutnya.

Demikian untuk Pembuktian Kesamaan Trigonometri sinx+siny=2 sin 1/2 (x+y)cos1/2 (x-y)
 Semoga bermanfaat

Pembuktian 0!=1 dan 1!=1 dengan ! Notasi Faktorial

Pembuktian 0!=1 dan 1!=1 dengan ! Notasi Faktorial

Saya teringat dengan dosen saya yang mengatakan bahwa pembuktian itu paling banyak nanti di semester 5, terutama pada matakuliah Struktur Aljabar atau Analisis Real. Namun, untuk hal-hal semacam ini, penting untuk dapat melakukan pembuktiannya sehingga ketika ada yang bertanya, mengapa $0!=1$, $1!=1$, $0!=1!$, kita bisa mengemukakan alasan matematisnya, bukan menjawabnya dengan alasan klasik "karena memangnya" (senyum).

Dalam matematika salah satu kemampuan yang harus kita miliki adalah kemampuan membuktikan kebenaran suatu teorema atau rumus-rumus dalam matematika. Kalau membuktikan $0!=1$ saja tak mampu, maka dipastikan kita gak akan mampu membuktikan yang lebih dari itu.

Pada kesempatan ini, kita akan membuktikan $0!=1$. Untuk dapat membuktikannya, pertama kita harus tahu apa itu faktorial yang dinotasikan dengan (!). Misalkan n bilangan asli maka $n!$ didefinisikan sebagai:

Berdasarkan definisi bahwa:
$n! = n ( n - 1 )!$
  • Untuk membuktikan 1!=1 ambil n=2 maka:
n! = n ( n - 1 )!
2! = 2 ( 2 - 1 )!
2 . 1 = 2 . 1!
2 = 2 . 1!
2/2 = 1!
1 = 1 !
(Terbukti bahwa 1! = 1)
  • Untuk membuktikan 0!=1 ambil n=1 maka
n! = n ( n - 1 )!
1! = 1 ( 1 - 1 )!
1 = 0!
(Terbukti bahwa 0! = 1)

Dengan demikian terbukti benar bahwa $ 0! = 1! =1$. Mudah bukan? Semoga bermanfaat!

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