Aplikasi Selisih Kubik pada Soal Identitas Trigonometri

Rumus selisih kubik adalah a³ – b³ = (a – b)(a² + ab + b²). Sebelumnya saya akan buktikan bagaimana rumus ini diperoleh. Dari ekspansi binomial pangkat tiga (a + b)³ = a³ + 3a²b + 3ab² + b³, maka:

a³ + b³ = (a + b)³ – 3a²b – 3ab²

a³ + b³ = (a + b)³ – 3ab (a + b)

a³ + b³ = (a + b)[(a + b)² – 3ab]

a³ + b³ = (a + b)(a² + 2ab + b² – 3ab)

a³ + b³ = (a + b)(a² – ab + b²) ← Rumus jumlah kubik.

Kita hanya mengganti b dengan -b untuk menurunkannya. Diperoleh:

a³ – b³ = [a + (-b)] [a² – a(-b) + (-b)²]

a³ – b³ = (a – b)(a² + ab + b²)

 

1.  Buktikan bahwa \displaystyle \frac{\tan \theta }{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=\csc \theta ~\sec \theta +1=\tan \theta+\cot \theta+1=\tan \theta~(\cot^2 \theta+\cot \theta+1)

Jawab:

Perhatikan penggunaannya pada baris ke 7 !

\displaystyle \frac{\tan \theta }{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}\\ =\frac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}}+\frac{\frac{\cos \theta}{\sin \theta}}{1-\frac{\sin \theta}{\cos \theta}}\\ =\frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta-\cos \theta}{\sin \theta}}+\frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta-\sin \theta}{\cos \theta}}\\ =\frac{\sin^2 \theta }{\cos \theta~(\sin \theta-\cos \theta)}-\frac{\cos^2 \theta }{\sin \theta~(\sin \theta-\cos \theta)}\\ =\frac{1}{\sin \theta-\cos \theta}\left ( \frac{\sin^2 \theta }{\cos \theta }-\frac{\cos^2 \theta }{\sin \theta } \right )\\ =\frac{1}{\sin \theta-\cos \theta}~\frac{\sin^3 \theta -\cos^3 \theta }{\cos \theta ~\sin \theta }\\=\frac{1}{\sin \theta-\cos \theta}~\frac{(\sin \theta -\cos \theta)(\sin^2 \theta +\sin \theta~ \cos \theta+\cos^2 \theta ) }{\cos \theta ~\sin \theta }\\ =\frac{1+\sin \theta ~\cos \theta }{\cos \theta ~\sin \theta}\\ =\csc \theta ~\sec \theta +1

Dari baris ke 7 kita bisa ubah ke bentuk (a + b + c)/d ke a/d + b/d + c/d, diperoleh:

\displaystyle \frac{\sin^2 \theta +\sin \theta ~\cos \theta +\cos^2 \theta }{\cos \theta ~\sin \theta }\\ =\frac{\sin^2 \theta }{\cos \theta ~\sin \theta}+\frac{\sin \theta ~\cos \theta }{\cos \theta ~\sin \theta}+\frac{\cos^2 \theta }{\cos \theta ~\sin \theta}\\ =\tan \theta +\cot \theta +1

Dari baris ke 8 kita ubah bentuk (a + b)/b = a/b + 1 dan menggunakan manipulasi, diperoleh:

\displaystyle \frac{1+\sin \theta ~\cos \theta }{\sin \theta ~\cos \theta}\\ =\frac{1}{\sin \theta ~\cos \theta}+1\\ =\frac{\sin \theta }{\cos \theta }\left ( \frac{1}{\sin^2 \theta }+\frac{\cos \theta }{\sin \theta } \right )\\ =\tan \theta~ (\csc^2 \theta +\cot \theta )\\ =\tan \theta~ (\cot^2 \theta +\cot \theta +1)

 

2. Buktikan bahwa \displaystyle \sec^6 x-\tan^6 x=1+3\tan^2 x \sec^2 x

Jawab:

Soal ini bentuknya selisih pangkat enam. Kita gunakan rumus (a^p)^q = a^(p + q). Jadi: a^6 – b^6 = (a² – b²)(a^4 + a²b² + b^4)

\displaystyle \sec^6 x-\tan^6 x\\ =(\sec^2 x-\tan^2 x)(\sec^4 x+\sec^2 x~\tan^2 x+\tan^4 x)\\ =1\left ( \frac{1}{\cos^4 x}+\frac{\sin^2 x}{\cos^2 x~\cos^2 x}+\frac{\sin^4 x}{\cos^4 x} \right )\\ =\frac{\sin^2 x+\cos^2 x+\sin^2 x+\sin^4 x}{\cos^4 x}\\ =\frac{2\sin^2 x+\cos^2 x+\sin^2 x~(\sin^2 x)}{\cos^4 x}\\ =\frac{2\sin^2 x+\cos^2 x+\sin^2 x~(1-\cos^2 x)}{\cos^4 x}\\ =\frac{3\sin^2 x+\cos^2 x~(1-\sin^2 x)}{\cos^4 x}\\=\frac{3\sin^2 x+\cos^4 x}{\cos^4 x}\\ =3~\frac{\sin^2 x}{\cos^2 x~\cos^2 x}+1\\ =1+3\tan^2 x~\sec^2 x
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