Identitas Trigonometri

Identitas dasar yang merupakan hubungan kebalikan
Setelah mengetahui dasar-dasar dari trigonometri, maka selanjutnya adalah identitas dasar trigonometri yang merupakan hubungan kebalikan.

  • \large cosec\: \alpha =\frac{1}{sin\: \alpha }
  • \large sec\: \alpha =\frac{1}{cos\: \alpha }
  • \large cotan\: \alpha =\frac{1}{tan\: \alpha }

Identitas dasar yang merupakan hubungan perbandingan

  • \large tan \: \alpha = \frac{sin \: \alpha }{cos \: \alpha }\Rightarrow cotan \: \alpha =\frac{cos \: \alpha }{sin\: \alpha }

 Identitas dasar yang merupakan hubungan Pythagoras

  • \large sin^{2}\: \alpha +cos^{2}\: \alpha =1

Pembuktian \large sin^{2}\: \alpha +cos^{2}\: \alpha =1 dengan menggunakan pythagoras

 

Berdasarkan teorema Pythagoras diketahui bahwa:

\large c^{2}=a^{2}+b^{2}

 

\large sin^{2} \alpha = \frac{a}{c} \rightarrow sin^{2}\alpha = \left ( \frac{b}{c} \right )^{2}\rightarrow sin^{2}\alpha =\frac{a^{2}}{c^{2}}

 

\large cos^{2} \alpha = \frac{a}{c} \rightarrow cos^{2}\alpha = \left ( \frac{b}{c} \right )^{2}\rightarrow cos^{2}\alpha =\frac{a^{2}}{c^{2}}

 

\large sin^{2}\alpha +cos^{2}\alpha = \frac{a^{2}}{c^{2}} + \frac{b^{2}}{c^{2}}

 

\rightarrow \frac{a^{2}+b^{2}}{c^{2}} \rightarrow Ingat \: Bahwa \; c^{2}=a^{2}+b^{2}

 

\rightarrow \frac{c^{2}}{c^{2}}= 1

 

  • \large sec^{2}\: \alpha - tan^{2}\: \alpha =1
  • \large cosec^{2}\: \alpha +cotan^{2}\: \alpha =1

Dari identitas trigonometri di atas, maka diperoleh identitas trigonometri lainnya sebagai berikut:
1. \large sin^{2}\: \alpha +cos^{2}\: \alpha =1

  • \large sin^{2}\: \alpha = 1-cos^{2}\: \alpha
  • \large cos^{2}\: \alpha = 1-sin^{2}\: \alpha

2. \large sec^{2}\: \alpha +cos^{2}\alpha =1 \large \Rightarrow sin^{2}\: \alpha +cos^{2}\alpha =1   (masing-masing ruas dikalikan dengan \large \frac{1}{cos^{2}\: \alpha } )

\large \Rightarrow \frac{sin^{2}\: \alpha }{cos^{2}\: \alpha }+\frac{sin^{2}\: \alpha }{cos^{2}\: \alpha }=\frac{1}{cos^{2}\: \alpha } \large \Rightarrow \frac{sin^{2}\: \alpha }{cos^{2}\: \alpha }+1=\frac{1}{cos^{2}\: \alpha } \large \Rightarrow tan^{2}\: \alpha +1 = sec^{2}\: \alpha \large \Rightarrow sec^{2}\: \alpha =1 +tan^{2}\: \alpha

 

3. \large cosec^{2}\: \alpha =1 + cotan^{2}\: \alpha \large \Rightarrow sec^{2}\: \alpha - tan^{2}\: \alpha = 1 (masing-masing ruas dikalikan dengan \large \frac{1}{tan^{2}\: \alpha } )

\large \Rightarrow \frac{sec^{2}\: \alpha }{tan^{2}\: \alpha }-\frac{tan^{2}\: \alpha}{tan^{2}\: \alpha}=\frac{1}{tan^{2}\: \alpha}

 

\large \Rightarrow \frac{1}{sin^{2}\alpha } - 1=\frac{1}{tan^{2}\alpha }

 

\large \Rightarrow cosec^{2}\alpha -1=cotan^{2}\alpha

 

\large \Rightarrow cosec^{2}\alpha =1+cotan^{2}\alpha

 

4. \large tan \: \alpha =\frac{sin\: \alpha }{cos\: \alpha }

5. \large cotan \: \alpha =\frac{cos\: \alpha }{tan\: \alpha }

6. \large sec\: \alpha =\frac{1 }{tan\: \alpha }

7. \large cosec\: \alpha =\frac{1 }{sin\: \alpha }

8.\large tan\: \alpha =\frac{1 }{cotan\: \alpha }

Contoh Soal
1. Buktikan identitas berikut
a. \large cos \alpha \: cotan \: \alpha = cosec \alpha - sin \alpha

b. \large \frac{1+cot^{2\alpha }}{cot \: \alpha \: sec^{2}\alpha }= cot \: \alpha

2. Sederhanakanlah bentuk berikut dengan menggunakan identitas trigonometri.

a. \large tan \: A - \frac{sec^{2}A}{tan \: A}

b. \large \left ( cosec \: B - cotan \: B \right ) \left ( 1+ cos \: B \right )

c. \large \frac{sin\: C}{1+ cos\: C}+\frac{sin\: C}{1- cos\: C} Jawab:
1.Pembuktian

a. \large cos \alpha \: cotan \: \alpha = cosec \alpha - sin \alpha

 

\large \Rightarrow cos \: \alpha \cdot \frac{cos \: \alpha }{sin \: \alpha }

 

\large \Rightarrow \frac{cos^{2} \: \alpha }{sin \: \alpha }

 

\large \Rightarrow \frac{1-sin^{2} \: \alpha }{sin \: \alpha }

 

\large \Rightarrow \frac{1}{sin \: \alpha }- \frac{sin^{2}\alpha }{sin \: \alpha }

 

\large \Rightarrow cosec \: \alpha - sin \: \alpha

 

Jadi, terbukti bahwa \large cos \alpha \: cotan \: \alpha = cosec \alpha - sin \alpha

 

b. \large \frac{1+cot^{2\alpha }}{cot \: \alpha \: sec^{2}\alpha }= cot \: \alpha

 

\large \Rightarrow \frac{cosec^{2}\alpha }{\frac{cos\: \alpha }{sin \: \alpha}\cdot \frac{1}{cos^{2}\alpha }}

 

\large \Rightarrow \frac{\frac{1}{sin^{2}\alpha }}{\frac{cos\: \alpha }{sin \: \alpha}\cdot \frac{1}{cos^{2}\alpha }}

 

\large \Rightarrow \frac{\frac{1}{sin^{2}\alpha }}{\frac{cos\: \alpha }{sin \: \alpha \: cos^{2}\alpha }}

 

\large \Rightarrow \frac{1}{sin^{2}\alpha }\cdot \frac{sin\: \alpha \: cos^{2}\alpha }{cos\: \alpha }

 

\large \Rightarrow \frac{cos\: \alpha }{sin \: \alpha }

 

\large cot\: \alpha

 

Jadi, terbukti bahwa \large \frac{1+cot^{2\alpha }}{cot \: \alpha \: sec^{2}\alpha }= cot \: \alpha

 

2. Menyederhanakan:

a. \large tan \: A - \frac{sec^{2}A}{tan \: A}

 

\large \Rightarrow \large tan \: A - \frac{1+tan^{2}A}{tan \: A}

 

\large \Rightarrow \large \frac{tan^{2}A -(1+tan^{2}A)}{tan \: A}

 

\large \Rightarrow \large \frac{tan^{2}A -1-tan^{2}A)}{tan \: A}

 

\large \Rightarrow \large -\frac{1}{tan \: A}

 

\large \Rightarrow \large cotan \: A

 

b. \large \left ( cosec \: B - cotan \: B \right ) \left ( 1+ cos \: B \right ) \large \Rightarrow Ubah ke dalam bentuk sin B dan cos B

\large \Rightarrow \frac{1}{sin\: B}+ \frac{1}{sin\: B}\cdot cos\: B- \frac{cos\: B}{sin\: B}-\frac{cos\: B}{sin\: B}\cdot cos\: B

 

\large \Rightarrow \frac{1}{sin\: B}+ \frac{ cos\: B}{sin\: B}- \frac{cos\: B}{sin\: B}-\frac{cos^{2} B}{sin\: B}

 

\large \Rightarrow \frac{1}{sin\: B}-\frac{cos^{2} B}{sin\: B}

 

\large \Rightarrow \frac{1-cos^{2} B}{sin\: B}

 

\large \Rightarrow \frac{sin^{2} B}{sin\: B}

 

\large \Rightarrow \frac{1}{sin\: B}

 

\large \Rightarrow cosec\: B

 

c. \large \frac{sin\: C}{1+ cos\: C}+\frac{sin\: C}{1- cos\: C}

 

\large \Rightarrow \frac{sin\: C (1-cos\: C)+sin \: (1+cos\: C)}{1-cos^{2} C}

 

\large \Rightarrow \frac{sin\: C (1-cos\: C)+sin \: (1+cos\: C)}{sin^{2} C}

 

\large \Rightarrow \frac{1-cos\: C+1+cos\: C}{sin\: C}

 

\large \Rightarrow \frac{2}{sin\: C}

 

\large \Rightarrow 2\frac{1}{sin\: C}

 

\large \Rightarrow 2cosec\: C

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