Identitas Trigonometri

Identitas trigonometri adalah persamaan-persamaan yang memuat fungsi trigonometri dan selalu bernilai benar untuk setiap nilai variabel dimana kedua ruas persamaan tersebut terdefinisi.

Berikut identitas-identitas dasar trigonometri yang merupakan acuan dalam menentukan atau membuktikan identitas-identitas trigonometri lainnya.

Identitas Kebalikan (Reciprocal Identity)

sin θ =

\frac{1}{c s c θ}

$\mathrm{\frac{1}{csc\,\theta}}$
cos θ =

\frac{1}{s e c θ}

$\mathrm{\frac{1}{sec\,\theta}}$
tan θ =

\frac{1}{c o t θ}

$\mathrm{\frac{1}{cot\,\theta}}$

Identitas Rasio (Quotient Identity)

tan θ =

\frac{s i n θ}{c o s θ}

$\mathrm{\frac{sin\,\theta}{cos\,\theta}}$
cot θ =

\frac{c o s θ}{s i n θ}

$\mathrm{\frac{cos\,\theta}{sin\,\theta}}$

Identitas Pythagoras (Pythagorean Identity)

sin²θ + cos²θ = 1
tan²θ + 1 = sec²θ
1 + cot²θ = csc²θ

Pembuktian identitas trigonometri dilakukan dengan cara mengubah bentuk persamaan pada ruas kiri tanpa mengubah nilainya, sehingga diperoleh bentuk yang persis sama dengan persamaan pada ruas kanan atau berlaku juga sebaliknya.

Mengubah bentuk suatu ruas dapat dilakukan dengan cara :

Substitusi identitas trigonometri : identitas kebalikan, rasio dan Phythagoras.
Manipulasi aljabar : menyederhanakan bentuk pecahan, mengalikan dengan bentuk sekawan, pemfaktoran dan lain-lain.

Berikut beberapa bentuk pemfaktoran yang sering digunakan :
a² − b² = (a + b)(a − b)
a⁴ − b⁴ = (a² + b²)(a² − b²)
a³ − b³ = (a − b)(a² + b² + ab)
a³ + b³ = (a + b)(a² + b² − ab)

Contoh 1
Buktikan :

c s c^{2} x s e c x = s e c x + c o t x c s c x

$\mathrm{csc^{2}x\,sec\,x=sec\,x+cot\,x\,csc\,x}$
Jawab :
csc²x sec x = (1 + cot²x) sec x
csc²x sec x = sec x + cot²x sec x
csc²x sec x = sec x + cot x . cot x . sec x
csc²x sec x = sec x + cot x

\cdot \frac{c o / s x}{s i n x} \cdot \frac{1}{c o / s x}

$\mathrm{\cdot \frac{co{\color{red}\not}s\,x}{sin\,x }\cdot \frac{1}{co{\color{red}\not}s\,x}}$
csc²x sec x = sec x + cot x

\cdot \frac{1}{s i n x}

$\mathrm{\cdot \frac{1}{sin\,x}}$
csc²x sec x = sec x + cot x csc x

Contoh 2
Buktikan : sec⁴t − sec²t = tan⁴t + tan²t
Jawab :
sec⁴t − sec²t = (sec²t)² − sec²t
sec⁴t − sec²t = (1 + tan²t)² − (1 + tan²t)
sec⁴t − sec²t = 1 + 2tan²t + tan⁴t − 1 − tan²t
sec⁴t − sec²t = tan⁴t + tan²t

Contoh 3
Buktikan : sin⁴x − cos⁴x = 1 − 2cos²x
Jawab :
sin⁴x − cos⁴x = (sin²x + cos²x)(sin²x − cos²x)
sin⁴x − cos⁴x = 1 . (sin²x − cos²x)
sin⁴x − cos⁴x = sin²x − cos²x
sin⁴x − cos⁴x = 1 − cos²x − cos²x
sin⁴x − cos⁴x = 1 − 2cos²x

Contoh 4
Buktikan :

(s e c x - t a n x)^{2} = \frac{1 - s i n x}{1 + s i n x}

$\mathrm{(sec\,x-tan\,x)^{2}=\frac{1-sin\,x}{1+sin\,x}}$
Jawab :

\begin{matrix} (s e c x - t a n x)^{2} & = {(\frac{1}{c o s x} - \frac{s i n x}{c o s x})}^{2} = {(\frac{1 - s i n x}{c o s x})}^{2} = \frac{(1 - s i n x)^{2}}{c o s^{2} x} = \frac{(1 - s i n x)^{2}}{1 - s i n^{2} x} = \frac{(1 - s / i n x) (1 - s i n x)}{(1 - s / i n x) (1 + s i n x)} = \frac{1 - s i n x}{1 + s i n x} \end{matrix}

$\begin{align} \mathrm{(sec\,x-tan\,x)^{2}} & =\mathrm{\left ( \frac{1}{cos\,x}-\frac{sin\,x}{cos\,x} \right )^{2}} \\ & = \mathrm{\left ( \frac{1-sin\,x}{cos\,x} \right )^{2}} \\ & = \mathrm{\frac{(1-sin\,x)^{2}}{cos^{2}x}} \\ & = \mathrm{\frac{(1-sin\,x)^{2}}{1-sin^{2}x}} \\ & = \mathrm{\frac{(1-s{\color{red}\not}in\,x)(1-sin\,x)}{(1-s{\color{red}\not}in\,x)(1+sin\,x)}} \\ & = \mathrm{\frac{1-sin\,x}{1+sin\,x}} \end{align}$

Contoh 5
Buktikan :

\frac{2 - s e c^{2} x}{s e c^{2} x} = 1 - 2 s i n^{2} x

$\mathrm{\frac{2-sec^{2}x}{sec^{2}x}=1-2sin^{2}x}$
Jawab :

\begin{matrix} \frac{2 - s e c^{2} x}{s e c^{2} x} & = \frac{2}{s e c^{2} x} - \frac{s e c^{2} x}{s e c^{2} x} = 2 c o s^{2} x - 1 = 2 (1 - s i n^{2} x) - 1 = 2 - 2 s i n^{2} x - 1 = 1 - 2 s i n^{2} x \end{matrix}

$\begin{align} \mathrm{\frac{2-sec^{2}x}{sec^{2}x}} & = \mathrm{\frac{2}{sec^{2}x}-\frac{sec^{2}x}{sec^{2}x}} \\ & = \mathrm{2\,cos^{2}x-1} \\ & = \mathrm{2(1-sin^{2}x)-1} \\ & = \mathrm{2-2\,sin^{2}x-1} \\ & = \mathrm{1-2\,sin^{2}x} \end{align}$

Contoh 6
Buktikan :

\frac{1 - c o s x}{s i n x} = \frac{1}{c s c x + c o t x}

$\mathrm{\frac{1-cos\,x}{sin\,x}=\frac{1}{csc\,x+cot\,x}}$
Jawab :

\begin{matrix} \frac{1 - c o s x}{s i n x} & = \frac{1}{s i n x} - \frac{c o s x}{s i n x} = c s c x - c o t x = (c s c x - c o t x) \cdot \frac{c s c x + c o t x}{c s c x + c o t x} = \frac{(c s c x - c o t x) (c s c x + c o t x)}{c s c x + c o t x} = \frac{c s c^{2} x - c o t^{2} x}{c s c x + c o t x} = \frac{1}{c s c x + c o t x} \end{matrix}

$\begin{align} \mathrm{\frac{1-cos\,x}{sin\,x}} & = \mathrm{\frac{1}{sin\,x}-\frac{cos\,x}{sin\,x}} \\ & = \mathrm{csc\,x-cot\,x} \\ & = \mathrm{(csc\,x-cot\,x)\cdot \frac{csc\,x+cot\,x}{csc\,x+cot\,x}} \\ & = \mathrm{\frac{(csc\,x-cot\,x)(csc\,x+cot\,x)}{csc\,x+cot\,x}} \\ & = \mathrm{\frac{csc^{2}x-cot^{2}x}{csc\,x+cot\,x}} \\ & = \mathrm{\frac{1}{csc\,x+cot\,x}} \end{align}$

Contoh 7
Buktikan :

\frac{s i n x}{1 + c o s x} + \frac{1 + c o s x}{s i n x} = 2 c s c x

$\mathrm{\frac{sin\,x}{1+cos\,x}+\frac{1+cos\,x}{sin\,x}=2csc\,x}$
Jawab :

\begin{matrix} \frac{s i n x}{1 + c o s x} + \frac{1 + c o s x}{s i n x} & = \frac{1 - c o s x}{1 - c o s x} \cdot \frac{s i n x}{1 + c o s x} + \frac{1 + c o s x}{s i n x} = \frac{(1 - c o s x) s i n x}{1 - c o s^{2} x} + \frac{1 + c o s x}{s i n x} = \frac{(1 - c o s x) s i / n x}{s i n^{/ 2} x} + \frac{1 + c o s x}{s i n x} = \frac{1 - c o s x}{s i n x} + \frac{1 + c o s x}{s i n x} = \frac{2}{s i n x} = 2 c s c x \end{matrix}

$\begin{align} \mathrm{\frac{sin\,x}{1+cos\,x}+\frac{1+cos\,x}{sin\,x}} & = \mathrm{\frac{1-cos\,x}{1-cos\,x}\cdot \frac{sin\,x}{1+cos\,x}+\frac{1+cos\,x}{sin\,x}} \\ & = \mathrm{\frac{(1-cos\,x)sin\,x}{1-cos^{2}x}+\frac{1+cos\,x}{sin\,x}} \\ & = \mathrm{\frac{(1-cos\,x)si{\color{red}\not}n\,x}{sin^{{\color{red}\not}2}x}+\frac{1+cos\,x}{sin\,x}} \\ & = \mathrm{\frac{1-cos\,x}{sin\,x}+\frac{1+cos\,x}{sin\,x}} \\ & = \mathrm{\frac{2}{sin\,x}} \\ & = \mathrm{2\,csc\,x} \end{align}$

Contoh 8
Buktikan :

\frac{c s c x - 1}{c o t x} = \frac{c o t x}{c s c x + 1}

$\mathrm{\frac{csc\,x-1}{cot\,x}=\frac{cot\,x}{csc\,x+1}}$
Jawab :

\begin{matrix} \frac{c s c x - 1}{c o t x} & = \frac{c s c x - 1}{c o t x} \cdot \frac{c s c x + 1}{c s c x + 1} = \frac{c s c^{2} x - 1}{c o t x (c s c x + 1)} = \frac{c o t^{/ 2} x}{c / o t x (c s c x + 1)} = \frac{c o t x}{c s c x + 1} \end{matrix}

$\begin{align} \mathrm{\frac{csc\,x-1}{cot\,x}} & = \mathrm{\frac{csc\,x-1}{cot\,x}\cdot \frac{csc\,x+1}{csc\,x+1}} \\ & = \mathrm{\frac{csc^{2}x-1}{cot\,x\,(csc\,x+1)}} \\ & = \mathrm{\frac{cot^{{\color{red}\not}2}x}{c{\color{red}\not}ot\,x\,(csc\,x+1)}} \\ & = \mathrm{\frac{cot\,x}{csc\,x+1}} \end{align}$

Contoh 9
Buktikan :

\frac{c o s^{2} x - s i n^{2} x}{1 - t a n^{2} x} = c o s^{2} x

$\mathrm{\frac{cos^{2}x-sin^{2}x}{1-tan^{2}x}=cos^{2}x}$
Jawab :

\begin{matrix} \frac{c o s^{2} x - s i n^{2} x}{1 - t a n^{2} x} & = \frac{c o s^{2} x - s i n^{2} x}{1 - t a n^{2} x} \cdot \frac{s e c^{2} x}{s e c^{2} x} = \frac{c o s^{2} x \cdot s e c^{2} x - s i n^{2} x \cdot s e c^{2} x}{(1 - t a n^{2} x) s e c^{2} x} = \frac{c o s^{2} x \cdot \frac{1}{c o s^{2} x} - s i n^{2} x \cdot \frac{1}{c o s^{2} x}}{(1 - t a n^{2} x) s e c^{2} x} = \frac{1 - / t a n^{2} x}{(1 - / t a n^{2} x) s e c^{2} x} = \frac{1}{s e c^{2} x} = c o s^{2} x \end{matrix}

$\begin{align} \mathrm{\frac{cos^{2}x-sin^{2}x}{1-tan^{2}x}} & = \mathrm{\frac{cos^{2}x-sin^{2}x}{1-tan^{2}x}\cdot \frac{sec^{2}x}{sec^{2}x}} \\ & = \mathrm{\frac{cos^{2}x\cdot sec^{2}x-sin^{2}x\cdot sec^{2}x}{(1-tan^{2}x)\,sec^{2}x}} \\ & = \mathrm{\frac{cos^{2}x\cdot \frac{1}{cos^{2}x}-sin^{2}x\cdot \frac{1}{cos^{2}x}}{(1-tan^{2}x)\,sec^{2}x}} \\ & = \mathrm{\frac{1-{\color{red}\not}tan^{2}x}{(1-{\color{red}\not}tan^{2}x)\,sec^{2}x}} \\ & = \mathrm{\frac{1}{sec^{2}x}} \\ & = \mathrm{cos^{2}x} \end{align}$

Contoh 10
Buktikan :

\frac{c o s^{2} x + 2 c o s x + 1}{c o s x + 1} = \frac{1 + s e c x}{s e c x}

$\mathrm{\frac{cos^{2}x+2cos\,x+1}{cos\,x+1}=\frac{1+sec\,x}{sec\,x}}$
Jawab :

\begin{matrix} \frac{c o s^{2} x + 2 c o s x + 1}{c o s x + 1} & = \frac{(c o s x + 1)^{/ 2}}{(c o s / x + 1)} = c o s x + 1 = (c o s x + 1) \cdot \frac{s e c x}{s e c x} = \frac{c o s x s e c x + s e c x}{s e c x} = \frac{c o s x \cdot \frac{1}{c o s x} + s e c x}{s e c x} = \frac{1 + s e c x}{s e c x} \end{matrix}

$\begin{align} \mathrm{\frac{cos^{2}x+2cos\,x+1}{cos\,x+1}} & = \mathrm{\frac{(cos\,x+1)^{{\color{red}\not}2}}{(cos{\color{red}\not}\,x+1)}} \\ & = \mathrm{cos\,x+1} \\ & = \mathrm{(cos\,x+1)\cdot \frac{sec\,x}{sec\,x}} \\ & = \mathrm{\frac{cos\,x\,sec\,x+sec\,x}{sec\,x}} \\ & = \mathrm{\frac{cos\,x\cdot \frac{1}{cos\,x}+sec\,x}{sec\,x}} \\ & = \mathrm{\frac{1+sec\,x}{sec\,x}} \end{align}$

Contoh 11
Buktikan :

\sqrt{\frac{1 + c o s x}{1 - c o s x}} = c s c x + c o t x

$\mathrm{\sqrt{ \frac{1+cos\,x}{1-cos\,x}}=csc\,x+cot\,x}$
Jawab :

\begin{matrix} \sqrt{\frac{1 + c o s x}{1 - c o s x}} & = \sqrt{\frac{1 + c o s x}{1 - c o s x} \cdot \frac{1 + c o s x}{1 + c o s x}} = \sqrt{\frac{(1 + c o s x)^{2}}{1 - c o s^{2} x}} = \sqrt{\frac{(1 + c o s x)^{2}}{s i n^{2} x}} = \frac{1 + c o s x}{s i n x} = \frac{1}{s i n x} + \frac{c o s x}{s i n x} = c s c x + c o t x \end{matrix}

$\begin{align} \mathrm{\sqrt{ \frac{1+cos\,x}{1-cos\,x}}} & =\mathrm{\sqrt{ \frac{1+cos\,x}{1-cos\,x}\cdot \frac{1+cos\,x}{1+cos\,x}}} \\ & = \mathrm{\sqrt{ \frac{(1+cos\,x)^{2}}{1-cos^{2}x}}} \\ & = \mathrm{\sqrt{ \frac{(1+cos\,x)^{2}}{sin^{2}x}}} \\ & = \mathrm{\frac{1+cos\,x}{sin\,x}} \\ & = \mathrm{\frac{1}{sin\,x}+\frac{cos\,x}{sin\,x}} \\ & = \mathrm{csc\,x+cot\,x} \end{align}$

Contoh 12
Buktikan :

\frac{s i n^{3} x + c o s^{3} x}{s i n x + c o s x} = 1 - s i n x c o s x

$\mathrm{\frac{sin^{3}x+cos^{3}x}{sin\,x+cos\,x}=1-sin\,x\,cos\,x}$
Jawab :

\begin{matrix} \frac{s i n^{3} x + c o s^{3} x}{s i n x + c o s x} & = \frac{(s i n x / + c o s x) (s i n^{2} x + c o s^{2} x - s i n x c o s x)}{(s i n x / + c o s x)} = s i n^{2} x + c o s^{2} x - s i n x c o s x = 1 - s i n x c o s x \end{matrix}

$\begin{align} \mathrm{\frac{sin^{3}x+cos^{3}x}{sin\,x+cos\,x}} & = \mathrm{\frac{(sin\,x{\color{red}\not}+cos\,x)(sin^{2}x+cos^{2}x-sin\,x\,cos\,x)}{(sin\,x{\color{red}\not}+cos\,x)}} \\ & = \mathrm{sin^{2}x+cos^{2}x-sin\,x\,cos\,x} \\ & = \mathrm{1-sin\,x\,cos\,x} \\ \end{align}$

Perlu diketahui bahwa pembuktian identitas trigonometri dapat dilakukan dengan cara yang berbeda-beda, tergantung strategi yang digunakan.

Semakin banyak pembuktian yang dilakukan, kita akan menemukan strategi terbaik dalam menentukan konsep apa yang digunakan dan konsep mana yang didahulukan, sehingga pembukitan menjadi lebih efisien.

Latihan Soal Identitas Trigonometri

Latihan 1
Buktikan setiap identitas berikut!
a. tan x + cot x = sec x csc x
b. sec x − tan x sin x = cos x
c. cos x(csc x + tan x) = cot x + sin x

Latihan 2
Buktikan setiap identitas berikut!
a.

\frac{1 - s i n x}{c o s x} = s e c x - t a n x

$\mathrm{\frac{1-sin\,x}{cos\,x}=sec\,x-tan\,x}$
b.

\frac{s e c x - 1}{s e c x} = 1 - c o s x

$\mathrm{\frac{sec\,x-1}{sec\,x}=1-cos\,x}$
c.

\frac{s i n x - s e c x}{t a n x} = c o s x - c s c x

$\mathrm{\frac{sin\,x-sec\,x}{tan\,x}=cos\,x-csc\,x}$

Latihan 3
Buktikan setiap identitas berikut!
a.

s i n^{4} x - c o s^{4} x = s i n^{2} x - c o s^{2} x

$\mathrm{sin^{4}x-cos^{4}x=sin^{2}x-cos^{2}x}$
b.

s e c^{4} x - t a n^{4} x = s e c^{2} x + t a n^{2} x

$\mathrm{sec^{4}x-tan^{4}x=sec^{2}x+tan^{2}x}$
c.

c s c^{4} x - c o t^{4} x = c s c^{2} x + c o t^{2} x

$\mathrm{csc^{4}x-cot^{4}x=csc^{2}x+cot^{2}x}$

Latihan 4
Buktikan setiap identitas berikut!
a.

c o s^{2} x - s i n^{2} x = 2 c o s^{2} x - 1

$\mathrm{cos^{2}x-sin^{2}x=2cos^{2}x-1}$
b.

c o s^{2} x - s i n^{2} x = 1 - 2 s i n^{2} x

$\mathrm{cos^{2}x-sin^{2}x=1-2sin^{2}x}$
c.

1 - t a n^{2} x = 2 - s e c^{2} x

$\mathrm{1-tan^{2}x=2-sec^{2}x}$
d.

c o t^{2} x - 1 = c s c^{2} x - 2

$\mathrm{cot^{2}x-1=csc^{2}x-2}$

Latihan 5
Buktikan setiap identitas berikut!
a.

\frac{s i n^{2} x - c o s^{2} x}{s i n x + c o s x} = s i n x - c o s x

$\mathrm{\frac{sin^{2}x-cos^{2}x}{sin\,x+cos\,x}=sin\,x-cos\,x}$
b.

\frac{1 - t a n^{2} x}{1 + t a n x} = 1 - t a n x

$\mathrm{\frac{1-tan^{2}x}{1+tan\,x}=1-tan\,x}$
c.

\frac{c o s^{4} x - s i n^{4} x}{c o s x + s i n x} = c o s x - s i n x

$\mathrm{\frac{cos^{4}x-sin^{4}x}{cos\,x+sin\,x}=cos\,x-sin\,x}$

Latihan 6
Buktikan setiap identitas berikut!
a.

$\mathrm{\frac{cos\,x}{1-sin\,x}=\frac{1+sin\,x}{cos\,x}}$
b.

$\mathrm{\frac{cos\,x}{1-sin\,x}=\frac{1}{sec\,x-tan\,x}}$
c.

$\mathrm{\frac{1+sin\,x}{cos\,x}=\frac{1}{sec\,x-tan\,x}}$

Latihan 7
Buktikan setiap identitas berikut!
a. tan²x sin²x = tan²x − sin²x
b. sec²x + cot²x = tan²x + csc²x
c. cos⁴x − sin⁴θ + 1 = 2cos²θ
d. cos⁴x − cos²x = sin⁴x − sin²x
e.

$\mathrm{\frac{sec^{2}x-tan^{2}x}{csc\,x}=sin\,x}$
f.

$\mathrm{\frac{1+tan^{2}x}{1+cot^{2}x}=tan^{2}x}$
g.

$\mathrm{\frac{cos^{2}x-1}{cos\,x}=-tan\,x\,sin\,x}$
h.

$\mathrm{\frac{cos^{4}x-sin^{4}x}{sin^{2}x}=cot^{2}x-1}$
i.

$\mathrm{\frac{1-tan^{2}x}{1+tan^{2}x}=2cos^{2}x-1}$
j.

$\mathrm{1-\frac{sin^{2}x}{1+cos\,x}=cos\,x}$
k.

$\mathrm{\frac{cos\,x}{1-sin\,x}=tan\,x+sec\,x}$
l.

$\mathrm{\frac{1+tan\,x}{1+cot\,x}=tan\,x}$
m.

$\mathrm{\frac{1}{tan\,x}+tan\,x=sec\,x\,csc\,x}$
n.

$\mathrm{\frac{csc\,x-1}{1-sin\,x}=csc\,x}$
o.

$\mathrm{\frac{csc\,x}{cot\,x+tan\,x}=cos\,x}$
p.

$\mathrm{\frac{sec\,t\;sin\,t}{tan\,t+cot\,t}=sin^{2}t}$
q.

$\mathrm{\frac{1+cos\,x}{sin\,x}=\frac{sin\,x}{1-cos\,x}}$
r.

$\mathrm{\frac{sin\,x-cos\,x}{sin\,x}+\frac{cos\,x-sin\,x}{cos\,x}=2-sec\,x\,csc\,x}$
s.

$\mathrm{\frac{cos\,x}{1-sin\,x}+\frac{1-sin\,x}{cos\,x}=2sec\,x}$
t.

$\mathrm{\frac{1}{csc\,x-cot\,x}-csc\,x=csc\,x-\frac{1}{csc\,x+cot\,x}}$
u.

$\mathrm{\frac{cos\,x}{1-tan\,x}+\frac{sin\,x}{1-cot\,x}=sin\,x+cos\,x}$
v.

$\mathrm{\left ( csc\,x-cot\,x \right )^{2}=\frac{sec\,x-1}{sec\,x+1}}$
w.

$\mathrm{\frac{cos\,x\,cot\,x}{1-sin\,x}-1=csc\,x}$
x.

$\mathrm{\frac{sin^{6}x+cos^{6}x}{1-sin^{2}x}=sec^{2}x-3sin^{2}x}$
y.

$\mathrm{\frac{tan\,x+sec\,x-1}{tan\,x-sec\,x+1}=tan\,x+sec\,x}$
z.

$\mathrm{\sqrt{ \frac{csc\,x-1}{csc\,x+1}}=\frac{cos\,x}{sin\,x+1}}$