Tag: Kotangen

Identitas yang digunakan:
$$\boxed{{\displaystyle \begin{array}{rlrl}\tan x& ={\displaystyle \frac{\sin x}{\cos x}}& & (\text{Identitas Perbandingan})\\ \cot x& ={\displaystyle \frac{\cos x}{\sin x}}& & (\text{Identitas Perbandingan})\\ {\sin }^{2}x+{\cos }^{2}x& =1& & (\text{Identitas Pyt}\text{hagoras})\end{array}}}$$Pembuktian dari ruas kiri:
$\begin{array}{rl}& \tan A{\cos }^{4}A+\cot A{\sin }^{4}A\\ =& {\displaystyle \frac{\sin A}{\cos A}}\cdot {\cos }^{4}A+{\displaystyle \frac{\cos A}{\sin A}}\cdot {\sin }^{4}A\\ =& \sin A{\cos }^{3}A+\cos A{\sin }^{3}A\\ =& \sin A\cos A({\cos }^{2}A+{\sin }^{2}A)\\ =& \sin A\cos A(1)=\sin A\cos A\end{array}$
Jadi, terbukti bahwa $\tan A{\cos }^{4}A+\cot A{\sin }^{4}A$ $=\sin A\cos A.$

Identitas trigonometri yang digunakan:
$$\boxed{{\displaystyle \begin{array}{rlrl}\csc x& ={\displaystyle \frac{1}{\sin x}}& & (\text{Identitas Kebalikan})\\ \sec x& ={\displaystyle \frac{1}{\cos x}}& & (\text{Identitas Kebalikan})\\ {\sin }^{2}x+{\cos }^{2}x& =1& & (\text{Identitas Pyt}\text{hagoras})\end{array}}}$$Pembuktian dari ruas kiri:
$$\begin{array}{rl}& {\displaystyle \frac{{\sin }^{2}A}{{\cos }^{2}A}}-{\displaystyle \frac{{\cos }^{2}A}{{\sin }^{2}A}}\\ & ={\displaystyle \frac{{\sin }^{4}A-{\cos }^{4}A}{{\cos }^{2}A{\sin }^{2}A}}\\ & ={\displaystyle \frac{({\sin }^{2}A-{\cos }^{2}A)({\sin }^{2}A+{\cos }^{2}A)}{{\cos }^{2}A{\sin }^{2}A}}\\ & ={\displaystyle \frac{({\sin }^{2}A-{\cos }^{2}A)(1)}{{\cos }^{2}A{\sin }^{2}A}}\\ & ={\displaystyle \frac{\cancel{{\sin }^{2}A}}{{\cos }^{2}A\cancel{{\sin }^{2}A}}}-{\displaystyle \frac{\bcancel{{\cos }^{2}A}}{\bcancel{{\cos }^{2}A}{\sin }^{2}A}}\\ & ={\displaystyle \frac{1}{{\cos }^{2}A}}-{\displaystyle \frac{1}{{\sin }^{2}A}}\\ & ={\sec }^{2}A-{\csc }^{2}A\end{array}$$Jadi, terbukti bahwa ${\displaystyle \frac{{\sin }^{2}A}{{\cos }^{2}A}}-{\displaystyle \frac{{\cos }^{2}A}{{\sin }^{2}A}}={\sec }^{2}A-{\csc }^{2}A.$

Identitas trigonometri yang digunakan:
$$\boxed{{\displaystyle \begin{array}{rlrl}\tan x& ={\displaystyle \frac{\sin x}{\cos x}}& & (\text{Identitas Perbandingan})\\ \sec x& ={\displaystyle \frac{1}{\cos x}}& & (\text{Identitas Kebalikan})\\ {\sec }^{2}x& =1+{\tan }^{2}x& & (\text{Identitas Pyt}\text{hagoras})\end{array}}}$$Pembuktian dari ruas kiri:
$$\begin{array}{rl}& {\displaystyle \frac{{\sin }^{2}A-{\sin }^{2}B}{{\cos }^{2}A{\cos }^{2}B}}\\ & ={\displaystyle \frac{{\sin }^{2}A}{{\cos }^{2}A{\cos }^{2}B}}-{\displaystyle \frac{{\sin }^{2}B}{{\cos }^{2}A{\cos }^{2}B}}\\ & ={\displaystyle \frac{{\sin }^{2}A}{{\cos }^{2}A}}\cdot {\displaystyle \frac{1}{{\cos }^{2}B}}-{\displaystyle \frac{{\sin }^{2}B}{{\cos }^{2}B}}\cdot {\displaystyle \frac{1}{{\cos }^{2}A}}\\ & ={\tan }^{2}A({\sec }^{2}B)-{\tan }^{2}B({\sec }^{2}A)\\ & ={\tan }^{2}A(1+{\tan }^{2}B)-{\tan }^{2}B(1+{\tan }^{2}A)\\ & ={\tan }^{2}A+\cancel{{\tan }^{2}A{\tan }^{2}B}-{\tan }^{2}B-\cancel{{\tan }^{2}A{\tan }^{2}B}\\ & ={\tan }^{2}A-{\tan }^{2}B\end{array}$$Jadi, terbukti bahwa ${\displaystyle \frac{{\sin }^{2}A-{\sin }^{2}B}{{\cos }^{2}A{\cos }^{2}B}}={\tan }^{2}A-{\tan }^{2}B.$

Identitas trigonometri yang digunakan:
$$\boxed{{\displaystyle \begin{array}{rlrl}1+{\cot }^{2}x& ={\csc }^{2}x& & (\text{Identitas Pyt}\text{hagoras})\\ \csc x& ={\displaystyle \frac{1}{\sin x}}& & (\text{Identitas Kebalikan})\\ \cot x& ={\displaystyle \frac{\cos x}{\sin x}}& & (\text{Identitas Perbandingan})\end{array}}}$$Pembuktian dari ruas kiri:
$\begin{array}{rl}& {\cos }^{2}A+{\cot }^{2}A{\cos }^{2}A\\ & ={\cos }^{2}A(1+{\cot }^{2}A)\\ & ={\cos }^{2}A({\csc }^{2}A)\\ & ={\cos }^{2}A\cdot {\displaystyle \frac{1}{{\sin }^{2}A}}\\ & ={\cot }^{2}A\end{array}$
Jadi, terbukti bahwa ${\cos }^{2}A+{\cot }^{2}A{\cos }^{2}A={\cot }^{2}A.$

Identitas trigonometri yang digunakan:
$$\boxed{{\displaystyle \begin{array}{rlrl}\sec x& ={\displaystyle \frac{1}{\cos x}}& & (\text{Identitas Kebalikan})\\ \tan x& ={\displaystyle \frac{\sin x}{\cos x}}& & (\text{Identitas Perbandingan})\end{array}}}$$Pembuktian dari ruas kiri: 
$\begin{array}{rl}{\displaystyle \frac{1+\sec A}{\tan A+\sin A}}& ={\displaystyle \frac{1+\frac{1}{\cos A}}{\frac{\sin A}{\cos A}+\sin A}}\\ & ={\displaystyle \frac{\cos A+1}{\sin A+\sin A\cos A}}\\ & ={\displaystyle \frac{\cancel{\cos A+1}}{\sin A(\cancel{1+\cos A})}}\\ & ={\displaystyle \frac{1}{\sin A}}=\csc A\end{array}$
Jadi, terbukti bahwa ${\displaystyle \frac{1+\sec A}{\tan A+\sin A}}=\csc A.$

Identitas trigonometri yang digunakan:
$$\boxed{{\displaystyle \begin{array}{rlrl}{\sin }^{2}x+{\cos }^{2}x& =1& & (\text{Identitas Pyt}\text{hagoras})\\ \sec x& ={\displaystyle \frac{1}{\cos x}}& & (\text{Identitas Kebalikan})\\ \tan x& ={\displaystyle \frac{\sin x}{\cos x}}& & (\text{Identitas Perbandingan})\end{array}}}$$Pembuktian dari ruas kiri:
$$\begin{array}{rl}{\displaystyle \frac{1+\sin x}{1-\sin x}}& ={\displaystyle \frac{1+\sin x}{1-\sin x}}\times {\displaystyle \frac{1+\sin x}{1+\sin x}}\\ & ={\displaystyle \frac{1+2\sin x+{\sin }^{2}x}{1-{\sin }^{2}x}}\\ & ={\displaystyle \frac{1+2\sin x+{\sin }^{2}x}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{1}{{\cos }^{2}x}}+{\displaystyle \frac{2\sin x}{{\cos }^{2}x}}+{\displaystyle \frac{{\sin }^{2}x}{{\cos }^{2}x}}\\ & ={\sec }^{2}x+2\sec x\tan x+{\tan }^{2}x\\ & =(\sec x+\tan x{)}^2\end{array}$$Jadi, terbukti bahwa ${\displaystyle \frac{1+\sin x}{1-\sin x}}=(\sec x+\tan x{)}^2.$

Gunakan identitas trigonometri berikut. 
$\boxed{{\displaystyle \begin{array}{rl}\sin 2x& =2\sin x\cos x\\ {\tan }^{2}x+1& ={\sec }^{2}x\end{array}}}$
Pembuktian dari ruas kanan:
$\begin{array}{rl}& {\displaystyle \frac{1}{2}}\sin 2A(1+{\tan }^{2}A)\\ & ={\displaystyle \frac{1}{2}}(2\sin A\cos A)(1+{\tan }^{2}A)\\ & =(\sin A\cos A)(1+{\tan }^{2}A)\\ & =(\sin A\cos A)({\sec }^{2}A)\\ & ={\displaystyle \frac{\sin A\cancel{\cos A}}{{\cancel{{\cos }^{2}A}}^{\cos A}}}\\ & ={\displaystyle \frac{\sin A}{\cos A}}=\tan A\end{array}$
Jadi, terbukti bahwa $\tan A={\displaystyle \frac{1}{2}}\sin 2A(1+{\tan }^{2}A).$

Gunakan identitas trigonometri berikut. 
$\boxed{{\displaystyle \begin{array}{rl}1+{\tan }^{2}x& ={\sec }^{2}x\\ \cos 2x& =1-2{\sin }^{2}x\\ \sec x& ={\displaystyle \frac{1}{\cos x}}\end{array}}}$
Pembuktian dari ruas kanan:
$\begin{array}{rl}& 1–\cos 2A(1+{\tan }^{2}A)\\ & =1-\cos 2A({\sec }^{2}A)\\ & =1-{\displaystyle \frac{1-2{\sin }^{2}A}{{\cos }^{2}A}}\\ & =1-{\displaystyle \frac{1}{{\cos }^{2}A}}+{\displaystyle \frac{2{\sin }^{2}A}{{\cos }^{2}A}}\\ & =1-{\sec }^{2}A+2{\tan }^{2}A\\ & =1-(1+{\tan }^{2}A)+2{\tan }^{2}A\\ & ={\tan }^{2}A\end{array}$
Jadi, terbukti bahwa ${\tan }^{2}A=1-\cos 2A(1+{\tan }^{2}A).$

Gunakan identitas trigonometri berikut. 
$\boxed{{\displaystyle {\tan }^{2}x+1={\sec }^{2}x}}$
Pembuktian dari ruas kanan:
$$\begin{array}{rl}& {\tan }^{4}A\cdot {\sec }^{2}A-{\tan }^{2}A\cdot {\sec }^{2}A+{\sec }^{2}A-1\\ & ={\tan }^{4}A\cdot {\sec }^{2}A-{\tan }^{2}A\cdot {\sec }^{2}A+{\tan }^{2}A\\ & ={\tan }^{2}A({\tan }^{2}A{\sec }^{2}A-{\sec }^{2}A+1)\\ & ={\tan }^{2}A(({\sec }^{2}A)({\tan }^{2}A-1)+1)\\ & ={\tan }^{2}A(({\tan }^{2}A+1)({\tan }^{2}A-1)+1)\\ & ={\tan }^{2}A({\tan }^{4}A-1+1)\\ & ={\tan }^{6}A\end{array}$$Jadi, terbukti bahwa $${\tan }^{6}A={\tan }^{4}A\cdot {\sec }^{2}A-{\tan }^{2}A\cdot {\sec }^{2}A+{\sec }^{2}A-1.$$

Gunakan identitas trigonometri berikut. 
$$\boxed{{\displaystyle \begin{array}{rl}\sin x-\sin y& =2\cos {\displaystyle \frac{1}{2}}(x+y)\sin {\displaystyle \frac{1}{2}}(x-y)\\ \cos x+\cos y& =2\cos {\displaystyle \frac{1}{2}}(x+y)\cos {\displaystyle \frac{1}{2}}(x-y)\\ \tan x& ={\displaystyle \frac{\sin x}{\cos x}}\end{array}}}$$Pembuktian dari ruas kanan:
$$\begin{array}{rl}{\displaystyle \frac{\sin 2A-\sin 2B}{\cos 2A+\cos 2B}}& ={\displaystyle \frac{2\cos {\displaystyle \frac{1}{2}}(2A+2B)\sin {\displaystyle \frac{1}{2}}(2A-2B)}{2\cos {\displaystyle \frac{1}{2}}(2A+2B)\cos {\displaystyle \frac{1}{2}}(2A-2B)}}\\ & ={\displaystyle \frac{\cancel{2\cos (A+B)}\sin (A-B)}{\cancel{2\cos (A+B)}\cos (A-B)}}\\ & ={\displaystyle \frac{\sin (A-B)}{\cos (A-B)}}=\tan (A-B)\end{array}$$Jadi, terbukti bahwa $\tan (A-B)={\displaystyle \frac{\sin 2A-\sin 2B}{\cos 2A+\cos 2B}}.$

Gunakan identitas trigonometri berikut. 
$$\boxed{{\displaystyle \begin{array}{rl}\sin x-\sin y& =2\cos {\displaystyle \frac{1}{2}}(x+y)\sin {\displaystyle \frac{1}{2}}(x-y)\\ \cos x-\cos y& =-2\sin {\displaystyle \frac{1}{2}}(x+y)\sin {\displaystyle \frac{1}{2}}(x-y)\\ \sin (-x)& =-\sin x\\ \cot x& ={\displaystyle \frac{\cos x}{\sin x}}\end{array}}}$$Pembuktian dari ruas kanan:
$$\begin{array}{rl}{\displaystyle \frac{\sin A-\sin 3A}{\cos 3A-\cos A}}& ={\displaystyle \frac{2\cos {\displaystyle \frac{1}{2}}(A+3A)\sin {\displaystyle \frac{1}{2}}(A-3A)}{-2\sin {\displaystyle \frac{1}{2}}(3A+A)\sin {\displaystyle \frac{1}{2}}(3A-A)}}\\ & ={\displaystyle \frac{2\cos 2A\sin (-A)}{-2\sin 2A\sin A}}\\ & ={\displaystyle \frac{\cancel{-2\sin A}\cos 2A}{\cancel{-2\sin A}\sin 2A}}\\ & ={\displaystyle \frac{\cos 2A}{\sin 2A}}=\cot 2A\end{array}$$Jadi, terbukti bahwa $\cot 2A={\displaystyle \frac{\sin A-\sin 3A}{\cos 3A-\cos A}}.$

Identitas trigonometri yang digunakan:
$$\boxed{{\displaystyle \begin{array}{rlrl}\cos 2x& ={\cos }^{2}x–{\sin }^{2}x& & (\text{Identitas Su}\text{dut Ganda})\\ \sin 2x& =2\sin x\cos x& & (\text{Identitas Su}\text{dut Ganda})\\ {\sin }^{2}x+{\cos }^{2}x& =1& & (\text{Identitas Pyt}\text{hagoras})\\ \tan x& ={\displaystyle \frac{\sin x}{\cos x}}& & (\text{Identitas Perbandingan})\end{array}}}$$Pembuktian dari ruas kiri:
$$\begin{array}{rl}{\displaystyle \frac{1-\cos 2A+\sin A}{\sin 2A+\cos A}}& ={\displaystyle \frac{1-({\cos }^{2}A-{\sin }^{2}A)+\sin A}{(2\sin A\cos A)+\cos A}}\\ & ={\displaystyle \frac{(1-{\cos }^{2}A)+{\sin }^{2}A+\sin A}{\cos A(2\sin A+1)}}\\ & ={\displaystyle \frac{{\sin }^{2}A+{\sin }^{2}A+\sin A}{\cos A(2\sin A+1)}}\\ & ={\displaystyle \frac{\sin A\cancel{(2\sin A+1)}}{\cos A\cancel{(2\sin A+1)}}}\\ & ={\displaystyle \frac{\sin A}{\cos A}}=\tan A\end{array}$$Jadi, terbukti bahwa ${\displaystyle \frac{1-\cos 2A+\sin A}{\sin 2A+\cos A}}=\tan A.$

Identitas trigonometri yang digunakan:
$$\boxed{{\displaystyle \begin{array}{rlrl}\cos 2x& ={\cos }^{2}x-{\sin }^{2}x& & (\text{Identitas Su}\text{dut Ganda})\\ \sin 2x& =2\sin x\cos x& & (\text{Identitas Su}\text{dut Ganda})\\ {\sin }^{2}x+{\cos }^{2}x& =1& & (\text{Identitas Pyt}\text{hagoras})\\ \csc x& ={\displaystyle \frac{1}{\sin x}}& & (\text{Identitas Kebalikan})\\ \cot x& ={\displaystyle \frac{\cos x}{\sin x}}& & (\text{Identitas Perbandingan})\end{array}}}$$Pembuktian dari ruas kiri:
$$\begin{array}{rl}\csc 2A+\cot 2A& ={\displaystyle \frac{1}{\sin 2A}}+{\displaystyle \frac{\cos 2A}{\sin 2A}}\\ & ={\displaystyle \frac{1+({\cos }^{2}A-{\sin }^{2}A)}{2\sin A\cos A}}\\ & ={\displaystyle \frac{(1-{\sin }^{2}A)+{\cos }^{2}A}{2\sin A\cos A}}\\ & ={\displaystyle \frac{2{\cos }^{2}A}{2\sin A\cos A}}\\ & ={\displaystyle \frac{\cos A}{\sin A}}=\cot A\end{array}$$Jadi, terbukti bahwa $\csc 2A+\cot 2A=\cot A.$

Ingat rumus jumlah dan selisih sudut:
$\boxed{{\displaystyle \begin{array}{rl}\sin (x\pm y)& =\sin x\cos y\pm \cos x\sin y\\ \cos (x\pm y)& =\cos x\cos y\mp \sin x\sin y\end{array}}}$
Identitas trigonometri yang digunakan:
$\cot x={\displaystyle \frac{\cos x}{\sin x}}$
Pembuktian dari ruas kiri:
$$\begin{array}{rl}& {\displaystyle \frac{2\sin (A-B)}{\cos (A-B)-\cos (A+B)}}\\ & ={\displaystyle \frac{2(\sin A\cos B-\sin B\cos A)}{(\cancel{\cos A\cos B}+\sin A\sin B)-(\cancel{\cos A\cos B}-\sin A\sin B)}}\\ & ={\displaystyle \frac{\cancel{2}(\sin A\cos B-\sin B\cos A)}{\cancel{2}\sin A\sin B}}\\ & ={\displaystyle \frac{\sin A\cos B-\sin B\cos A}{\sin A\sin B}}\\ & ={\displaystyle \frac{\cancel{\sin A}\cos B}{\cancel{\sin A}\sin B}}-{\displaystyle \frac{\cancel{\sin B}\cos A}{\sin A\cancel{\sin B}}}\\ & ={\displaystyle \frac{\cos B}{\sin B}}-{\displaystyle \frac{\cos A}{\sin A}}\\ & =\cot B-\cot A\end{array}$$Jadi, terbukti bahwa $${\displaystyle \frac{2\sin (A-B)}{\cos (A-B)-\cos (A+B)}}=\cot B-\cot A.$$

Identitas trigonometri yang digunakan:
$$\boxed{{\displaystyle \begin{array}{rlrl}\tan x& ={\displaystyle \frac{\sin x}{\cos x}}& & (\text{Identitas Perbandingan})\\ \cot x& ={\displaystyle \frac{\cos x}{\sin x}}& & (\text{Identitas Perbandingan})\\ {\sin }^{2}x+{\cos }^{2}x& =1& & (\text{Identitas Pyt}\text{hagoras})\end{array}}}$$Pembuktian dari ruas kiri:
$$\begin{array}{rl}\tan A-\cot A& ={\displaystyle \frac{\sin A}{\cos A}}-{\displaystyle \frac{\cos A}{\sin A}}\\ & ={\displaystyle \frac{{\sin }^{2}A-{\cos }^{2}A}{\sin A\cos A}}\\ & ={\displaystyle \frac{(1-{\cos }^{2}A)-{\cos }^{2}A}{\sin A\cos A}}\\ & ={\displaystyle \frac{1-2{\cos }^{2}A}{\sin A\cos A}}\end{array}$$Jadi, terbukti bahwa $\tan A-\cot A={\displaystyle \frac{1-2{\cos }^{2}A}{\sin A\cos A}}.$

Identitas trigonometri yang digunakan:
$$\boxed{{\displaystyle \begin{array}{rlrl}\cos 2x& ={\cos }^{2}x-{\sin }^{2}x& & (\text{Identitas Su}\text{dut Ganda})\\ \sec x& ={\displaystyle \frac{1}{\cos x}}& & (\text{Identitas Kebalikan})\\ \tan x& ={\displaystyle \frac{\sin x}{\cos x}}& & (\text{Identitas Perbandingan})\end{array}}}$$Pembuktian dari ruas kanan:
$$\begin{array}{rl}\cos 2A{\sec }^{2}A& =({\cos }^{2}A-{\sin }^{2}A)\cdot {\displaystyle \frac{1}{{\cos }^{2}A}}\\ & =1-{\displaystyle \frac{{\sin }^{2}A}{{\cos }^{2}A}}\\ & =1-{\tan }^{2}A\end{array}$$Jadi, terbukti bahwa $1-{\tan }^{2}A=\cos 2A{\sec }^{2}A.$