Question

How do you simplify $\dfrac{{\cos (x + 2\pi )}}{{\sin x}}?$

Answer 1

In the given problem first we try to reduce the numerator in the simplest form by eliminating $2\pi$ such that we are able to simplify it by using trigonometric formula.
Formula used:
i). $\cos (2\pi + x) = \cos x$
ii). $\dfrac{{\cos x}}{{\sin x}} = \cot x$

Complete step-by-step solution:
First writing the given expression as follows,
$= \dfrac{{\cos (x + 2\pi )}}{{\sin x}}$
Arranging numerator and writing it as following,
$= \dfrac{{\cos (2\pi + x)}}{{\sin x}}$
By using above given formula, we can write numerator as following,
$= \dfrac{{\cos x}}{{\sin x}}$
Again, by applying above given trigonometric formula, we get
$= \dfrac{{\cos x}}{{\sin x}}$
$= \cot x$

Note: We have to keep in mind that when we add $2\pi$ or its multiple to any trigonometric ratio does not change its value. For example,
$\sin (x + 2n\pi ) = \sin x$ where, $n = 0,1,2,3,4,.....,\infty$
$\cos (x + 2n\pi ) = \cos x$ where, $n = 0,1,2,3,4,.....,\infty$
$\tan (x + 2n\pi ) = \tan x$ where, $n = 0,1,2,3,4,.....,\infty$
$\cos ec(x + 2n\pi ) = \cos ecx$ where, $n = 0,1,2,3,4,.....,\infty$
$\sec (x + 2n\pi ) = \sec x$ where, $n = 0,1,2,3,4,.....,\infty$
$\cot (x + 2n\pi ) = \cot x$ where, $n = 0,1,2,3,4,.....,\infty$
For the above used directly trigonometric formula can be understood by using right-angle triangle and naming its sides as perpendicular, base, hypotenuse. In a right-angle triangle in which one angle is 90 degree and the side which exists in front of that 90-degree is named as hypotenuse which is also the longest side of the triangle. Rest of the two sides are named as base and perpendicular.
Reduction of $\dfrac{{\cos x}}{{\sin x}}$ to $\cot x$ as following,
$\dfrac{{\cos x}}{{\sin x}}$ can be written in the form of perpendicular, base and hypotenuse as following,
= $$$$\dfrac{{\cos x}}{{\sin x}}
$\cos x$ can be written as following,
= $$$$\cos x
= $$$$\dfrac{{base}}{{hypotenuse}}
Or $\dfrac{b}{h}$ where, (b=base, h=hypotenuse)
$\sin x$ can be written as following,
= $$$$\sin x
= $$$$\dfrac{{perpendicular}}{{hypotenuse}}
Or $\dfrac{p}{h}$ where, (p=perpendicular, h=hypotenuse)
Now, $\dfrac{{\cos x}}{{\sin x}}$ can be written as following,
$= \dfrac{{\cos x}}{{\sin x}}$
Replacing $\cos x$ by (b=base, h=hypotenuse) and writing it as,
$= \dfrac{{\dfrac{b}{h}}}{{\sin x}}$
Replacing $\sin x$ by (p=perpendicular, h=hypotenuse) and writing it as,
$= \dfrac{{\dfrac{b}{h}}}{{\dfrac{p}{h}}}$
Cancelling denominator and writing it as,
$= \dfrac{b}{p}$
Now we know that $\cot x$ is equal to $\dfrac{b}{p}$.
Therefore, $\dfrac{b}{p}$ can be also written as following,
$\Rightarrow \dfrac{b}{p} = \cot x$
Hence, $\dfrac{{\cos x}}{{\sin x}}$ can be written as $\cot x$.
Therefore, \dfrac{{\cos x}}{{\sin x}} = $$$$\cot x.