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Question: Find the value of \(\cot \left( { - \dfrac{{15\pi }}{4}} \right)\)...

Find the value of cot(15π4)\cot \left( { - \dfrac{{15\pi }}{4}} \right)

Explanation

Solution

Note that, cot(x)=cotx{\text{cot}}( - x) = - \cot x
Again, we know that the function y= cotx has a period ofπ\pi or 180.180^\circ ., i.e. the value of cotx repeats after an interval of π\pi or 180°.
Therefore write 15π4\dfrac{{15\pi }}{4} as 4ππ44\pi - \dfrac{\pi }{4} and proceed.

Complete step-by-step answer:
We know that the function y= cotx has a period of π\pi or 180.180^\circ ., i.e. the value of cotx repeats after an interval of π\pi or 180.180^\circ ..
Therefore,
cot(15π4)\cot \left( { - \dfrac{{15\pi }}{4}} \right)
Since, cot(x)=cotx{\text{cot}}( - x) = - \cot x,
=cot(15π4)= - \cot \left( {\dfrac{{15\pi }}{4}} \right)
On expanding the numerator we get,
=cot(16ππ4)= - \cot \left( {\dfrac{{16\pi - \pi }}{4}} \right)
On simplification we get,
=cot(4ππ4)= - \cot \left( {4\pi - \dfrac{\pi }{4}} \right)
Since, (15π4) \left( {\dfrac{{15\pi }}{4}} \right){\text{ }} lies in the fourth quadrant, therefore cot(15π4) \cot \left( {\dfrac{{15\pi }}{4}} \right){\text{ }} will be negative
=(cotπ3)= - \left( { - \cot \dfrac{\pi }{3}} \right)
=cotπ3= \cot \dfrac{\pi }{3}
As, cotπ3=13\cot \dfrac{\pi }{3} = \dfrac{1}{{\sqrt 3 }}
=13= \dfrac{1}{{\sqrt 3 }}
Therefore the value of cot(15π4)\cot \left( { - \dfrac{{15\pi }}{4}} \right)is 13\dfrac{1}{{\sqrt 3 }}.

Note: Note the following important formulae:
1.cosx=1secx\cos x = \dfrac{1}{{\sec x}} , sinx=1cosecx\sin x = \dfrac{1}{{\cos ecx}} , tanx=1cotx\tan x = \dfrac{1}{{\cot x}}
2.sin2x+cos2x=1{\sin ^2}x + {\cos ^2}x = 1
3.sec2xtan2x=1{\sec ^2}x - {\tan ^2}x = 1
4.cosec2xcot2x=1{\operatorname{cosec} ^2}x - {\cot ^2}x = 1
5.sin(x)=sinx\sin ( - x) = - \sin x
6.cos(x)=cosx\cos ( - x) = \cos x
7.tan(x)=tanx\tan ( - x) = - \tan x
8.sin(2nπ±x)=sinx , period 2π or 360\sin \left( {2n\pi \pm x} \right) = \sin x{\text{ , period 2}}\pi {\text{ or 3}}{60^ \circ }
9.cos(2nπ±x)=cosx , period 2π or 360\cos \left( {2n\pi \pm x} \right) = \cos x{\text{ , period 2}}\pi {\text{ or 3}}{60^ \circ }
10.tan(nπ±x)=tanx , period π or 180\tan \left( {n\pi \pm x} \right) = \tan x{\text{ , period }}\pi {\text{ or 18}}{0^ \circ }
Sign convention:

Also, the trigonometric ratios of the standard angles are given by

| 00^\circ | 3030^\circ | 4545^\circ | 6060^\circ | 9090^\circ
---|---|---|---|---|---
Sinx\operatorname{Sin} x| 0| 12\dfrac{1}{2} | 12\dfrac{1}{{\sqrt 2 }} | 32\dfrac{{\sqrt 3 }}{2} | 1
Cosx\operatorname{Cos} x| 1| 32\dfrac{{\sqrt 3 }}{2}| 12\dfrac{1}{{\sqrt 2 }}| 12\dfrac{1}{2}| 0
Tanx\operatorname{Tan} x| 0| 13\dfrac{1}{{\sqrt 3 }} | 1| 3\sqrt 3 | Undefined
CotxCotx| undefined| 3\sqrt 3 | 1| 13\dfrac{1}{{\sqrt 3 }}| 0
cosecx\cos ecx| undefined| 2| 2\sqrt 2 | 23\dfrac{2}{{\sqrt 3 }}| 1
Secx\operatorname{Sec} x| 1| 23\dfrac{2}{{\sqrt 3 }}| 2\sqrt 2 | 2| Undefined