
Prove that $\cot \dfrac{\pi }{4} = \sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 $
Answer
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Hint: Here in this question we must know the following properties and identities of trigonometric functions.
They are mentioned below: -
$\cot A = \dfrac{{(1 + \cos 2A)}}{{\sin 2A}}$
Conversion of radian into degree is done by multiplying that angle by $\dfrac{{180}}{\pi }$
$\cos (A - B) = \cos A\cos B + \sin A\sin B$
Complete step-by-step answer:
First of all we will convert the angle which is given in radian into degree so that the calculation part can become a little less confusing.
$\cot \dfrac{\pi }{4} = \cot (\dfrac{\pi }{4} \times \dfrac{{180}}{\pi })$
$ \Rightarrow \cot \dfrac{{180}}{4} = \cot 7.5 = \cot \dfrac{{15}}{2}$
Now we will apply $\cot A = \dfrac{{(1 + \cos 2A)}}{{\sin 2A}}$ identity so that fractional part can be eliminated. $\cot \dfrac{{15}}{2} = \dfrac{{(1 + \cos 2 \times \dfrac{{15}}{2})}}{{\sin 2 \times \dfrac{{15}}{2}}}$
$ \Rightarrow \dfrac{{(1 + \cos 15)}}{{\sin 15}}$ (Cancelling 2 from numerator and denominator)
For finding the value of $\cos 15$ we know that $\cos 15 = \cos (45 - 30)$ so applying identity $\cos (A - B) = \cos A\cos B + \sin A\sin B$
$ \Rightarrow \cos (45 - 30) = \cos 45\cos 30 + \sin 45\sin 30$ (Here A=45 and B=30)
$ \Rightarrow \dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2} + \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2}$ (Putting values of trigonometric functions)
$ \Rightarrow \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$ (Taking L.C.M)
$ \Rightarrow \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }}$ (Rationalising by multiplying and dividing $\sqrt 2 $)
$ \Rightarrow \dfrac{{\sqrt 2 (\sqrt 3 + 1)}}{{2 \times 2}} = \dfrac{{\sqrt 2 (\sqrt 3 + 1)}}{4}$
$\therefore \cos 15 = \dfrac{{\sqrt 2 (\sqrt 3 + 1)}}{4}$
Now putting value of $\cos 15$in $\dfrac{{(1 + \cos 15)}}{{\sin 15}}$
$ \Rightarrow \dfrac{{(1 + \dfrac{{\sqrt 2 (\sqrt 3 + 1)}}{4})}}{{\sin 15}}$
For $\sin 15$ we will use $\cos 15$ because $\cos \theta = \dfrac{b}{h}$and \[\sin \theta = \dfrac{p}{h}\]where p=perpendicular, b=base and h=hypotenuse.
$\therefore \cos 15 = \dfrac{{\sqrt 2 (\sqrt 3 + 1)}}{4} = \dfrac{b}{h}$0
\[h = \sqrt {{p^2} + {b^2}} \](By Pythagoras theorem)
\[ \Rightarrow p = \sqrt {{h^2} - {b^2}} \]
\[ \Rightarrow p = \sqrt {{4^2} - {{[\sqrt 2 (\sqrt 3 + 1])}^2}} \]
Further simplifying we will get
\[ \Rightarrow p = \sqrt {16 - [2{{(3 + 1 + 2\sqrt 3 ]}^2})} \] (Expanding\[{(a + b)^2} = {a^2} + {b^2} + 2ab\])
\[ \Rightarrow p = \sqrt {16 - [2(4 + 2\sqrt 3 ])} \]
\[ \Rightarrow p = \sqrt {16 - [8 + 4\sqrt 3 ]} \] (Multiplying 2 inside)
\[ \Rightarrow p = \sqrt {16 - 8 - 4\sqrt 3 } \]
\[ \Rightarrow p = \sqrt {8 - 4\sqrt 3 } \] (Taking 4 outside the root)
\[\therefore p = 2\sqrt {2 - \sqrt 3 } \]
\[ \Rightarrow p = \dfrac{{2\sqrt {2 - \sqrt 3 } }}{4}\]Or $\dfrac{{\sqrt 2 (\sqrt 3 - 1)}}{4}$ because square root of \[\sqrt {2 - \sqrt 3 } \] is $\dfrac{{(\sqrt 3 - 1)}}{{\sqrt 2 }}$
You can directly write sine value from cosine value without doing so much of calculation by just making a sign opposite.
Now putting values of $\sin 15$ in $\dfrac{{(1 + \dfrac{{\sqrt 2 (\sqrt 3 + 1)}}{4})}}{{\sin 15}}$ we will get
$ \Rightarrow \dfrac{{(1 + \dfrac{{\sqrt 2 (\sqrt 3 + 1)}}{4})}}{{\dfrac{{\sqrt 2 (\sqrt 3 - 1)}}{4}}}$
$ \Rightarrow \dfrac{{(\dfrac{{4 + \sqrt 2 (\sqrt 3 + 1)}}{4})}}{{\dfrac{{\sqrt 2 (\sqrt 3 - 1)}}{4}}}$ (Cancelling 4 from both denominator terms)\[ \Rightarrow \dfrac{{4 + \sqrt 2 (\sqrt 3 + 1)}}{{\sqrt 2 (\sqrt 3 - 1)}} = \dfrac{{4 + \sqrt 6 + \sqrt 2 }}{{\sqrt 6 - \sqrt 2 }}\]
Now rationalising the denominator irrational term
\[ \Rightarrow \dfrac{{4 + \sqrt 6 + \sqrt 2 }}{{\sqrt 6 - \sqrt 2 }} \times \dfrac{{\sqrt 6 + \sqrt 2 }}{{\sqrt 6 + \sqrt 2 }}\]\[ \Rightarrow \dfrac{{(4 + \sqrt 6 + \sqrt 2 )(\sqrt 6 + \sqrt 2 )}}{{6 - 2}} = \dfrac{{4\sqrt 6 + 4\sqrt 2 + 6 + \sqrt 6 \times \sqrt 2 + \sqrt 6 \times \sqrt 2 + 2}}{4}\]\[ \Rightarrow \dfrac{{4\sqrt 6 + 4\sqrt 2 + 8 + 2(\sqrt {12} )}}{4} = \dfrac{{4\sqrt 6 + 4\sqrt 2 + 8 + 2(2\sqrt 3 )}}{4}\]\[ \Rightarrow \dfrac{{4\sqrt 6 + 4\sqrt 2 + 8 + 4\sqrt 3 }}{4} = \sqrt 6 + \sqrt 2 + 2 + \sqrt 3 \]
\[\therefore \sqrt 6 + \sqrt 2 + \sqrt 4 + \sqrt 3 \] (2 can be written as $\sqrt 4 $)
Therefore $\cot \dfrac{\pi }{4} = \sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 $
Hence proved.
Note: Students should know basic values of trigonometric functions which can boost their calculation part some of the values are as follows: -
$\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2},\sin {30^ \circ } = \dfrac{1}{2}$
$\cos {60^ \circ } = \dfrac{1}{2},\sin {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$
$\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }},\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$
\[\cos {90^ \circ } = 0,\sin {90^ \circ } = 1\]
$\cos {180^ \circ } = 1,\sin {180^ \circ } = 0$
Also while doing rationalisation some students make mistakes by multiplying with the same term but that is not correct. The same term with opposite signs should be multiplied and divided.
They are mentioned below: -
$\cot A = \dfrac{{(1 + \cos 2A)}}{{\sin 2A}}$
Conversion of radian into degree is done by multiplying that angle by $\dfrac{{180}}{\pi }$
$\cos (A - B) = \cos A\cos B + \sin A\sin B$
Complete step-by-step answer:
First of all we will convert the angle which is given in radian into degree so that the calculation part can become a little less confusing.
$\cot \dfrac{\pi }{4} = \cot (\dfrac{\pi }{4} \times \dfrac{{180}}{\pi })$
$ \Rightarrow \cot \dfrac{{180}}{4} = \cot 7.5 = \cot \dfrac{{15}}{2}$
Now we will apply $\cot A = \dfrac{{(1 + \cos 2A)}}{{\sin 2A}}$ identity so that fractional part can be eliminated. $\cot \dfrac{{15}}{2} = \dfrac{{(1 + \cos 2 \times \dfrac{{15}}{2})}}{{\sin 2 \times \dfrac{{15}}{2}}}$
$ \Rightarrow \dfrac{{(1 + \cos 15)}}{{\sin 15}}$ (Cancelling 2 from numerator and denominator)
For finding the value of $\cos 15$ we know that $\cos 15 = \cos (45 - 30)$ so applying identity $\cos (A - B) = \cos A\cos B + \sin A\sin B$
$ \Rightarrow \cos (45 - 30) = \cos 45\cos 30 + \sin 45\sin 30$ (Here A=45 and B=30)
$ \Rightarrow \dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2} + \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2}$ (Putting values of trigonometric functions)
$ \Rightarrow \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$ (Taking L.C.M)
$ \Rightarrow \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }}$ (Rationalising by multiplying and dividing $\sqrt 2 $)
$ \Rightarrow \dfrac{{\sqrt 2 (\sqrt 3 + 1)}}{{2 \times 2}} = \dfrac{{\sqrt 2 (\sqrt 3 + 1)}}{4}$
$\therefore \cos 15 = \dfrac{{\sqrt 2 (\sqrt 3 + 1)}}{4}$
Now putting value of $\cos 15$in $\dfrac{{(1 + \cos 15)}}{{\sin 15}}$
$ \Rightarrow \dfrac{{(1 + \dfrac{{\sqrt 2 (\sqrt 3 + 1)}}{4})}}{{\sin 15}}$
For $\sin 15$ we will use $\cos 15$ because $\cos \theta = \dfrac{b}{h}$and \[\sin \theta = \dfrac{p}{h}\]where p=perpendicular, b=base and h=hypotenuse.
$\therefore \cos 15 = \dfrac{{\sqrt 2 (\sqrt 3 + 1)}}{4} = \dfrac{b}{h}$0
\[h = \sqrt {{p^2} + {b^2}} \](By Pythagoras theorem)
\[ \Rightarrow p = \sqrt {{h^2} - {b^2}} \]
\[ \Rightarrow p = \sqrt {{4^2} - {{[\sqrt 2 (\sqrt 3 + 1])}^2}} \]
Further simplifying we will get
\[ \Rightarrow p = \sqrt {16 - [2{{(3 + 1 + 2\sqrt 3 ]}^2})} \] (Expanding\[{(a + b)^2} = {a^2} + {b^2} + 2ab\])
\[ \Rightarrow p = \sqrt {16 - [2(4 + 2\sqrt 3 ])} \]
\[ \Rightarrow p = \sqrt {16 - [8 + 4\sqrt 3 ]} \] (Multiplying 2 inside)
\[ \Rightarrow p = \sqrt {16 - 8 - 4\sqrt 3 } \]
\[ \Rightarrow p = \sqrt {8 - 4\sqrt 3 } \] (Taking 4 outside the root)
\[\therefore p = 2\sqrt {2 - \sqrt 3 } \]
\[ \Rightarrow p = \dfrac{{2\sqrt {2 - \sqrt 3 } }}{4}\]Or $\dfrac{{\sqrt 2 (\sqrt 3 - 1)}}{4}$ because square root of \[\sqrt {2 - \sqrt 3 } \] is $\dfrac{{(\sqrt 3 - 1)}}{{\sqrt 2 }}$
You can directly write sine value from cosine value without doing so much of calculation by just making a sign opposite.
Now putting values of $\sin 15$ in $\dfrac{{(1 + \dfrac{{\sqrt 2 (\sqrt 3 + 1)}}{4})}}{{\sin 15}}$ we will get
$ \Rightarrow \dfrac{{(1 + \dfrac{{\sqrt 2 (\sqrt 3 + 1)}}{4})}}{{\dfrac{{\sqrt 2 (\sqrt 3 - 1)}}{4}}}$
$ \Rightarrow \dfrac{{(\dfrac{{4 + \sqrt 2 (\sqrt 3 + 1)}}{4})}}{{\dfrac{{\sqrt 2 (\sqrt 3 - 1)}}{4}}}$ (Cancelling 4 from both denominator terms)\[ \Rightarrow \dfrac{{4 + \sqrt 2 (\sqrt 3 + 1)}}{{\sqrt 2 (\sqrt 3 - 1)}} = \dfrac{{4 + \sqrt 6 + \sqrt 2 }}{{\sqrt 6 - \sqrt 2 }}\]
Now rationalising the denominator irrational term
\[ \Rightarrow \dfrac{{4 + \sqrt 6 + \sqrt 2 }}{{\sqrt 6 - \sqrt 2 }} \times \dfrac{{\sqrt 6 + \sqrt 2 }}{{\sqrt 6 + \sqrt 2 }}\]\[ \Rightarrow \dfrac{{(4 + \sqrt 6 + \sqrt 2 )(\sqrt 6 + \sqrt 2 )}}{{6 - 2}} = \dfrac{{4\sqrt 6 + 4\sqrt 2 + 6 + \sqrt 6 \times \sqrt 2 + \sqrt 6 \times \sqrt 2 + 2}}{4}\]\[ \Rightarrow \dfrac{{4\sqrt 6 + 4\sqrt 2 + 8 + 2(\sqrt {12} )}}{4} = \dfrac{{4\sqrt 6 + 4\sqrt 2 + 8 + 2(2\sqrt 3 )}}{4}\]\[ \Rightarrow \dfrac{{4\sqrt 6 + 4\sqrt 2 + 8 + 4\sqrt 3 }}{4} = \sqrt 6 + \sqrt 2 + 2 + \sqrt 3 \]
\[\therefore \sqrt 6 + \sqrt 2 + \sqrt 4 + \sqrt 3 \] (2 can be written as $\sqrt 4 $)
Therefore $\cot \dfrac{\pi }{4} = \sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 $
Hence proved.
Note: Students should know basic values of trigonometric functions which can boost their calculation part some of the values are as follows: -
$\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2},\sin {30^ \circ } = \dfrac{1}{2}$
$\cos {60^ \circ } = \dfrac{1}{2},\sin {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$
$\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }},\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$
\[\cos {90^ \circ } = 0,\sin {90^ \circ } = 1\]
$\cos {180^ \circ } = 1,\sin {180^ \circ } = 0$
Also while doing rationalisation some students make mistakes by multiplying with the same term but that is not correct. The same term with opposite signs should be multiplied and divided.
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