
Calculate $ \sin (2\arctan 2) $
Answer
545.1k+ views
Hint: Here the question is related to the trigonometry where it contains the sine and tan function. The question is exactly related to inverse trigonometry because we see “arctan” which means tan inverse. Hence by applying the inverse trigonometry we can find the result.
Complete step-by-step answer:
The question is given as $ \sin (2\arctan 2) $
Now let us consider the $ x = \arctan 2 $ --- (1)
We know that the inverse is also called an arc. Therefore the equation (1) can be written as
$ x = {\tan ^{ - 1}}(2) $ --------- (2)
The equation (2) can be written as
$ \tan x = 2 $ ---------- (3)
By the considering the triangle we can define the $ \tan x $
Let us consider the right angled triangle ABC
$ \tan x = \dfrac{{opposite}}{{adjacent}} = \dfrac{{AC}}{{BC}} $
$ \Rightarrow \dfrac{2}{1} = \dfrac{{AC}}{{BC}} $
Therefore the value of AC=2 and BC=1
By applying the Pythagoras theorem we have
$
A{B^2} = A{C^2} + B{C^2} \\
\Rightarrow A{B^2} = {(2)^2} + {(1)^2} \\
\Rightarrow A{B^2} = 4 + 1 \\
\Rightarrow A{B^2} = 5 \\
\Rightarrow AB = \sqrt 5 \;
$
Hence the length of the AB= $ \sqrt 5 $
Now we have to find the value of $ \sin 2x $
We know the trigonometry formula that is $ \sin 2x = 2.\sin x.cosx $ ----(4)
By considering the triangle we can find the value of sine and cosine.
So we have
$ \sin x = \dfrac{{opposite}}{{hypothesis}} = \dfrac{{AC}}{{AB}} $
As we know the value. Substituting we have
$ \Rightarrow \sin x = \dfrac{2}{{\sqrt 5 }} $ ---- (5)
$ \cos x = \dfrac{{adjacent}}{{hypothesis}} = \dfrac{{BC}}{{AB}} $
As we know the value. Substituting we have
$ \Rightarrow \cos x = \dfrac{1}{{\sqrt 5 }} $ ---- (6)
Substituting the equation (5) and equation (6) in equation (4) we have
$
\sin 2x = 2\sin x.\cos x \\
\Rightarrow \sin 2x = 2 \times \dfrac{2}{{\sqrt 5 }} \times \dfrac{1}{{\sqrt 5 }} \\
\Rightarrow \sin 2x = \dfrac{4}{5} \;
$
Hence we have obtained the required result.
Therefore, the value of $ \sin (2\arctan 2) = \dfrac{4}{5} $
So, the correct answer is “ $ \dfrac{4}{5} $ ”.
Note: The inverse is also called an arc. This word is used in question very rare. In most questions they will ask directly like find the inverse. The sine function, cosine function and tan function are defined by considering the right angled triangle. Hence we know about the Pythagoras theorem and to which type of triangle it is applicable.
Complete step-by-step answer:
The question is given as $ \sin (2\arctan 2) $
Now let us consider the $ x = \arctan 2 $ --- (1)
We know that the inverse is also called an arc. Therefore the equation (1) can be written as
$ x = {\tan ^{ - 1}}(2) $ --------- (2)
The equation (2) can be written as
$ \tan x = 2 $ ---------- (3)
By the considering the triangle we can define the $ \tan x $
Let us consider the right angled triangle ABC
$ \tan x = \dfrac{{opposite}}{{adjacent}} = \dfrac{{AC}}{{BC}} $
$ \Rightarrow \dfrac{2}{1} = \dfrac{{AC}}{{BC}} $
Therefore the value of AC=2 and BC=1
By applying the Pythagoras theorem we have
$
A{B^2} = A{C^2} + B{C^2} \\
\Rightarrow A{B^2} = {(2)^2} + {(1)^2} \\
\Rightarrow A{B^2} = 4 + 1 \\
\Rightarrow A{B^2} = 5 \\
\Rightarrow AB = \sqrt 5 \;
$
Hence the length of the AB= $ \sqrt 5 $
Now we have to find the value of $ \sin 2x $
We know the trigonometry formula that is $ \sin 2x = 2.\sin x.cosx $ ----(4)
By considering the triangle we can find the value of sine and cosine.
So we have
$ \sin x = \dfrac{{opposite}}{{hypothesis}} = \dfrac{{AC}}{{AB}} $
As we know the value. Substituting we have
$ \Rightarrow \sin x = \dfrac{2}{{\sqrt 5 }} $ ---- (5)
$ \cos x = \dfrac{{adjacent}}{{hypothesis}} = \dfrac{{BC}}{{AB}} $
As we know the value. Substituting we have
$ \Rightarrow \cos x = \dfrac{1}{{\sqrt 5 }} $ ---- (6)
Substituting the equation (5) and equation (6) in equation (4) we have
$
\sin 2x = 2\sin x.\cos x \\
\Rightarrow \sin 2x = 2 \times \dfrac{2}{{\sqrt 5 }} \times \dfrac{1}{{\sqrt 5 }} \\
\Rightarrow \sin 2x = \dfrac{4}{5} \;
$
Hence we have obtained the required result.
Therefore, the value of $ \sin (2\arctan 2) = \dfrac{4}{5} $
So, the correct answer is “ $ \dfrac{4}{5} $ ”.
Note: The inverse is also called an arc. This word is used in question very rare. In most questions they will ask directly like find the inverse. The sine function, cosine function and tan function are defined by considering the right angled triangle. Hence we know about the Pythagoras theorem and to which type of triangle it is applicable.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

