
If $ \sin x = \cos x $ and x is acute state the value of x in degrees
Answer
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Hint: We have been given a trigonometric equation, by making certain changes to it and using various trigonometric identities, we can easily obtain the required value of angle x. This angle x is given to be acute, which means its value is less than 90°.
Trigonometric identities to be used:
$
{\sin ^2}x + {\cos ^2}x = 1 \\
\sin 2x = 2\sin x\cos x \;
$
Complete step-by-step answer:
We have been given an equation:
$ \sin x = \cos x $ and we need to find the value of angle x in degrees.
This equation can be written as:
$ \sin x - \cos x = 0 $
Squaring both the sides, we get:
$
{\left( {\sin x - \cos x} \right)^2} = 0 \\
{\sin ^2}x + {\cos ^2}x - 2\sin x\cos x = 0 \\
1 - 2\sin x\cos x = 0\left( {\because {{\sin }^2}x + {{\cos }^2}x = 1} \right) \;
1 = 2\sin x\cos x \;
$
The value of double sine angle is given as:
$ \sin 2x = 2\sin x\cos x $
Substituting this value, we get:
$ \sin 2x = 1 $
The value of sin 90° is 1, so the above equation can be written as:
$
\sin 2x = \sin {90^\circ } \\
\Rightarrow 2x = {90^\circ } \\
\therefore x = {45^\circ } \;\
$
It is given that the angle is acute i.e. less than 90° which is also true for the angle obtained.
Therefore, if $ \sin x = \cos x $ and x is acute then the value of x is 45 degrees
So, the correct answer is “ 45° ”.
Note: We can also find the value of angle x by the following method, using the basic formula for tanx i.e. \[\dfrac{{\sin x}}{{\cos x}} = \tan x\]
Given equation: $ \sin x = \cos x $
Dividing both the sides by $ \cos x $ , we get:
\[
\dfrac{{\sin x}}{{\cos x}} = \dfrac{{\cos x}}{{\cos x}} \\
\tan x = 1 \\
\left( {\because \dfrac{{\sin x}}{{\cos x}} = \tan x} \right) \\
\]
The value of tan is 1 when the angle is equal to 45°, x can be calculated mathematically as:
$
\tan x = 1 \\
\Rightarrow x = {\tan ^{ - 1}}(1) \\
\therefore x = {45^\circ }\left( {\because \tan {{45}^\circ } = 1} \right) \;
$
Thus, we get the value of x as 45° by following every method.
The angles less than 90° are called acute angles and greater than that are called obtuse.
Trigonometric identities to be used:
$
{\sin ^2}x + {\cos ^2}x = 1 \\
\sin 2x = 2\sin x\cos x \;
$
Complete step-by-step answer:
We have been given an equation:
$ \sin x = \cos x $ and we need to find the value of angle x in degrees.
This equation can be written as:
$ \sin x - \cos x = 0 $
Squaring both the sides, we get:
$
{\left( {\sin x - \cos x} \right)^2} = 0 \\
{\sin ^2}x + {\cos ^2}x - 2\sin x\cos x = 0 \\
1 - 2\sin x\cos x = 0\left( {\because {{\sin }^2}x + {{\cos }^2}x = 1} \right) \;
1 = 2\sin x\cos x \;
$
The value of double sine angle is given as:
$ \sin 2x = 2\sin x\cos x $
Substituting this value, we get:
$ \sin 2x = 1 $
The value of sin 90° is 1, so the above equation can be written as:
$
\sin 2x = \sin {90^\circ } \\
\Rightarrow 2x = {90^\circ } \\
\therefore x = {45^\circ } \;\
$
It is given that the angle is acute i.e. less than 90° which is also true for the angle obtained.
Therefore, if $ \sin x = \cos x $ and x is acute then the value of x is 45 degrees
So, the correct answer is “ 45° ”.
Note: We can also find the value of angle x by the following method, using the basic formula for tanx i.e. \[\dfrac{{\sin x}}{{\cos x}} = \tan x\]
Given equation: $ \sin x = \cos x $
Dividing both the sides by $ \cos x $ , we get:
\[
\dfrac{{\sin x}}{{\cos x}} = \dfrac{{\cos x}}{{\cos x}} \\
\tan x = 1 \\
\left( {\because \dfrac{{\sin x}}{{\cos x}} = \tan x} \right) \\
\]
The value of tan is 1 when the angle is equal to 45°, x can be calculated mathematically as:
$
\tan x = 1 \\
\Rightarrow x = {\tan ^{ - 1}}(1) \\
\therefore x = {45^\circ }\left( {\because \tan {{45}^\circ } = 1} \right) \;
$
Thus, we get the value of x as 45° by following every method.
The angles less than 90° are called acute angles and greater than that are called obtuse.
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