Question

If $\sin \theta = \cos \theta $, then $\theta = ?$
A.${45^0}$
B.${90^0}$
C.${0^0}$
D.${30^0}$

Answer 1

Hint: The trigonometric function is the function that relates the ratio of the length of two sides with the angles of the right-angled triangle widely used in navigation, oceanography, the theory of periodic functions, and projectiles. Commonly used trigonometric functions are the sine, the cosine, and the tangent, whereas the cosecant, the secant, the cotangent are their reciprocal, respectively.
In this question, we need to determine the angle $\theta $ such that $\sin \theta = \cos \theta $. For this we will use a general trigonometric identity which is given as $\cos \theta = \sin \left( {\dfrac{\pi }{2} - \theta } \right)$.

Complete step-by-step answer:
The given equation in the question is $\sin \theta = \cos \theta - - - - (i)$.
Using $\cos \theta = \sin \left( {\dfrac{\pi }{2} - \theta } \right)$ in the equation (i), we get
$
  \sin \theta = \cos \theta \\
  \sin \theta = \sin \left( {\dfrac{\pi }{2} - \theta } \right) - - - - (ii) \\
 $
If an equation has defined trigonometric function to both sides of an equation, then it can be eliminated leaving behind only the corresponding angles.
Here, in equation (ii) we can see that sin trigonometric function is common to both the sides of the equation so,
$\theta = \left( {\dfrac{\pi }{2} - \theta } \right) - - - - (iii)$
Now, solving equation (iii) for the value of $\theta $ as
$
  \theta = \left( {\dfrac{\pi }{2} - \theta } \right) \\
  \theta + \theta = \dfrac{\pi }{2} \\
  2\theta = \dfrac{\pi }{2} \\
  \theta = \dfrac{\pi }{4} \\
 $
Now, substitute the value of $\theta $ as 180 degrees for converting the result obtained in radians to degrees as:
$
  \theta = \dfrac{\pi }{4} \\
   = \dfrac{{{{180}^0}}}{4} \\
   = {45^0} \\
 $
Hence, if $\sin \theta = \cos \theta $ then, the value of $\theta $ is ${45^0}$
So, the correct answer is “Option A”.

Note: Alternatively, following the defined table for the trigonometric terms.

Function	$\sin \theta $	$\cos \theta $	$\tan \theta $
0	0	1	0
${30^0}$	$\dfrac{1}{2}$	$\dfrac{{\sqrt 3 }}{2}$	$\dfrac{1}{{\sqrt 3 }}$
${45^0}$	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{1}{{\sqrt 2 }}$	1
${60^0}$	$\dfrac{{\sqrt 3 }}{2}$	$\dfrac{1}{2}$	$\sqrt 3 $
${90^0}$	1	0	Indeterminate

We can see that the value of sin and cosine terms are equal only at $\theta = \dfrac{\pi }{4} = {45^0}$.