Question

How do you evaluate ${\sin ^2}\left( {\dfrac{\pi }{6}} \right)$?

Answer 1

We write the given square of trigonometric function as square of complete term and then substitute the value of sine of the angle. Square the value inside the bracket in the end.

• ${\sin ^2}\left( x \right) = {\left( {\sin x} \right)^2}$

Complete step-by-step answer:
We have to find the value of ${\sin ^2}\left( {\dfrac{\pi }{6}} \right)$.
Here the function is the square of sine function and the angle is $\dfrac{\pi }{6}$which means ${30^ \circ }$.
We can write the value of the given function at given angle as
$\Rightarrow {\sin ^2}\left( {\dfrac{\pi }{6}} \right) = {\left[ {\sin \left( {\dfrac{\pi }{6}} \right)} \right]^2}$
Now we substitute the value of sine at the angle $\dfrac{\pi }{6}$ on the right hand side of the equation. Put $\sin \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{2}$.
$\Rightarrow {\sin ^2}\left( {\dfrac{\pi }{6}} \right) = {\left[ {\dfrac{1}{2}} \right]^2}$
Now square the term inside the bracket, i.e. multiply the term with itself to obtain the square value
$\Rightarrow {\sin ^2}\left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{2} \times \dfrac{1}{2}$
Multiply numerator with numerator and denominator with denominator
$\Rightarrow {\sin ^2}\left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{4}$

$\therefore$The value of ${\sin ^2}\left( {\dfrac{\pi }{6}} \right)$is $\dfrac{1}{4}$.

Note:
Many students make the mistake of writing the value of a function by substituting the value of sine at the given angle but they don’t square it as they think sine square is another angle and it cannot be written as square of complete function. Keep in mind if ‘x’ is any angle then we can write that ${\sin ^2}\left( x \right) = {\left( {\sin x} \right)^2}$, be it any trigonometric function and any angle.
Also, many students who don’t remember the value of sine of angle obtained at the end can take help of the table that gives values of some common trigonometric functions at a few angles.

Angles (in degrees)	${0^ \circ }$	${30^ \circ }$	${45^ \circ }$	${60^ \circ }$	${90^ \circ }$
sin	0	$\dfrac{1}{2}$	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{{\sqrt 3 }}{2}$	$1$
cos	1	$\dfrac{{\sqrt 3 }}{2}$	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{1}{2}$	0
tan	0	$\dfrac{1}{{\sqrt 3 }}$	1	$\sqrt 3$	Not defined
cosec	Not defined	2	$\sqrt 2$	$\dfrac{2}{{\sqrt 3 }}$	1
sec	1	$\dfrac{2}{{\sqrt 3 }}$	$\sqrt 2$	2	Not defined
cot	Not defined	$\sqrt 3$	1	$\dfrac{1}{{\sqrt 3 }}$	0