Question

How do you evaluate $\sin (\dfrac{\pi }{6})$

Answer 1

Hint: This sum can be solved directly as it is a common value of $\sin $ function. In order to solve such a type of sum with values not common for example $\sin (120)$a student has to make use of the formula of $\sin $ to evaluate. To evaluate the above sum the student can substitute the value directly or can use the formula $\sin (A - B) = \sin A\cos B - \cos A\sin B$ as we know the value of $\sin (90)\& \sin (60)$. Another way to solve such a type of sum is by proving it. A student can take an equilateral triangle and use the properties of hypotenuse and similarity of triangles to evaluate $\sin (\dfrac{\pi }{6})$.

Complete step-by-step answer:
1st Method is directly using the value and noting down the answer. Since $\pi $ stands for ${180^ \circ }$ and it is divided by $6$, $\sin (\dfrac{\pi }{6})$ is equivalent to $\sin ({30^ \circ })$.
$\Rightarrow \sin ({30^ \circ }) = \dfrac{1}{2}$
2nd Method is to use the formula $\sin (A - B) = \sin A\cos B - \cos A\sin B$ and find the answer for $\sin ({30^ \circ })$.
We can write $\sin ({30^ \circ })$ as $\sin ({90^ \circ } - {60^ \circ })$.
Using the formula :
$\sin ({90^ \circ } - {60^ \circ }) = \sin 90 \times \cos 60 - \sin 60 \times \cos 90..........(1)$
As we know that value of $\cos 90$ is $0$,
$\sin ({90^ \circ } - {60^ \circ }) = \sin 90 \times \cos 60..........(2)$
Making use of the standard values , we get the value of $\sin ({30^ \circ })$
$\sin ({90^ \circ } - {60^ \circ }) = 1 \times \dfrac{1}{2}.........(3)$
$\Rightarrow \sin ({30^ \circ }) = \dfrac{1}{2}$

Note: In order to solve trigonometric numerical the student should know the standard values of all the trigonometric functions which are ${0^ \circ },{30^ \circ },{45^ \circ },{60^ \circ },{90^ \circ }$. These values would not only be useful for trigonometric related problems but also help in calculating the values of sides in triangles, useful in heights and distances chapter. Also they are useful while solving chapters like integration, derivatives.